To test: The hypothesis that the mean IQs of the various educational levels of the subjects are equal or not.

Question

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Answer 1

Question

Chapter 12, Problem 2DA

To determine

To test: The hypothesis that the mean IQs of the various educational levels of the subjects are equal or not.

Expert Solution & Answer

Answer to Problem 2DA

There is significant difference between the means $X¯1$ and $X¯4$ , $X¯2$ and $X¯4$ , and $X¯3$ and $X¯4$ . Thus, it can be concluding that there is significant difference between the means of “No high school degree and Graduate degree”, “High school degree and Graduate degree”, “College graduate and Graduate degree”.

Explanation of Solution

Given info:

The databank shows the IQs of the various educational levels of the subjects.

Calculation:

Answers may vary. One of the possible answers is as follows:

Select random samples from the databank. The samples are given below.

No high school degree	High school degree	College graduate	Graduate degree
103	106	110	121
101	98	106	129
103	99	116	126
100	122	100	128
116	109	113	131
103	100	114

The hypotheses are given below:

Null hypothesis:

$H0:μ1=μ2=μ3=μ4$

Alternative hypothesis:

$H1:$ At least one mean is different from the others

Here, the difference in the mean percentage of voters in different places is tested. Hence, the claim is that,there is a difference in the mean percentage of voters in different places.

The number of samples k is 4, the sample sizes $n1$ , $n2$ , $n3$ and $n4$ are 6, 6, 6 and 5.

The degrees of freedom are $d.f.N=k−1 and d.f.D=N−k$ .

Where

$N=n1+n2+n3+n4=6+6+6+5=23$

Substitute 4 for k in $d.f.N$

$d.f.N=k−1 =4-1=3$

Substitute 23 for N and 4 for k in $d.f.D$

$d.f.D=N−k=23−4=19$

Critical value:

The critical F-value is obtained using the Table H: The F-Distribution with the level of significance $α=0.05$ .

Procedure:

Locate 19 in the degrees of freedom, denominator row of the Table H.
Obtain the value in the corresponding degrees of freedom, numerator column below 3.

That is, the critical value is 3.13.

Rejection region:

The null hypothesis would be rejected if $F>3.13$ .

Software procedure:

Step-by-step procedure to obtain the test statistic using the MINITAB software:

Choose Stat > ANOVA > One-Way.
In Response, enter the IQ.
In Factor, enter the Factor.
Click OK.

Output using the MINITAB software is given below:

ALEKS 360 ELEM STATISTICS, Chapter 12, Problem 2DA

From the MINITAB output, the test valueF is 13.33.

Conclusion:

From the results, the test value is 13.33.

Here, the F-statistic value is greater than the critical value.

That is, $13.33>3.13$ .

Thus, it can be concluding that, the null hypothesis is rejected.

Hence, there is evidence to reject the claim thatthe mean IQs of the various educational levels of the subjectsare equal. So use Scheffe test for where the difference exists.

Consider, $X¯1$ , $X¯2$ , $X¯3$ and $X¯4$ represents the means of No high school degree, High school degree, College graduate and Graduate degree, $s12,s22, s32 and s42$ represents the variances of samples of No high school degree, High school degree, College graduate and Graduate degree.

From the Minitab output, the sample sizes $n1,n2,n3 and n4$ are 6, 6, 6 and 5.

The means are $X¯1,X¯2, X¯3 and X¯4$ are 104.33, 105.67, 109.83 and 127.00.

The sample variances are $s12,s22, s32 and s42$ are 34.2225, 82.6281, 35.4025 and 14.5161.

Critical value:

The formula for critical value F¹ for the Scheffe test is,

$F1=(k−1)(Critical value)$

Here, the critical value of F test is 3.13.

Substitute 3.13 for critical value is of F and 3 for k-1 in $F1$

$F1=3(3.13)=9.39$

Comparison of the means:

The formula for finding $sW2$ is,

$sW2=∑(ni−1)si2∑(ni−1)$

That is,

$sW2=(6−1)34.2225+(6−1)82.6281+(6−1)35.4025+(5−1)14.5161(6−1)+(6−1)+(6−1)+(5−1)=171.1125+413.1405+177.0125+58.064419=819.329919=43.1226$

Comparison between the means $X¯1$ and $X¯2$ :

The hypotheses are given below:

Null hypothesis:

$H0:$ There is no significant difference between $X¯1$ and $X¯2$ .

Alternative hypothesis:

$H1:$ There is significant difference between $X¯1$ and $X¯2$ .

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

The formula for comparing the means $X¯1$ and $X¯2$ is,

$Fs=(X¯1−X¯2)2sW2[1n1+1n2]$

Substitute 104.33 and 105.67 for $X¯1$ and $X¯2$ , 6 for n_1, 6 for n₂ and 43.1226 for $sW2$

$Fs=(104.33−105.67)243.1226[16+16]=1.795643.1226[0.1667+0.1667]=1.795614.3771=0.125$

Thus, the value of $Fs$ is 0.125.

Conclusion:

The value of $Fs$ is 0.125.

Here, the value of $Fs$ is lesser than the critical value.

That is, $0.125<9.39$ .

Thus, the null hypothesis is not rejected.

Hence, there is no significant difference between the means $X¯1$ and $X¯2$ .

Comparison between the means $X¯1$ and $X¯3$ :

The hypotheses are given below:

Null hypothesis:

$H0:$ There is no significant difference between $X¯1$ and $X¯3$ .

Alternative hypothesis:

$H1:$ There is significant difference between $X¯1$ and $X¯3$ .

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

The formula for comparing the means $X¯1$ and $X¯3$ .is,

$Fs=(X¯1−X¯3)2sW2[1n1+1n3]$

Substitute 104.33 and 109.83 for $X¯1$ and $X¯3$ , 6 for n_1, 6 for n₃ and 43.1226 for $sW2$

$Fs=(104.33−109.83)243.1226[16+16]=30.2543.1226[0.1667+0.1667]=30.2514.3771=2.104$

Thus, the value of $Fs$ is 2.104.

Conclusion:

The value of $Fs$ is 2.104.

Here, the value of $Fs$ is lesser than the critical value.

That is, $2.104<9.39$ .

Thus, the null hypothesis is not rejected.

Hence, there is no significant difference between the means $X¯1$ and $X¯3$ .

Comparison between the means $X¯2$ and $X¯3$ :

The hypotheses are given below:

Null hypothesis:

$H0:$ There is no significant difference between $X¯2$ and $X¯3$ .

Alternative hypothesis:

$H1:$ There is significant difference between $X¯2$ and $X¯3$ .

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

The formula for comparing the means $X¯2$ and $X¯3$ .is,

$Fs=(X¯2−X¯3)2sW2[1n2+1n3]$

Substitute 105.67 and 109.83 for $X¯2$ and $X¯3$ , 6 for n_2, 6 for n₃ and 43.1226 for $sW2$

$Fs=(105.67−109.83)243.1226[16+16]=17.305643.1226[0.1667+0.1667]=17.305614.3771=1.204$

Thus, the value of $Fs$ is 1.204.

Conclusion:

The value of $Fs$ is 1.204.

Here, the value of $Fs$ is lesser than the critical value.

That is, $1.204<9.39$ .

Thus, the null hypothesis is not rejected.

Hence, there is no significant difference between the means $X¯2$ and $X¯3$ .

Comparison between the means $X¯1$ and $X¯4$ :

The hypotheses are given below:

Null hypothesis:

$H0:$ There is no significant difference between $X¯1$ and $X¯4$ .

Alternative hypothesis:

$H1:$ There is significant difference between $X¯1$ and $X¯4$ .

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

The formula for comparing the means $X¯1$ and $X¯4$ .is,

$Fs=(X¯1−X¯4)2sW2[1n2+1n4]$

Substitute 104.33 and 127.00 for $X¯1$ and $X¯4$ , 6 for n_1, 5 for n₄ and 43.1226 for $sW2$

$Fs=(104.33−127.00)243.1226[16+15]=513.928943.1226[0.1667+0.2]=513.928915.8131=32.5002$

Thus, the value of $Fs$ is 32.5002.

Conclusion:

The value of $Fs$ is 32.5002.

Here, the value of $Fs$ is greater than the critical value.

That is, $32.5002>9.39$ .

Thus, the null hypothesis is rejected.

Hence, there is significant difference between the means $X¯1$ and $X¯4$ .

Comparison between the means $X¯2$ and $X¯4$ :

The hypotheses are given below:

Null hypothesis:

$H0:$ There is no significant difference between $X¯2$ and $X¯4$ .

Alternative hypothesis:

$H1:$ There is significant difference between $X¯2$ and $X¯4$ .

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

The formula for comparing the means $X¯2$ and $X¯4$ .is,

$Fs=(X¯2−X¯4)2sW2[1n2+1n4]$

Substitute 105.67 and 127.00 for $X¯2$ and $X¯4$ , 6 for n_2, 5 for n₄ and 43.1226 for $sW2$

$Fs=(105.67−127.00)243.1226[16+15]=454.968943.1226[0.1667+0.2]=454.968915.8131=28.7716$

Thus, the value of $Fs$ is 28.7716.

Conclusion:

The value of $Fs$ is 28.7716.

Here, the value of $Fs$ is greater than the critical value.

That is, $28.7716>9.39$ .

Thus, the null hypothesis is rejected.

Hence, there is significant difference between the means $X¯2$ and $X¯4$ .

Comparison between the means $X¯3$ and $X¯4$ :

The hypotheses are given below:

Null hypothesis:

$H0:$ There is no significant difference between $X¯3$ and $X¯4$ .

Alternative hypothesis:

$H1:$ There is significant difference between $X¯3$ and $X¯4$ .

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

The formula for comparing the means $X¯3$ and $X¯4$ .is,

$Fs=(X¯3−X¯4)2sW2[1n3+1n4]$

Substitute 109.83 and 127.00 for $X¯3$ and $X¯4$ , 6 for n_2, 5 for n₄ and 43.1226 for $sW2$

$Fs=(109.83−127.00)243.1226[16+15]=294.808943.1226[0.1667+0.2]=294.808915.8131=18.6433$

Thus, the value of $Fs$ is 18.6433.

Conclusion:

The value of $Fs$ is 18.6433.

Here, the value of $Fs$ is greater than the critical value.

That is, $18.6433>9.39$ .

Thus, the null hypothesis is rejected.

Hence, there is significant difference between the means $X¯3$ and $X¯4$ .

Justification:

Here, there is significant difference between the means $X¯1$ and $X¯4$ , $X¯2$ and $X¯4$ , and $X¯3$ and $X¯4$ . Thus, it can be concluding that there is significant difference between the means of “No high school degree and Graduate degree”, “High school degree and Graduate degree”, “College graduate and Graduate degree”.

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Answer 2

Question

Chapter 12, Problem 2DA

To determine

To test: The hypothesis that the mean IQs of the various educational levels of the subjects are equal or not.

Expert Solution & Answer

Answer to Problem 2DA

There is significant difference between the means $X¯1$ and $X¯4$ , $X¯2$ and $X¯4$ , and $X¯3$ and $X¯4$ . Thus, it can be concluding that there is significant difference between the means of “No high school degree and Graduate degree”, “High school degree and Graduate degree”, “College graduate and Graduate degree”.

Explanation of Solution

Given info:

The databank shows the IQs of the various educational levels of the subjects.

Calculation:

Answers may vary. One of the possible answers is as follows:

Select random samples from the databank. The samples are given below.

No high school degree	High school degree	College graduate	Graduate degree
103	106	110	121
101	98	106	129
103	99	116	126
100	122	100	128
116	109	113	131
103	100	114

The hypotheses are given below:

Null hypothesis:

$H0:μ1=μ2=μ3=μ4$

Alternative hypothesis:

$H1:$ At least one mean is different from the others

Here, the difference in the mean percentage of voters in different places is tested. Hence, the claim is that,there is a difference in the mean percentage of voters in different places.

The number of samples k is 4, the sample sizes $n1$ , $n2$ , $n3$ and $n4$ are 6, 6, 6 and 5.

The degrees of freedom are $d.f.N=k−1 and d.f.D=N−k$ .

Where

$N=n1+n2+n3+n4=6+6+6+5=23$

Substitute 4 for k in $d.f.N$

$d.f.N=k−1 =4-1=3$

Substitute 23 for N and 4 for k in $d.f.D$

$d.f.D=N−k=23−4=19$

Critical value:

The critical F-value is obtained using the Table H: The F-Distribution with the level of significance $α=0.05$ .

Procedure:

Locate 19 in the degrees of freedom, denominator row of the Table H.
Obtain the value in the corresponding degrees of freedom, numerator column below 3.

That is, the critical value is 3.13.

Rejection region:

The null hypothesis would be rejected if $F>3.13$ .

Software procedure:

Step-by-step procedure to obtain the test statistic using the MINITAB software:

Choose Stat > ANOVA > One-Way.
In Response, enter the IQ.
In Factor, enter the Factor.
Click OK.

Output using the MINITAB software is given below:

ALEKS 360 ELEM STATISTICS, Chapter 12, Problem 2DA

From the MINITAB output, the test valueF is 13.33.

Conclusion:

From the results, the test value is 13.33.

Here, the F-statistic value is greater than the critical value.

That is, $13.33>3.13$ .

Thus, it can be concluding that, the null hypothesis is rejected.

Hence, there is evidence to reject the claim thatthe mean IQs of the various educational levels of the subjectsare equal. So use Scheffe test for where the difference exists.

Consider, $X¯1$ , $X¯2$ , $X¯3$ and $X¯4$ represents the means of No high school degree, High school degree, College graduate and Graduate degree, $s12,s22, s32 and s42$ represents the variances of samples of No high school degree, High school degree, College graduate and Graduate degree.

From the Minitab output, the sample sizes $n1,n2,n3 and n4$ are 6, 6, 6 and 5.

The means are $X¯1,X¯2, X¯3 and X¯4$ are 104.33, 105.67, 109.83 and 127.00.

The sample variances are $s12,s22, s32 and s42$ are 34.2225, 82.6281, 35.4025 and 14.5161.

Critical value:

The formula for critical value F¹ for the Scheffe test is,

$F1=(k−1)(Critical value)$

Here, the critical value of F test is 3.13.

Substitute 3.13 for critical value is of F and 3 for k-1 in $F1$

$F1=3(3.13)=9.39$

Comparison of the means:

The formula for finding $sW2$ is,

$sW2=∑(ni−1)si2∑(ni−1)$

That is,

$sW2=(6−1)34.2225+(6−1)82.6281+(6−1)35.4025+(5−1)14.5161(6−1)+(6−1)+(6−1)+(5−1)=171.1125+413.1405+177.0125+58.064419=819.329919=43.1226$

Comparison between the means $X¯1$ and $X¯2$ :

The hypotheses are given below:

Null hypothesis:

$H0:$ There is no significant difference between $X¯1$ and $X¯2$ .

Alternative hypothesis:

$H1:$ There is significant difference between $X¯1$ and $X¯2$ .

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

The formula for comparing the means $X¯1$ and $X¯2$ is,

$Fs=(X¯1−X¯2)2sW2[1n1+1n2]$

Substitute 104.33 and 105.67 for $X¯1$ and $X¯2$ , 6 for n_1, 6 for n₂ and 43.1226 for $sW2$

$Fs=(104.33−105.67)243.1226[16+16]=1.795643.1226[0.1667+0.1667]=1.795614.3771=0.125$

Thus, the value of $Fs$ is 0.125.

Conclusion:

The value of $Fs$ is 0.125.

Here, the value of $Fs$ is lesser than the critical value.

That is, $0.125<9.39$ .

Thus, the null hypothesis is not rejected.

Hence, there is no significant difference between the means $X¯1$ and $X¯2$ .

Comparison between the means $X¯1$ and $X¯3$ :

The hypotheses are given below:

Null hypothesis:

$H0:$ There is no significant difference between $X¯1$ and $X¯3$ .

Alternative hypothesis:

$H1:$ There is significant difference between $X¯1$ and $X¯3$ .

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

The formula for comparing the means $X¯1$ and $X¯3$ .is,

$Fs=(X¯1−X¯3)2sW2[1n1+1n3]$

Substitute 104.33 and 109.83 for $X¯1$ and $X¯3$ , 6 for n_1, 6 for n₃ and 43.1226 for $sW2$

$Fs=(104.33−109.83)243.1226[16+16]=30.2543.1226[0.1667+0.1667]=30.2514.3771=2.104$

Thus, the value of $Fs$ is 2.104.

Conclusion:

The value of $Fs$ is 2.104.

Here, the value of $Fs$ is lesser than the critical value.

That is, $2.104<9.39$ .

Thus, the null hypothesis is not rejected.

Hence, there is no significant difference between the means $X¯1$ and $X¯3$ .

Comparison between the means $X¯2$ and $X¯3$ :

The hypotheses are given below:

Null hypothesis:

$H0:$ There is no significant difference between $X¯2$ and $X¯3$ .

Alternative hypothesis:

$H1:$ There is significant difference between $X¯2$ and $X¯3$ .

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

The formula for comparing the means $X¯2$ and $X¯3$ .is,

$Fs=(X¯2−X¯3)2sW2[1n2+1n3]$

Substitute 105.67 and 109.83 for $X¯2$ and $X¯3$ , 6 for n_2, 6 for n₃ and 43.1226 for $sW2$

$Fs=(105.67−109.83)243.1226[16+16]=17.305643.1226[0.1667+0.1667]=17.305614.3771=1.204$

Thus, the value of $Fs$ is 1.204.

Conclusion:

The value of $Fs$ is 1.204.

Here, the value of $Fs$ is lesser than the critical value.

That is, $1.204<9.39$ .

Thus, the null hypothesis is not rejected.

Hence, there is no significant difference between the means $X¯2$ and $X¯3$ .

Comparison between the means $X¯1$ and $X¯4$ :

The hypotheses are given below:

Null hypothesis:

$H0:$ There is no significant difference between $X¯1$ and $X¯4$ .

Alternative hypothesis:

$H1:$ There is significant difference between $X¯1$ and $X¯4$ .

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

The formula for comparing the means $X¯1$ and $X¯4$ .is,

$Fs=(X¯1−X¯4)2sW2[1n2+1n4]$

Substitute 104.33 and 127.00 for $X¯1$ and $X¯4$ , 6 for n_1, 5 for n₄ and 43.1226 for $sW2$

$Fs=(104.33−127.00)243.1226[16+15]=513.928943.1226[0.1667+0.2]=513.928915.8131=32.5002$

Thus, the value of $Fs$ is 32.5002.

Conclusion:

The value of $Fs$ is 32.5002.

Here, the value of $Fs$ is greater than the critical value.

That is, $32.5002>9.39$ .

Thus, the null hypothesis is rejected.

Hence, there is significant difference between the means $X¯1$ and $X¯4$ .

Comparison between the means $X¯2$ and $X¯4$ :

The hypotheses are given below:

Null hypothesis:

$H0:$ There is no significant difference between $X¯2$ and $X¯4$ .

Alternative hypothesis:

$H1:$ There is significant difference between $X¯2$ and $X¯4$ .

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

The formula for comparing the means $X¯2$ and $X¯4$ .is,

$Fs=(X¯2−X¯4)2sW2[1n2+1n4]$

Substitute 105.67 and 127.00 for $X¯2$ and $X¯4$ , 6 for n_2, 5 for n₄ and 43.1226 for $sW2$

$Fs=(105.67−127.00)243.1226[16+15]=454.968943.1226[0.1667+0.2]=454.968915.8131=28.7716$

Thus, the value of $Fs$ is 28.7716.

Conclusion:

The value of $Fs$ is 28.7716.

Here, the value of $Fs$ is greater than the critical value.

That is, $28.7716>9.39$ .

Thus, the null hypothesis is rejected.

Hence, there is significant difference between the means $X¯2$ and $X¯4$ .

Comparison between the means $X¯3$ and $X¯4$ :

The hypotheses are given below:

Null hypothesis:

$H0:$ There is no significant difference between $X¯3$ and $X¯4$ .

Alternative hypothesis:

$H1:$ There is significant difference between $X¯3$ and $X¯4$ .

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

The formula for comparing the means $X¯3$ and $X¯4$ .is,

$Fs=(X¯3−X¯4)2sW2[1n3+1n4]$

Substitute 109.83 and 127.00 for $X¯3$ and $X¯4$ , 6 for n_2, 5 for n₄ and 43.1226 for $sW2$

$Fs=(109.83−127.00)243.1226[16+15]=294.808943.1226[0.1667+0.2]=294.808915.8131=18.6433$

Thus, the value of $Fs$ is 18.6433.

Conclusion:

The value of $Fs$ is 18.6433.

Here, the value of $Fs$ is greater than the critical value.

That is, $18.6433>9.39$ .

Thus, the null hypothesis is rejected.

Hence, there is significant difference between the means $X¯3$ and $X¯4$ .

Justification:

Here, there is significant difference between the means $X¯1$ and $X¯4$ , $X¯2$ and $X¯4$ , and $X¯3$ and $X¯4$ . Thus, it can be concluding that there is significant difference between the means of “No high school degree and Graduate degree”, “High school degree and Graduate degree”, “College graduate and Graduate degree”.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 12, Problem 2DA

To determine

To test: The hypothesis that the mean IQs of the various educational levels of the subjects are equal or not.

Expert Solution & Answer

Answer to Problem 2DA

There is significant difference between the means $X¯1$ and $X¯4$ , $X¯2$ and $X¯4$ , and $X¯3$ and $X¯4$ . Thus, it can be concluding that there is significant difference between the means of “No high school degree and Graduate degree”, “High school degree and Graduate degree”, “College graduate and Graduate degree”.

Explanation of Solution

Given info:

The databank shows the IQs of the various educational levels of the subjects.

Calculation:

Answers may vary. One of the possible answers is as follows:

Select random samples from the databank. The samples are given below.

No high school degree	High school degree	College graduate	Graduate degree
103	106	110	121
101	98	106	129
103	99	116	126
100	122	100	128
116	109	113	131
103	100	114

The hypotheses are given below:

Null hypothesis:

$H0:μ1=μ2=μ3=μ4$

Alternative hypothesis:

$H1:$ At least one mean is different from the others

Here, the difference in the mean percentage of voters in different places is tested. Hence, the claim is that,there is a difference in the mean percentage of voters in different places.

The number of samples k is 4, the sample sizes $n1$ , $n2$ , $n3$ and $n4$ are 6, 6, 6 and 5.

The degrees of freedom are $d.f.N=k−1 and d.f.D=N−k$ .

Where

$N=n1+n2+n3+n4=6+6+6+5=23$

Substitute 4 for k in $d.f.N$

$d.f.N=k−1 =4-1=3$

Substitute 23 for N and 4 for k in $d.f.D$

$d.f.D=N−k=23−4=19$

Critical value:

The critical F-value is obtained using the Table H: The F-Distribution with the level of significance $α=0.05$ .

Procedure:

Locate 19 in the degrees of freedom, denominator row of the Table H.
Obtain the value in the corresponding degrees of freedom, numerator column below 3.

That is, the critical value is 3.13.

Rejection region:

The null hypothesis would be rejected if $F>3.13$ .

Software procedure:

Step-by-step procedure to obtain the test statistic using the MINITAB software:

Choose Stat > ANOVA > One-Way.
In Response, enter the IQ.
In Factor, enter the Factor.
Click OK.

Output using the MINITAB software is given below:

ALEKS 360 ELEM STATISTICS, Chapter 12, Problem 2DA

From the MINITAB output, the test valueF is 13.33.

Conclusion:

From the results, the test value is 13.33.

Here, the F-statistic value is greater than the critical value.

That is, $13.33>3.13$ .

Thus, it can be concluding that, the null hypothesis is rejected.

Hence, there is evidence to reject the claim thatthe mean IQs of the various educational levels of the subjectsare equal. So use Scheffe test for where the difference exists.

Consider, $X¯1$ , $X¯2$ , $X¯3$ and $X¯4$ represents the means of No high school degree, High school degree, College graduate and Graduate degree, $s12,s22, s32 and s42$ represents the variances of samples of No high school degree, High school degree, College graduate and Graduate degree.

From the Minitab output, the sample sizes $n1,n2,n3 and n4$ are 6, 6, 6 and 5.

The means are $X¯1,X¯2, X¯3 and X¯4$ are 104.33, 105.67, 109.83 and 127.00.

The sample variances are $s12,s22, s32 and s42$ are 34.2225, 82.6281, 35.4025 and 14.5161.

Critical value:

The formula for critical value F¹ for the Scheffe test is,

$F1=(k−1)(Critical value)$

Here, the critical value of F test is 3.13.

Substitute 3.13 for critical value is of F and 3 for k-1 in $F1$

$F1=3(3.13)=9.39$

Comparison of the means:

The formula for finding $sW2$ is,

$sW2=∑(ni−1)si2∑(ni−1)$

That is,

$sW2=(6−1)34.2225+(6−1)82.6281+(6−1)35.4025+(5−1)14.5161(6−1)+(6−1)+(6−1)+(5−1)=171.1125+413.1405+177.0125+58.064419=819.329919=43.1226$

Comparison between the means $X¯1$ and $X¯2$ :

The hypotheses are given below:

Null hypothesis:

$H0:$ There is no significant difference between $X¯1$ and $X¯2$ .

Alternative hypothesis:

$H1:$ There is significant difference between $X¯1$ and $X¯2$ .

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

The formula for comparing the means $X¯1$ and $X¯2$ is,

$Fs=(X¯1−X¯2)2sW2[1n1+1n2]$

Substitute 104.33 and 105.67 for $X¯1$ and $X¯2$ , 6 for n_1, 6 for n₂ and 43.1226 for $sW2$

$Fs=(104.33−105.67)243.1226[16+16]=1.795643.1226[0.1667+0.1667]=1.795614.3771=0.125$

Thus, the value of $Fs$ is 0.125.

Conclusion:

The value of $Fs$ is 0.125.

Here, the value of $Fs$ is lesser than the critical value.

That is, $0.125<9.39$ .

Thus, the null hypothesis is not rejected.

Hence, there is no significant difference between the means $X¯1$ and $X¯2$ .

Comparison between the means $X¯1$ and $X¯3$ :

The hypotheses are given below:

Null hypothesis:

$H0:$ There is no significant difference between $X¯1$ and $X¯3$ .

Alternative hypothesis:

$H1:$ There is significant difference between $X¯1$ and $X¯3$ .

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

The formula for comparing the means $X¯1$ and $X¯3$ .is,

$Fs=(X¯1−X¯3)2sW2[1n1+1n3]$

Substitute 104.33 and 109.83 for $X¯1$ and $X¯3$ , 6 for n_1, 6 for n₃ and 43.1226 for $sW2$

$Fs=(104.33−109.83)243.1226[16+16]=30.2543.1226[0.1667+0.1667]=30.2514.3771=2.104$

Thus, the value of $Fs$ is 2.104.

Conclusion:

The value of $Fs$ is 2.104.

Here, the value of $Fs$ is lesser than the critical value.

That is, $2.104<9.39$ .

Thus, the null hypothesis is not rejected.

Hence, there is no significant difference between the means $X¯1$ and $X¯3$ .

Comparison between the means $X¯2$ and $X¯3$ :

The hypotheses are given below:

Null hypothesis:

$H0:$ There is no significant difference between $X¯2$ and $X¯3$ .

Alternative hypothesis:

$H1:$ There is significant difference between $X¯2$ and $X¯3$ .

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

The formula for comparing the means $X¯2$ and $X¯3$ .is,

$Fs=(X¯2−X¯3)2sW2[1n2+1n3]$

Substitute 105.67 and 109.83 for $X¯2$ and $X¯3$ , 6 for n_2, 6 for n₃ and 43.1226 for $sW2$

$Fs=(105.67−109.83)243.1226[16+16]=17.305643.1226[0.1667+0.1667]=17.305614.3771=1.204$

Thus, the value of $Fs$ is 1.204.

Conclusion:

The value of $Fs$ is 1.204.

Here, the value of $Fs$ is lesser than the critical value.

That is, $1.204<9.39$ .

Thus, the null hypothesis is not rejected.

Hence, there is no significant difference between the means $X¯2$ and $X¯3$ .

Comparison between the means $X¯1$ and $X¯4$ :

The hypotheses are given below:

Null hypothesis:

$H0:$ There is no significant difference between $X¯1$ and $X¯4$ .

Alternative hypothesis:

$H1:$ There is significant difference between $X¯1$ and $X¯4$ .

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

The formula for comparing the means $X¯1$ and $X¯4$ .is,

$Fs=(X¯1−X¯4)2sW2[1n2+1n4]$

Substitute 104.33 and 127.00 for $X¯1$ and $X¯4$ , 6 for n_1, 5 for n₄ and 43.1226 for $sW2$

$Fs=(104.33−127.00)243.1226[16+15]=513.928943.1226[0.1667+0.2]=513.928915.8131=32.5002$

Thus, the value of $Fs$ is 32.5002.

Conclusion:

The value of $Fs$ is 32.5002.

Here, the value of $Fs$ is greater than the critical value.

That is, $32.5002>9.39$ .

Thus, the null hypothesis is rejected.

Hence, there is significant difference between the means $X¯1$ and $X¯4$ .

Comparison between the means $X¯2$ and $X¯4$ :

The hypotheses are given below:

Null hypothesis:

$H0:$ There is no significant difference between $X¯2$ and $X¯4$ .

Alternative hypothesis:

$H1:$ There is significant difference between $X¯2$ and $X¯4$ .

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

The formula for comparing the means $X¯2$ and $X¯4$ .is,

$Fs=(X¯2−X¯4)2sW2[1n2+1n4]$

Substitute 105.67 and 127.00 for $X¯2$ and $X¯4$ , 6 for n_2, 5 for n₄ and 43.1226 for $sW2$

$Fs=(105.67−127.00)243.1226[16+15]=454.968943.1226[0.1667+0.2]=454.968915.8131=28.7716$

Thus, the value of $Fs$ is 28.7716.

Conclusion:

The value of $Fs$ is 28.7716.

Here, the value of $Fs$ is greater than the critical value.

That is, $28.7716>9.39$ .

Thus, the null hypothesis is rejected.

Hence, there is significant difference between the means $X¯2$ and $X¯4$ .

Comparison between the means $X¯3$ and $X¯4$ :

The hypotheses are given below:

Null hypothesis:

$H0:$ There is no significant difference between $X¯3$ and $X¯4$ .

Alternative hypothesis:

$H1:$ There is significant difference between $X¯3$ and $X¯4$ .

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

The formula for comparing the means $X¯3$ and $X¯4$ .is,

$Fs=(X¯3−X¯4)2sW2[1n3+1n4]$

Substitute 109.83 and 127.00 for $X¯3$ and $X¯4$ , 6 for n_2, 5 for n₄ and 43.1226 for $sW2$

$Fs=(109.83−127.00)243.1226[16+15]=294.808943.1226[0.1667+0.2]=294.808915.8131=18.6433$

Thus, the value of $Fs$ is 18.6433.

Conclusion:

The value of $Fs$ is 18.6433.

Here, the value of $Fs$ is greater than the critical value.

That is, $18.6433>9.39$ .

Thus, the null hypothesis is rejected.

Hence, there is significant difference between the means $X¯3$ and $X¯4$ .

Justification:

Here, there is significant difference between the means $X¯1$ and $X¯4$ , $X¯2$ and $X¯4$ , and $X¯3$ and $X¯4$ . Thus, it can be concluding that there is significant difference between the means of “No high school degree and Graduate degree”, “High school degree and Graduate degree”, “College graduate and Graduate degree”.

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Answer 4

Question

Chapter 12, Problem 2DA

To determine

To test: The hypothesis that the mean IQs of the various educational levels of the subjects are equal or not.

Expert Solution & Answer

Answer to Problem 2DA

There is significant difference between the means $X¯1$ and $X¯4$ , $X¯2$ and $X¯4$ , and $X¯3$ and $X¯4$ . Thus, it can be concluding that there is significant difference between the means of “No high school degree and Graduate degree”, “High school degree and Graduate degree”, “College graduate and Graduate degree”.

Explanation of Solution

Given info:

The databank shows the IQs of the various educational levels of the subjects.

Calculation:

Answers may vary. One of the possible answers is as follows:

Select random samples from the databank. The samples are given below.

No high school degree	High school degree	College graduate	Graduate degree
103	106	110	121
101	98	106	129
103	99	116	126
100	122	100	128
116	109	113	131
103	100	114

The hypotheses are given below:

Null hypothesis:

$H0:μ1=μ2=μ3=μ4$

Alternative hypothesis:

$H1:$ At least one mean is different from the others

Here, the difference in the mean percentage of voters in different places is tested. Hence, the claim is that,there is a difference in the mean percentage of voters in different places.

The number of samples k is 4, the sample sizes $n1$ , $n2$ , $n3$ and $n4$ are 6, 6, 6 and 5.

The degrees of freedom are $d.f.N=k−1 and d.f.D=N−k$ .

Where

$N=n1+n2+n3+n4=6+6+6+5=23$

Substitute 4 for k in $d.f.N$

$d.f.N=k−1 =4-1=3$

Substitute 23 for N and 4 for k in $d.f.D$

$d.f.D=N−k=23−4=19$

Critical value:

The critical F-value is obtained using the Table H: The F-Distribution with the level of significance $α=0.05$ .

Procedure:

Locate 19 in the degrees of freedom, denominator row of the Table H.
Obtain the value in the corresponding degrees of freedom, numerator column below 3.

That is, the critical value is 3.13.

Rejection region:

The null hypothesis would be rejected if $F>3.13$ .

Software procedure:

Step-by-step procedure to obtain the test statistic using the MINITAB software:

Choose Stat > ANOVA > One-Way.
In Response, enter the IQ.
In Factor, enter the Factor.
Click OK.

Output using the MINITAB software is given below:

ALEKS 360 ELEM STATISTICS, Chapter 12, Problem 2DA

From the MINITAB output, the test valueF is 13.33.

Conclusion:

From the results, the test value is 13.33.

Here, the F-statistic value is greater than the critical value.

That is, $13.33>3.13$ .

Thus, it can be concluding that, the null hypothesis is rejected.

Hence, there is evidence to reject the claim thatthe mean IQs of the various educational levels of the subjectsare equal. So use Scheffe test for where the difference exists.

Consider, $X¯1$ , $X¯2$ , $X¯3$ and $X¯4$ represents the means of No high school degree, High school degree, College graduate and Graduate degree, $s12,s22, s32 and s42$ represents the variances of samples of No high school degree, High school degree, College graduate and Graduate degree.

From the Minitab output, the sample sizes $n1,n2,n3 and n4$ are 6, 6, 6 and 5.

The means are $X¯1,X¯2, X¯3 and X¯4$ are 104.33, 105.67, 109.83 and 127.00.

The sample variances are $s12,s22, s32 and s42$ are 34.2225, 82.6281, 35.4025 and 14.5161.

Critical value:

The formula for critical value F¹ for the Scheffe test is,

$F1=(k−1)(Critical value)$

Here, the critical value of F test is 3.13.

Substitute 3.13 for critical value is of F and 3 for k-1 in $F1$

$F1=3(3.13)=9.39$

Comparison of the means:

The formula for finding $sW2$ is,

$sW2=∑(ni−1)si2∑(ni−1)$

That is,

$sW2=(6−1)34.2225+(6−1)82.6281+(6−1)35.4025+(5−1)14.5161(6−1)+(6−1)+(6−1)+(5−1)=171.1125+413.1405+177.0125+58.064419=819.329919=43.1226$

Comparison between the means $X¯1$ and $X¯2$ :

The hypotheses are given below:

Null hypothesis:

$H0:$ There is no significant difference between $X¯1$ and $X¯2$ .

Alternative hypothesis:

$H1:$ There is significant difference between $X¯1$ and $X¯2$ .

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

The formula for comparing the means $X¯1$ and $X¯2$ is,

$Fs=(X¯1−X¯2)2sW2[1n1+1n2]$

Substitute 104.33 and 105.67 for $X¯1$ and $X¯2$ , 6 for n_1, 6 for n₂ and 43.1226 for $sW2$

$Fs=(104.33−105.67)243.1226[16+16]=1.795643.1226[0.1667+0.1667]=1.795614.3771=0.125$

Thus, the value of $Fs$ is 0.125.

Conclusion:

The value of $Fs$ is 0.125.

Here, the value of $Fs$ is lesser than the critical value.

That is, $0.125<9.39$ .

Thus, the null hypothesis is not rejected.

Hence, there is no significant difference between the means $X¯1$ and $X¯2$ .

Comparison between the means $X¯1$ and $X¯3$ :

The hypotheses are given below:

Null hypothesis:

$H0:$ There is no significant difference between $X¯1$ and $X¯3$ .

Alternative hypothesis:

$H1:$ There is significant difference between $X¯1$ and $X¯3$ .

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

The formula for comparing the means $X¯1$ and $X¯3$ .is,

$Fs=(X¯1−X¯3)2sW2[1n1+1n3]$

Substitute 104.33 and 109.83 for $X¯1$ and $X¯3$ , 6 for n_1, 6 for n₃ and 43.1226 for $sW2$

$Fs=(104.33−109.83)243.1226[16+16]=30.2543.1226[0.1667+0.1667]=30.2514.3771=2.104$

Thus, the value of $Fs$ is 2.104.

Conclusion:

The value of $Fs$ is 2.104.

Here, the value of $Fs$ is lesser than the critical value.

That is, $2.104<9.39$ .

Thus, the null hypothesis is not rejected.

Hence, there is no significant difference between the means $X¯1$ and $X¯3$ .

Comparison between the means $X¯2$ and $X¯3$ :

The hypotheses are given below:

Null hypothesis:

$H0:$ There is no significant difference between $X¯2$ and $X¯3$ .

Alternative hypothesis:

$H1:$ There is significant difference between $X¯2$ and $X¯3$ .

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

The formula for comparing the means $X¯2$ and $X¯3$ .is,

$Fs=(X¯2−X¯3)2sW2[1n2+1n3]$

Substitute 105.67 and 109.83 for $X¯2$ and $X¯3$ , 6 for n_2, 6 for n₃ and 43.1226 for $sW2$

$Fs=(105.67−109.83)243.1226[16+16]=17.305643.1226[0.1667+0.1667]=17.305614.3771=1.204$

Thus, the value of $Fs$ is 1.204.

Conclusion:

The value of $Fs$ is 1.204.

Here, the value of $Fs$ is lesser than the critical value.

That is, $1.204<9.39$ .

Thus, the null hypothesis is not rejected.

Hence, there is no significant difference between the means $X¯2$ and $X¯3$ .

Comparison between the means $X¯1$ and $X¯4$ :

The hypotheses are given below:

Null hypothesis:

$H0:$ There is no significant difference between $X¯1$ and $X¯4$ .

Alternative hypothesis:

$H1:$ There is significant difference between $X¯1$ and $X¯4$ .

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

The formula for comparing the means $X¯1$ and $X¯4$ .is,

$Fs=(X¯1−X¯4)2sW2[1n2+1n4]$

Substitute 104.33 and 127.00 for $X¯1$ and $X¯4$ , 6 for n_1, 5 for n₄ and 43.1226 for $sW2$

$Fs=(104.33−127.00)243.1226[16+15]=513.928943.1226[0.1667+0.2]=513.928915.8131=32.5002$

Thus, the value of $Fs$ is 32.5002.

Conclusion:

The value of $Fs$ is 32.5002.

Here, the value of $Fs$ is greater than the critical value.

That is, $32.5002>9.39$ .

Thus, the null hypothesis is rejected.

Hence, there is significant difference between the means $X¯1$ and $X¯4$ .

Comparison between the means $X¯2$ and $X¯4$ :

The hypotheses are given below:

Null hypothesis:

$H0:$ There is no significant difference between $X¯2$ and $X¯4$ .

Alternative hypothesis:

$H1:$ There is significant difference between $X¯2$ and $X¯4$ .

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

The formula for comparing the means $X¯2$ and $X¯4$ .is,

$Fs=(X¯2−X¯4)2sW2[1n2+1n4]$

Substitute 105.67 and 127.00 for $X¯2$ and $X¯4$ , 6 for n_2, 5 for n₄ and 43.1226 for $sW2$

$Fs=(105.67−127.00)243.1226[16+15]=454.968943.1226[0.1667+0.2]=454.968915.8131=28.7716$

Thus, the value of $Fs$ is 28.7716.

Conclusion:

The value of $Fs$ is 28.7716.

Here, the value of $Fs$ is greater than the critical value.

That is, $28.7716>9.39$ .

Thus, the null hypothesis is rejected.

Hence, there is significant difference between the means $X¯2$ and $X¯4$ .

Comparison between the means $X¯3$ and $X¯4$ :

The hypotheses are given below:

Null hypothesis:

$H0:$ There is no significant difference between $X¯3$ and $X¯4$ .

Alternative hypothesis:

$H1:$ There is significant difference between $X¯3$ and $X¯4$ .

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

The formula for comparing the means $X¯3$ and $X¯4$ .is,

$Fs=(X¯3−X¯4)2sW2[1n3+1n4]$

Substitute 109.83 and 127.00 for $X¯3$ and $X¯4$ , 6 for n_2, 5 for n₄ and 43.1226 for $sW2$

$Fs=(109.83−127.00)243.1226[16+15]=294.808943.1226[0.1667+0.2]=294.808915.8131=18.6433$

Thus, the value of $Fs$ is 18.6433.

Conclusion:

The value of $Fs$ is 18.6433.

Here, the value of $Fs$ is greater than the critical value.

That is, $18.6433>9.39$ .

Thus, the null hypothesis is rejected.

Hence, there is significant difference between the means $X¯3$ and $X¯4$ .

Justification:

Here, there is significant difference between the means $X¯1$ and $X¯4$ , $X¯2$ and $X¯4$ , and $X¯3$ and $X¯4$ . Thus, it can be concluding that there is significant difference between the means of “No high school degree and Graduate degree”, “High school degree and Graduate degree”, “College graduate and Graduate degree”.

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Answer 5

Chapter 12, Problem 2DA

To determine

To test: The hypothesis that the mean IQs of the various educational levels of the subjects are equal or not.

Expert Solution & Answer

Answer to Problem 2DA

There is significant difference between the means $X¯1$ and $X¯4$ , $X¯2$ and $X¯4$ , and $X¯3$ and $X¯4$ . Thus, it can be concluding that there is significant difference between the means of “No high school degree and Graduate degree”, “High school degree and Graduate degree”, “College graduate and Graduate degree”.

Explanation of Solution

Given info:

The databank shows the IQs of the various educational levels of the subjects.

Calculation:

Answers may vary. One of the possible answers is as follows:

Select random samples from the databank. The samples are given below.

No high school degree	High school degree	College graduate	Graduate degree
103	106	110	121
101	98	106	129
103	99	116	126
100	122	100	128
116	109	113	131
103	100	114

The hypotheses are given below:

Null hypothesis:

$H0:μ1=μ2=μ3=μ4$

Alternative hypothesis:

$H1:$ At least one mean is different from the others

Here, the difference in the mean percentage of voters in different places is tested. Hence, the claim is that,there is a difference in the mean percentage of voters in different places.

The number of samples k is 4, the sample sizes $n1$ , $n2$ , $n3$ and $n4$ are 6, 6, 6 and 5.

The degrees of freedom are $d.f.N=k−1 and d.f.D=N−k$ .

Where

$N=n1+n2+n3+n4=6+6+6+5=23$

Substitute 4 for k in $d.f.N$

$d.f.N=k−1 =4-1=3$

Substitute 23 for N and 4 for k in $d.f.D$

$d.f.D=N−k=23−4=19$

Critical value:

The critical F-value is obtained using the Table H: The F-Distribution with the level of significance $α=0.05$ .

Procedure:

Locate 19 in the degrees of freedom, denominator row of the Table H.
Obtain the value in the corresponding degrees of freedom, numerator column below 3.

That is, the critical value is 3.13.

Rejection region:

The null hypothesis would be rejected if $F>3.13$ .

Software procedure:

Step-by-step procedure to obtain the test statistic using the MINITAB software:

Choose Stat > ANOVA > One-Way.
In Response, enter the IQ.
In Factor, enter the Factor.
Click OK.

Output using the MINITAB software is given below:

ALEKS 360 ELEM STATISTICS, Chapter 12, Problem 2DA

From the MINITAB output, the test valueF is 13.33.

Conclusion:

From the results, the test value is 13.33.

Here, the F-statistic value is greater than the critical value.

That is, $13.33>3.13$ .

Thus, it can be concluding that, the null hypothesis is rejected.

Hence, there is evidence to reject the claim thatthe mean IQs of the various educational levels of the subjectsare equal. So use Scheffe test for where the difference exists.

Consider, $X¯1$ , $X¯2$ , $X¯3$ and $X¯4$ represents the means of No high school degree, High school degree, College graduate and Graduate degree, $s12,s22, s32 and s42$ represents the variances of samples of No high school degree, High school degree, College graduate and Graduate degree.

From the Minitab output, the sample sizes $n1,n2,n3 and n4$ are 6, 6, 6 and 5.

The means are $X¯1,X¯2, X¯3 and X¯4$ are 104.33, 105.67, 109.83 and 127.00.

The sample variances are $s12,s22, s32 and s42$ are 34.2225, 82.6281, 35.4025 and 14.5161.

Critical value:

The formula for critical value F¹ for the Scheffe test is,

$F1=(k−1)(Critical value)$

Here, the critical value of F test is 3.13.

Substitute 3.13 for critical value is of F and 3 for k-1 in $F1$

$F1=3(3.13)=9.39$

Comparison of the means:

The formula for finding $sW2$ is,

$sW2=∑(ni−1)si2∑(ni−1)$

That is,

$sW2=(6−1)34.2225+(6−1)82.6281+(6−1)35.4025+(5−1)14.5161(6−1)+(6−1)+(6−1)+(5−1)=171.1125+413.1405+177.0125+58.064419=819.329919=43.1226$

Comparison between the means $X¯1$ and $X¯2$ :

The hypotheses are given below:

Null hypothesis:

$H0:$ There is no significant difference between $X¯1$ and $X¯2$ .

Alternative hypothesis:

$H1:$ There is significant difference between $X¯1$ and $X¯2$ .

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

The formula for comparing the means $X¯1$ and $X¯2$ is,

$Fs=(X¯1−X¯2)2sW2[1n1+1n2]$

Substitute 104.33 and 105.67 for $X¯1$ and $X¯2$ , 6 for n_1, 6 for n₂ and 43.1226 for $sW2$

$Fs=(104.33−105.67)243.1226[16+16]=1.795643.1226[0.1667+0.1667]=1.795614.3771=0.125$

Thus, the value of $Fs$ is 0.125.

Conclusion:

The value of $Fs$ is 0.125.

Here, the value of $Fs$ is lesser than the critical value.

That is, $0.125<9.39$ .

Thus, the null hypothesis is not rejected.

Hence, there is no significant difference between the means $X¯1$ and $X¯2$ .

Comparison between the means $X¯1$ and $X¯3$ :

The hypotheses are given below:

Null hypothesis:

$H0:$ There is no significant difference between $X¯1$ and $X¯3$ .

Alternative hypothesis:

$H1:$ There is significant difference between $X¯1$ and $X¯3$ .

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

The formula for comparing the means $X¯1$ and $X¯3$ .is,

$Fs=(X¯1−X¯3)2sW2[1n1+1n3]$

Substitute 104.33 and 109.83 for $X¯1$ and $X¯3$ , 6 for n_1, 6 for n₃ and 43.1226 for $sW2$

$Fs=(104.33−109.83)243.1226[16+16]=30.2543.1226[0.1667+0.1667]=30.2514.3771=2.104$

Thus, the value of $Fs$ is 2.104.

Conclusion:

The value of $Fs$ is 2.104.

Here, the value of $Fs$ is lesser than the critical value.

That is, $2.104<9.39$ .

Thus, the null hypothesis is not rejected.

Hence, there is no significant difference between the means $X¯1$ and $X¯3$ .

Comparison between the means $X¯2$ and $X¯3$ :

The hypotheses are given below:

Null hypothesis:

$H0:$ There is no significant difference between $X¯2$ and $X¯3$ .

Alternative hypothesis:

$H1:$ There is significant difference between $X¯2$ and $X¯3$ .

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

The formula for comparing the means $X¯2$ and $X¯3$ .is,

$Fs=(X¯2−X¯3)2sW2[1n2+1n3]$

Substitute 105.67 and 109.83 for $X¯2$ and $X¯3$ , 6 for n_2, 6 for n₃ and 43.1226 for $sW2$

$Fs=(105.67−109.83)243.1226[16+16]=17.305643.1226[0.1667+0.1667]=17.305614.3771=1.204$

Thus, the value of $Fs$ is 1.204.

Conclusion:

The value of $Fs$ is 1.204.

Here, the value of $Fs$ is lesser than the critical value.

That is, $1.204<9.39$ .

Thus, the null hypothesis is not rejected.

Hence, there is no significant difference between the means $X¯2$ and $X¯3$ .

Comparison between the means $X¯1$ and $X¯4$ :

The hypotheses are given below:

Null hypothesis:

$H0:$ There is no significant difference between $X¯1$ and $X¯4$ .

Alternative hypothesis:

$H1:$ There is significant difference between $X¯1$ and $X¯4$ .

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

The formula for comparing the means $X¯1$ and $X¯4$ .is,

$Fs=(X¯1−X¯4)2sW2[1n2+1n4]$

Substitute 104.33 and 127.00 for $X¯1$ and $X¯4$ , 6 for n_1, 5 for n₄ and 43.1226 for $sW2$

$Fs=(104.33−127.00)243.1226[16+15]=513.928943.1226[0.1667+0.2]=513.928915.8131=32.5002$

Thus, the value of $Fs$ is 32.5002.

Conclusion:

The value of $Fs$ is 32.5002.

Here, the value of $Fs$ is greater than the critical value.

That is, $32.5002>9.39$ .

Thus, the null hypothesis is rejected.

Hence, there is significant difference between the means $X¯1$ and $X¯4$ .

Comparison between the means $X¯2$ and $X¯4$ :

The hypotheses are given below:

Null hypothesis:

$H0:$ There is no significant difference between $X¯2$ and $X¯4$ .

Alternative hypothesis:

$H1:$ There is significant difference between $X¯2$ and $X¯4$ .

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

The formula for comparing the means $X¯2$ and $X¯4$ .is,

$Fs=(X¯2−X¯4)2sW2[1n2+1n4]$

Substitute 105.67 and 127.00 for $X¯2$ and $X¯4$ , 6 for n_2, 5 for n₄ and 43.1226 for $sW2$

$Fs=(105.67−127.00)243.1226[16+15]=454.968943.1226[0.1667+0.2]=454.968915.8131=28.7716$

Thus, the value of $Fs$ is 28.7716.

Conclusion:

The value of $Fs$ is 28.7716.

Here, the value of $Fs$ is greater than the critical value.

That is, $28.7716>9.39$ .

Thus, the null hypothesis is rejected.

Hence, there is significant difference between the means $X¯2$ and $X¯4$ .

Comparison between the means $X¯3$ and $X¯4$ :

The hypotheses are given below:

Null hypothesis:

$H0:$ There is no significant difference between $X¯3$ and $X¯4$ .

Alternative hypothesis:

$H1:$ There is significant difference between $X¯3$ and $X¯4$ .

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

The formula for comparing the means $X¯3$ and $X¯4$ .is,

$Fs=(X¯3−X¯4)2sW2[1n3+1n4]$

Substitute 109.83 and 127.00 for $X¯3$ and $X¯4$ , 6 for n_2, 5 for n₄ and 43.1226 for $sW2$

$Fs=(109.83−127.00)243.1226[16+15]=294.808943.1226[0.1667+0.2]=294.808915.8131=18.6433$

Thus, the value of $Fs$ is 18.6433.

Conclusion:

The value of $Fs$ is 18.6433.

Here, the value of $Fs$ is greater than the critical value.

That is, $18.6433>9.39$ .

Thus, the null hypothesis is rejected.

Hence, there is significant difference between the means $X¯3$ and $X¯4$ .

Justification:

Here, there is significant difference between the means $X¯1$ and $X¯4$ , $X¯2$ and $X¯4$ , and $X¯3$ and $X¯4$ . Thus, it can be concluding that there is significant difference between the means of “No high school degree and Graduate degree”, “High school degree and Graduate degree”, “College graduate and Graduate degree”.

Answer 6

Chapter 12, Problem 2DA

To determine

To test: The hypothesis that the mean IQs of the various educational levels of the subjects are equal or not.

Expert Solution & Answer

Answer to Problem 2DA

There is significant difference between the means $X¯1$ and $X¯4$ , $X¯2$ and $X¯4$ , and $X¯3$ and $X¯4$ . Thus, it can be concluding that there is significant difference between the means of “No high school degree and Graduate degree”, “High school degree and Graduate degree”, “College graduate and Graduate degree”.

Explanation of Solution

Given info:

The databank shows the IQs of the various educational levels of the subjects.

Calculation:

Answers may vary. One of the possible answers is as follows:

Select random samples from the databank. The samples are given below.

No high school degree	High school degree	College graduate	Graduate degree
103	106	110	121
101	98	106	129
103	99	116	126
100	122	100	128
116	109	113	131
103	100	114

The hypotheses are given below:

Null hypothesis:

$H0:μ1=μ2=μ3=μ4$

Alternative hypothesis:

$H1:$ At least one mean is different from the others

Here, the difference in the mean percentage of voters in different places is tested. Hence, the claim is that,there is a difference in the mean percentage of voters in different places.

The number of samples k is 4, the sample sizes $n1$ , $n2$ , $n3$ and $n4$ are 6, 6, 6 and 5.

The degrees of freedom are $d.f.N=k−1 and d.f.D=N−k$ .

Where

$N=n1+n2+n3+n4=6+6+6+5=23$

Substitute 4 for k in $d.f.N$

$d.f.N=k−1 =4-1=3$

Substitute 23 for N and 4 for k in $d.f.D$

$d.f.D=N−k=23−4=19$

Critical value:

The critical F-value is obtained using the Table H: The F-Distribution with the level of significance $α=0.05$ .

Procedure:

Locate 19 in the degrees of freedom, denominator row of the Table H.
Obtain the value in the corresponding degrees of freedom, numerator column below 3.

That is, the critical value is 3.13.

Rejection region:

The null hypothesis would be rejected if $F>3.13$ .

Software procedure:

Step-by-step procedure to obtain the test statistic using the MINITAB software:

Choose Stat > ANOVA > One-Way.
In Response, enter the IQ.
In Factor, enter the Factor.
Click OK.

Output using the MINITAB software is given below:

ALEKS 360 ELEM STATISTICS, Chapter 12, Problem 2DA

From the MINITAB output, the test valueF is 13.33.

Conclusion:

From the results, the test value is 13.33.

Here, the F-statistic value is greater than the critical value.

That is, $13.33>3.13$ .

Thus, it can be concluding that, the null hypothesis is rejected.

Hence, there is evidence to reject the claim thatthe mean IQs of the various educational levels of the subjectsare equal. So use Scheffe test for where the difference exists.

Consider, $X¯1$ , $X¯2$ , $X¯3$ and $X¯4$ represents the means of No high school degree, High school degree, College graduate and Graduate degree, $s12,s22, s32 and s42$ represents the variances of samples of No high school degree, High school degree, College graduate and Graduate degree.

From the Minitab output, the sample sizes $n1,n2,n3 and n4$ are 6, 6, 6 and 5.

The means are $X¯1,X¯2, X¯3 and X¯4$ are 104.33, 105.67, 109.83 and 127.00.

The sample variances are $s12,s22, s32 and s42$ are 34.2225, 82.6281, 35.4025 and 14.5161.

Critical value:

The formula for critical value F¹ for the Scheffe test is,

$F1=(k−1)(Critical value)$

Here, the critical value of F test is 3.13.

Substitute 3.13 for critical value is of F and 3 for k-1 in $F1$

$F1=3(3.13)=9.39$

Comparison of the means:

The formula for finding $sW2$ is,

$sW2=∑(ni−1)si2∑(ni−1)$

That is,

$sW2=(6−1)34.2225+(6−1)82.6281+(6−1)35.4025+(5−1)14.5161(6−1)+(6−1)+(6−1)+(5−1)=171.1125+413.1405+177.0125+58.064419=819.329919=43.1226$

Comparison between the means $X¯1$ and $X¯2$ :

The hypotheses are given below:

Null hypothesis:

$H0:$ There is no significant difference between $X¯1$ and $X¯2$ .

Alternative hypothesis:

$H1:$ There is significant difference between $X¯1$ and $X¯2$ .

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

The formula for comparing the means $X¯1$ and $X¯2$ is,

$Fs=(X¯1−X¯2)2sW2[1n1+1n2]$

Substitute 104.33 and 105.67 for $X¯1$ and $X¯2$ , 6 for n_1, 6 for n₂ and 43.1226 for $sW2$

$Fs=(104.33−105.67)243.1226[16+16]=1.795643.1226[0.1667+0.1667]=1.795614.3771=0.125$

Thus, the value of $Fs$ is 0.125.

Conclusion:

The value of $Fs$ is 0.125.

Here, the value of $Fs$ is lesser than the critical value.

That is, $0.125<9.39$ .

Thus, the null hypothesis is not rejected.

Hence, there is no significant difference between the means $X¯1$ and $X¯2$ .

Comparison between the means $X¯1$ and $X¯3$ :

The hypotheses are given below:

Null hypothesis:

$H0:$ There is no significant difference between $X¯1$ and $X¯3$ .

Alternative hypothesis:

$H1:$ There is significant difference between $X¯1$ and $X¯3$ .

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

The formula for comparing the means $X¯1$ and $X¯3$ .is,

$Fs=(X¯1−X¯3)2sW2[1n1+1n3]$

Substitute 104.33 and 109.83 for $X¯1$ and $X¯3$ , 6 for n_1, 6 for n₃ and 43.1226 for $sW2$

$Fs=(104.33−109.83)243.1226[16+16]=30.2543.1226[0.1667+0.1667]=30.2514.3771=2.104$

Thus, the value of $Fs$ is 2.104.

Conclusion:

The value of $Fs$ is 2.104.

Here, the value of $Fs$ is lesser than the critical value.

That is, $2.104<9.39$ .

Thus, the null hypothesis is not rejected.

Hence, there is no significant difference between the means $X¯1$ and $X¯3$ .

Comparison between the means $X¯2$ and $X¯3$ :

The hypotheses are given below:

Null hypothesis:

$H0:$ There is no significant difference between $X¯2$ and $X¯3$ .

Alternative hypothesis:

$H1:$ There is significant difference between $X¯2$ and $X¯3$ .

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

The formula for comparing the means $X¯2$ and $X¯3$ .is,

$Fs=(X¯2−X¯3)2sW2[1n2+1n3]$

Substitute 105.67 and 109.83 for $X¯2$ and $X¯3$ , 6 for n_2, 6 for n₃ and 43.1226 for $sW2$

$Fs=(105.67−109.83)243.1226[16+16]=17.305643.1226[0.1667+0.1667]=17.305614.3771=1.204$

Thus, the value of $Fs$ is 1.204.

Conclusion:

The value of $Fs$ is 1.204.

Here, the value of $Fs$ is lesser than the critical value.

That is, $1.204<9.39$ .

Thus, the null hypothesis is not rejected.

Hence, there is no significant difference between the means $X¯2$ and $X¯3$ .

Comparison between the means $X¯1$ and $X¯4$ :

The hypotheses are given below:

Null hypothesis:

$H0:$ There is no significant difference between $X¯1$ and $X¯4$ .

Alternative hypothesis:

$H1:$ There is significant difference between $X¯1$ and $X¯4$ .

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

The formula for comparing the means $X¯1$ and $X¯4$ .is,

$Fs=(X¯1−X¯4)2sW2[1n2+1n4]$

Substitute 104.33 and 127.00 for $X¯1$ and $X¯4$ , 6 for n_1, 5 for n₄ and 43.1226 for $sW2$

$Fs=(104.33−127.00)243.1226[16+15]=513.928943.1226[0.1667+0.2]=513.928915.8131=32.5002$

Thus, the value of $Fs$ is 32.5002.

Conclusion:

The value of $Fs$ is 32.5002.

Here, the value of $Fs$ is greater than the critical value.

That is, $32.5002>9.39$ .

Thus, the null hypothesis is rejected.

Hence, there is significant difference between the means $X¯1$ and $X¯4$ .

Comparison between the means $X¯2$ and $X¯4$ :

The hypotheses are given below:

Null hypothesis:

$H0:$ There is no significant difference between $X¯2$ and $X¯4$ .

Alternative hypothesis:

$H1:$ There is significant difference between $X¯2$ and $X¯4$ .

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

The formula for comparing the means $X¯2$ and $X¯4$ .is,

$Fs=(X¯2−X¯4)2sW2[1n2+1n4]$

Substitute 105.67 and 127.00 for $X¯2$ and $X¯4$ , 6 for n_2, 5 for n₄ and 43.1226 for $sW2$

$Fs=(105.67−127.00)243.1226[16+15]=454.968943.1226[0.1667+0.2]=454.968915.8131=28.7716$

Thus, the value of $Fs$ is 28.7716.

Conclusion:

The value of $Fs$ is 28.7716.

Here, the value of $Fs$ is greater than the critical value.

That is, $28.7716>9.39$ .

Thus, the null hypothesis is rejected.

Hence, there is significant difference between the means $X¯2$ and $X¯4$ .

Comparison between the means $X¯3$ and $X¯4$ :

The hypotheses are given below:

Null hypothesis:

$H0:$ There is no significant difference between $X¯3$ and $X¯4$ .

Alternative hypothesis:

$H1:$ There is significant difference between $X¯3$ and $X¯4$ .

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

The formula for comparing the means $X¯3$ and $X¯4$ .is,

$Fs=(X¯3−X¯4)2sW2[1n3+1n4]$

Substitute 109.83 and 127.00 for $X¯3$ and $X¯4$ , 6 for n_2, 5 for n₄ and 43.1226 for $sW2$

$Fs=(109.83−127.00)243.1226[16+15]=294.808943.1226[0.1667+0.2]=294.808915.8131=18.6433$

Thus, the value of $Fs$ is 18.6433.

Conclusion:

The value of $Fs$ is 18.6433.

Here, the value of $Fs$ is greater than the critical value.

That is, $18.6433>9.39$ .

Thus, the null hypothesis is rejected.

Hence, there is significant difference between the means $X¯3$ and $X¯4$ .

Justification:

Here, there is significant difference between the means $X¯1$ and $X¯4$ , $X¯2$ and $X¯4$ , and $X¯3$ and $X¯4$ . Thus, it can be concluding that there is significant difference between the means of “No high school degree and Graduate degree”, “High school degree and Graduate degree”, “College graduate and Graduate degree”.

Answer 7

Chapter 12, Problem 2DA

To determine

To test: The hypothesis that the mean IQs of the various educational levels of the subjects are equal or not.

Answer 8

Expert Solution & Answer

Answer to Problem 2DA

There is significant difference between the means $X¯1$ and $X¯4$ , $X¯2$ and $X¯4$ , and $X¯3$ and $X¯4$ . Thus, it can be concluding that there is significant difference between the means of “No high school degree and Graduate degree”, “High school degree and Graduate degree”, “College graduate and Graduate degree”.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given info:

The databank shows the IQs of the various educational levels of the subjects.

Calculation:

Answers may vary. One of the possible answers is as follows:

Select random samples from the databank. The samples are given below.

No high school degree	High school degree	College graduate	Graduate degree
103	106	110	121
101	98	106	129
103	99	116	126
100	122	100	128
116	109	113	131
103	100	114

The hypotheses are given below:

Null hypothesis:

$H0:μ1=μ2=μ3=μ4$

Alternative hypothesis:

$H1:$ At least one mean is different from the others

Here, the difference in the mean percentage of voters in different places is tested. Hence, the claim is that,there is a difference in the mean percentage of voters in different places.

The number of samples k is 4, the sample sizes $n1$ , $n2$ , $n3$ and $n4$ are 6, 6, 6 and 5.

The degrees of freedom are $d.f.N=k−1 and d.f.D=N−k$ .

Where

$N=n1+n2+n3+n4=6+6+6+5=23$

Substitute 4 for k in $d.f.N$

$d.f.N=k−1 =4-1=3$

Substitute 23 for N and 4 for k in $d.f.D$

$d.f.D=N−k=23−4=19$

Critical value:

The critical F-value is obtained using the Table H: The F-Distribution with the level of significance $α=0.05$ .

Procedure:

Locate 19 in the degrees of freedom, denominator row of the Table H.
Obtain the value in the corresponding degrees of freedom, numerator column below 3.

That is, the critical value is 3.13.

Rejection region:

The null hypothesis would be rejected if $F>3.13$ .

Software procedure:

Step-by-step procedure to obtain the test statistic using the MINITAB software:

Choose Stat > ANOVA > One-Way.
In Response, enter the IQ.
In Factor, enter the Factor.
Click OK.

Output using the MINITAB software is given below:

ALEKS 360 ELEM STATISTICS, Chapter 12, Problem 2DA

From the MINITAB output, the test valueF is 13.33.

Conclusion:

From the results, the test value is 13.33.

Here, the F-statistic value is greater than the critical value.

That is, $13.33>3.13$ .

Thus, it can be concluding that, the null hypothesis is rejected.

Hence, there is evidence to reject the claim thatthe mean IQs of the various educational levels of the subjectsare equal. So use Scheffe test for where the difference exists.

Consider, $X¯1$ , $X¯2$ , $X¯3$ and $X¯4$ represents the means of No high school degree, High school degree, College graduate and Graduate degree, $s12,s22, s32 and s42$ represents the variances of samples of No high school degree, High school degree, College graduate and Graduate degree.

From the Minitab output, the sample sizes $n1,n2,n3 and n4$ are 6, 6, 6 and 5.

The means are $X¯1,X¯2, X¯3 and X¯4$ are 104.33, 105.67, 109.83 and 127.00.

The sample variances are $s12,s22, s32 and s42$ are 34.2225, 82.6281, 35.4025 and 14.5161.

Critical value:

The formula for critical value F¹ for the Scheffe test is,

$F1=(k−1)(Critical value)$

Here, the critical value of F test is 3.13.

Substitute 3.13 for critical value is of F and 3 for k-1 in $F1$

$F1=3(3.13)=9.39$

Comparison of the means:

The formula for finding $sW2$ is,

$sW2=∑(ni−1)si2∑(ni−1)$

That is,

$sW2=(6−1)34.2225+(6−1)82.6281+(6−1)35.4025+(5−1)14.5161(6−1)+(6−1)+(6−1)+(5−1)=171.1125+413.1405+177.0125+58.064419=819.329919=43.1226$

Comparison between the means $X¯1$ and $X¯2$ :

The hypotheses are given below:

Null hypothesis:

$H0:$ There is no significant difference between $X¯1$ and $X¯2$ .

Alternative hypothesis:

$H1:$ There is significant difference between $X¯1$ and $X¯2$ .

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

The formula for comparing the means $X¯1$ and $X¯2$ is,

$Fs=(X¯1−X¯2)2sW2[1n1+1n2]$

Substitute 104.33 and 105.67 for $X¯1$ and $X¯2$ , 6 for n_1, 6 for n₂ and 43.1226 for $sW2$

$Fs=(104.33−105.67)243.1226[16+16]=1.795643.1226[0.1667+0.1667]=1.795614.3771=0.125$

Thus, the value of $Fs$ is 0.125.

Conclusion:

The value of $Fs$ is 0.125.

Here, the value of $Fs$ is lesser than the critical value.

That is, $0.125<9.39$ .

Thus, the null hypothesis is not rejected.

Hence, there is no significant difference between the means $X¯1$ and $X¯2$ .

Comparison between the means $X¯1$ and $X¯3$ :

The hypotheses are given below:

Null hypothesis:

$H0:$ There is no significant difference between $X¯1$ and $X¯3$ .

Alternative hypothesis:

$H1:$ There is significant difference between $X¯1$ and $X¯3$ .

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

The formula for comparing the means $X¯1$ and $X¯3$ .is,

$Fs=(X¯1−X¯3)2sW2[1n1+1n3]$

Substitute 104.33 and 109.83 for $X¯1$ and $X¯3$ , 6 for n_1, 6 for n₃ and 43.1226 for $sW2$

$Fs=(104.33−109.83)243.1226[16+16]=30.2543.1226[0.1667+0.1667]=30.2514.3771=2.104$

Thus, the value of $Fs$ is 2.104.

Conclusion:

The value of $Fs$ is 2.104.

Here, the value of $Fs$ is lesser than the critical value.

That is, $2.104<9.39$ .

Thus, the null hypothesis is not rejected.

Hence, there is no significant difference between the means $X¯1$ and $X¯3$ .

Comparison between the means $X¯2$ and $X¯3$ :

The hypotheses are given below:

Null hypothesis:

$H0:$ There is no significant difference between $X¯2$ and $X¯3$ .

Alternative hypothesis:

$H1:$ There is significant difference between $X¯2$ and $X¯3$ .

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

The formula for comparing the means $X¯2$ and $X¯3$ .is,

$Fs=(X¯2−X¯3)2sW2[1n2+1n3]$

Substitute 105.67 and 109.83 for $X¯2$ and $X¯3$ , 6 for n_2, 6 for n₃ and 43.1226 for $sW2$

$Fs=(105.67−109.83)243.1226[16+16]=17.305643.1226[0.1667+0.1667]=17.305614.3771=1.204$

Thus, the value of $Fs$ is 1.204.

Conclusion:

The value of $Fs$ is 1.204.

Here, the value of $Fs$ is lesser than the critical value.

That is, $1.204<9.39$ .

Thus, the null hypothesis is not rejected.

Hence, there is no significant difference between the means $X¯2$ and $X¯3$ .

Comparison between the means $X¯1$ and $X¯4$ :

The hypotheses are given below:

Null hypothesis:

$H0:$ There is no significant difference between $X¯1$ and $X¯4$ .

Alternative hypothesis:

$H1:$ There is significant difference between $X¯1$ and $X¯4$ .

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

The formula for comparing the means $X¯1$ and $X¯4$ .is,

$Fs=(X¯1−X¯4)2sW2[1n2+1n4]$

Substitute 104.33 and 127.00 for $X¯1$ and $X¯4$ , 6 for n_1, 5 for n₄ and 43.1226 for $sW2$

$Fs=(104.33−127.00)243.1226[16+15]=513.928943.1226[0.1667+0.2]=513.928915.8131=32.5002$

Thus, the value of $Fs$ is 32.5002.

Conclusion:

The value of $Fs$ is 32.5002.

Here, the value of $Fs$ is greater than the critical value.

That is, $32.5002>9.39$ .

Thus, the null hypothesis is rejected.

Hence, there is significant difference between the means $X¯1$ and $X¯4$ .

Comparison between the means $X¯2$ and $X¯4$ :

The hypotheses are given below:

Null hypothesis:

$H0:$ There is no significant difference between $X¯2$ and $X¯4$ .

Alternative hypothesis:

$H1:$ There is significant difference between $X¯2$ and $X¯4$ .

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

The formula for comparing the means $X¯2$ and $X¯4$ .is,

$Fs=(X¯2−X¯4)2sW2[1n2+1n4]$

Substitute 105.67 and 127.00 for $X¯2$ and $X¯4$ , 6 for n_2, 5 for n₄ and 43.1226 for $sW2$

$Fs=(105.67−127.00)243.1226[16+15]=454.968943.1226[0.1667+0.2]=454.968915.8131=28.7716$

Thus, the value of $Fs$ is 28.7716.

Conclusion:

The value of $Fs$ is 28.7716.

Here, the value of $Fs$ is greater than the critical value.

That is, $28.7716>9.39$ .

Thus, the null hypothesis is rejected.

Hence, there is significant difference between the means $X¯2$ and $X¯4$ .

Comparison between the means $X¯3$ and $X¯4$ :

The hypotheses are given below:

Null hypothesis:

$H0:$ There is no significant difference between $X¯3$ and $X¯4$ .

Alternative hypothesis:

$H1:$ There is significant difference between $X¯3$ and $X¯4$ .

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

The formula for comparing the means $X¯3$ and $X¯4$ .is,

$Fs=(X¯3−X¯4)2sW2[1n3+1n4]$

Substitute 109.83 and 127.00 for $X¯3$ and $X¯4$ , 6 for n_2, 5 for n₄ and 43.1226 for $sW2$

$Fs=(109.83−127.00)243.1226[16+15]=294.808943.1226[0.1667+0.2]=294.808915.8131=18.6433$

Thus, the value of $Fs$ is 18.6433.

Conclusion:

The value of $Fs$ is 18.6433.

Here, the value of $Fs$ is greater than the critical value.

That is, $18.6433>9.39$ .

Thus, the null hypothesis is rejected.

Hence, there is significant difference between the means $X¯3$ and $X¯4$ .

Justification:

Here, there is significant difference between the means $X¯1$ and $X¯4$ , $X¯2$ and $X¯4$ , and $X¯3$ and $X¯4$ . Thus, it can be concluding that there is significant difference between the means of “No high school degree and Graduate degree”, “High school degree and Graduate degree”, “College graduate and Graduate degree”.

Answer 12

Ch. 12.1 - Colors That Make You Smarter The following set of...Ch. 12.1 - What test is used to compare three or more means?Ch. 12.1 - State three reasons why multiple t tests cannot be...Ch. 12.1 - What are the assumptions for ANOVA?Ch. 12.1 - Define between-group variance and within-group...Ch. 12.1 - State the hypotheses used in the ANOVA test.Ch. 12.1 - When there is no significant difference among...Ch. 12.1 - For Exercises 7 through 20, assume that all...Ch. 12.1 - For Exercises 7 through 20, assume that all...Ch. 12.1 - For Exercises 7 through 20, assume that all...

To test: The hypothesis that the mean IQs of the various educational levels of the subjects are equal or not.

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To test: The hypothesis that the mean IQs of the various educational levels of the subjects are equal or not.

Answer to Problem 2DA

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