whether the two sample variances are reasonable to use common population variance

Question

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Answer 1

Question

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

Chapter 10.4, Problem 10.24E

(a)

To determine

To explain:whether the two sample variances are reasonable to use common population variance

(a)

Expert Solution

Answer to Problem 10.24E

The obtained value is less than 3; therefore it is reasonable to use common population variance.

Explanation of Solution

Given:

D	E	F
t-Test: Two-sample Assuming Equal Variances
	Sample 1	Sample 2
Mean	28.667	28.286
Variance	5.067	2.238
Observations	6	7
Pooled Variance	3.524
Hypothesized Mean Difference	0
df	11
t Stat	0.365
P(T<=t) one-tail	0.361
t Critical one-tail	1.796
P(T<=t) two-tail	0.722
t Critical two-tail	2.201

Calculation:

To test the common population variance is reasonable or not, we have the test statistics asLarger s2smaller s2>3

From the output by substituting the values we get

5.0672.238=2.264

This value is less than 3 so it is reasonable to use common population variance.

Conclusion: The obtained value is less than 3, therefore it is reasonable to use common population variance.

(b)

To determine

To find: the observed value of the test statistic andthe P-value associated with the test

(b)

Expert Solution

Answer to Problem 10.24E

The observed value for the test statistics is tSTAT=0.365 . The P-value associated with the test statistics for two tailed test is P−value=0.722 .

Explanation of Solution

Given:

D	E	F
t-Test: Two-sample Assuming Equal Variances
	Sample 1	Sample 2
Mean	28.667	28.286
Variance	5.067	2.238
Observations	6	7
Pooled Variance	3.524
Hypothesized Mean Difference	0
df	11
t Stat	0.365
P(T<=t) one-tail	0.361
t Critical one-tail	1.796
P(T<=t) two-tail	0.722
t Critical two-tail	2.201

Calculation:

From the above output the observed value for the test statistics is tSTAT=0.365 .

The P-value associated with the test statistics for two tailed test is P−value=0.722 .

The observed value for the test statistics is tSTAT=0.365 . The P-value associated with the test statistics for two tailed test is P−value=0.722 .

Conclusion:

Therefore, observed value for the test statistics is tSTAT=0.365 and the P-value associated with the test statistics for two tailed test is P−value=0.722 .

(c)

To determine

To find: the pooled estimate s2 of the population variance

(c)

Expert Solution

Answer to Problem 10.24E

The pooled estimate s2 of the population variance is 3.524 .

Explanation of Solution

Given:

D	E	F
t-Test: Two-sample Assuming Equal Variances
	Sample 1	Sample 2
Mean	28.667	28.286
Variance	5.067	2.238
Observations	6	7
Pooled Variance	3.524
Hypothesized Mean Difference	0
df	11
t Stat	0.365
P(T<=t) one-tail	0.361
t Critical one-tail	1.796
P(T<=t) two-tail	0.722
t Critical two-tail	2.201

Calculation:

From the table, the pooled estimate s2 of the population varianceis 3.524

Conclusion: Thus, the pooled estimate s2 of the population variance is 3.524 .

(d)

To determine

To state: conclusions about the difference in the two populations by using part b answer

(d)

Expert Solution

Answer to Problem 10.24E

From part b we can see that the test statistics value is not lie in the critical region we fail to reject the null hypothesis that the population means are equal.

Explanation of Solution

Calculation:

t-critical value for the two-tail is 2.201

From part b we can see that the test statistics value is not lie in the critical region we fail to reject the null hypothesis that the population means are equal.

Conclusion: therefore, from part b we can see that the test statistics value is not lie in the critical region we fail to reject the null hypothesis that the population means are equal.

(e)

To determine

To find: a 95% confidence interval for the difference in the population and explain whether the interval confirm the conclusion in part d

(e)

Expert Solution

Answer to Problem 10.24E

The 95% confidence interval for the difference in the population mean is

(−1.890,2.698) . From the interval we can see that the interval contains 0 value, so the interval conforms the conclusion in part d.

Explanation of Solution

Given:

D	E	F
t-Test: Two-sample Assuming Equal Variances
	Sample 1	Sample 2
Mean	28.667	28.286
Variance	5.067	2.238
Observations	6	7
Pooled Variance	3.524
Hypothesized Mean Difference	0
df	11
t Stat	0.365
P(T<=t) one-tail	0.361
t Critical one-tail	1.796
P(T<=t) two-tail	0.722
t Critical two-tail	2.201

Calculation:

The 95% confidence interval for the difference in the population mean is

((x1¯−x2¯)±tαs2( 1 n 1 + 1 n 2 ))

From the above output by substituting the value, we get

((x1¯−x2¯)±tαs2( 1 n 1 + 1 n 2 ))

((28.667−28.268)±2.2013.524( 1 6 + 1 7 ))

(−1.890,2.698)

From the interval we can see that the interval contains 0 value so the interval conforms the conclusion in part d.

Conclusion: Therefore, 95% confidence interval for the difference in the population mean is

(−1.890,2.698) . From the interval we can see that the interval contains 0 value, so the interval conforms the conclusion in part d.

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Answer 2

Question

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

Chapter 10.4, Problem 10.24E

(a)

To determine

To explain:whether the two sample variances are reasonable to use common population variance

(a)

Expert Solution

Answer to Problem 10.24E

The obtained value is less than 3; therefore it is reasonable to use common population variance.

Explanation of Solution

Given:

D	E	F
t-Test: Two-sample Assuming Equal Variances
	Sample 1	Sample 2
Mean	28.667	28.286
Variance	5.067	2.238
Observations	6	7
Pooled Variance	3.524
Hypothesized Mean Difference	0
df	11
t Stat	0.365
P(T<=t) one-tail	0.361
t Critical one-tail	1.796
P(T<=t) two-tail	0.722
t Critical two-tail	2.201

Calculation:

To test the common population variance is reasonable or not, we have the test statistics asLarger s2smaller s2>3

From the output by substituting the values we get

5.0672.238=2.264

This value is less than 3 so it is reasonable to use common population variance.

Conclusion: The obtained value is less than 3, therefore it is reasonable to use common population variance.

(b)

To determine

To find: the observed value of the test statistic andthe P-value associated with the test

(b)

Expert Solution

Answer to Problem 10.24E

The observed value for the test statistics is tSTAT=0.365 . The P-value associated with the test statistics for two tailed test is P−value=0.722 .

Explanation of Solution

Given:

D	E	F
t-Test: Two-sample Assuming Equal Variances
	Sample 1	Sample 2
Mean	28.667	28.286
Variance	5.067	2.238
Observations	6	7
Pooled Variance	3.524
Hypothesized Mean Difference	0
df	11
t Stat	0.365
P(T<=t) one-tail	0.361
t Critical one-tail	1.796
P(T<=t) two-tail	0.722
t Critical two-tail	2.201

Calculation:

From the above output the observed value for the test statistics is tSTAT=0.365 .

The P-value associated with the test statistics for two tailed test is P−value=0.722 .

The observed value for the test statistics is tSTAT=0.365 . The P-value associated with the test statistics for two tailed test is P−value=0.722 .

Conclusion:

Therefore, observed value for the test statistics is tSTAT=0.365 and the P-value associated with the test statistics for two tailed test is P−value=0.722 .

(c)

To determine

To find: the pooled estimate s2 of the population variance

(c)

Expert Solution

Answer to Problem 10.24E

The pooled estimate s2 of the population variance is 3.524 .

Explanation of Solution

Given:

D	E	F
t-Test: Two-sample Assuming Equal Variances
	Sample 1	Sample 2
Mean	28.667	28.286
Variance	5.067	2.238
Observations	6	7
Pooled Variance	3.524
Hypothesized Mean Difference	0
df	11
t Stat	0.365
P(T<=t) one-tail	0.361
t Critical one-tail	1.796
P(T<=t) two-tail	0.722
t Critical two-tail	2.201

Calculation:

From the table, the pooled estimate s2 of the population varianceis 3.524

Conclusion: Thus, the pooled estimate s2 of the population variance is 3.524 .

(d)

To determine

To state: conclusions about the difference in the two populations by using part b answer

(d)

Expert Solution

Answer to Problem 10.24E

From part b we can see that the test statistics value is not lie in the critical region we fail to reject the null hypothesis that the population means are equal.

Explanation of Solution

Calculation:

t-critical value for the two-tail is 2.201

From part b we can see that the test statistics value is not lie in the critical region we fail to reject the null hypothesis that the population means are equal.

Conclusion: therefore, from part b we can see that the test statistics value is not lie in the critical region we fail to reject the null hypothesis that the population means are equal.

(e)

To determine

To find: a 95% confidence interval for the difference in the population and explain whether the interval confirm the conclusion in part d

(e)

Expert Solution

Answer to Problem 10.24E

The 95% confidence interval for the difference in the population mean is

(−1.890,2.698) . From the interval we can see that the interval contains 0 value, so the interval conforms the conclusion in part d.

Explanation of Solution

Given:

D	E	F
t-Test: Two-sample Assuming Equal Variances
	Sample 1	Sample 2
Mean	28.667	28.286
Variance	5.067	2.238
Observations	6	7
Pooled Variance	3.524
Hypothesized Mean Difference	0
df	11
t Stat	0.365
P(T<=t) one-tail	0.361
t Critical one-tail	1.796
P(T<=t) two-tail	0.722
t Critical two-tail	2.201

Calculation:

The 95% confidence interval for the difference in the population mean is

((x1¯−x2¯)±tαs2( 1 n 1 + 1 n 2 ))

From the above output by substituting the value, we get

((x1¯−x2¯)±tαs2( 1 n 1 + 1 n 2 ))

((28.667−28.268)±2.2013.524( 1 6 + 1 7 ))

(−1.890,2.698)

From the interval we can see that the interval contains 0 value so the interval conforms the conclusion in part d.

Conclusion: Therefore, 95% confidence interval for the difference in the population mean is

(−1.890,2.698) . From the interval we can see that the interval contains 0 value, so the interval conforms the conclusion in part d.

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Answer 3

Question

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

Chapter 10.4, Problem 10.24E

(a)

To determine

To explain:whether the two sample variances are reasonable to use common population variance

(a)

Expert Solution

Answer to Problem 10.24E

The obtained value is less than 3; therefore it is reasonable to use common population variance.

Explanation of Solution

Given:

D	E	F
t-Test: Two-sample Assuming Equal Variances
	Sample 1	Sample 2
Mean	28.667	28.286
Variance	5.067	2.238
Observations	6	7
Pooled Variance	3.524
Hypothesized Mean Difference	0
df	11
t Stat	0.365
P(T<=t) one-tail	0.361
t Critical one-tail	1.796
P(T<=t) two-tail	0.722
t Critical two-tail	2.201

Calculation:

To test the common population variance is reasonable or not, we have the test statistics asLarger s2smaller s2>3

From the output by substituting the values we get

5.0672.238=2.264

This value is less than 3 so it is reasonable to use common population variance.

Conclusion: The obtained value is less than 3, therefore it is reasonable to use common population variance.

(b)

To determine

To find: the observed value of the test statistic andthe P-value associated with the test

(b)

Expert Solution

Answer to Problem 10.24E

The observed value for the test statistics is tSTAT=0.365 . The P-value associated with the test statistics for two tailed test is P−value=0.722 .

Explanation of Solution

Given:

D	E	F
t-Test: Two-sample Assuming Equal Variances
	Sample 1	Sample 2
Mean	28.667	28.286
Variance	5.067	2.238
Observations	6	7
Pooled Variance	3.524
Hypothesized Mean Difference	0
df	11
t Stat	0.365
P(T<=t) one-tail	0.361
t Critical one-tail	1.796
P(T<=t) two-tail	0.722
t Critical two-tail	2.201

Calculation:

From the above output the observed value for the test statistics is tSTAT=0.365 .

The P-value associated with the test statistics for two tailed test is P−value=0.722 .

The observed value for the test statistics is tSTAT=0.365 . The P-value associated with the test statistics for two tailed test is P−value=0.722 .

Conclusion:

Therefore, observed value for the test statistics is tSTAT=0.365 and the P-value associated with the test statistics for two tailed test is P−value=0.722 .

(c)

To determine

To find: the pooled estimate s2 of the population variance

(c)

Expert Solution

Answer to Problem 10.24E

The pooled estimate s2 of the population variance is 3.524 .

Explanation of Solution

Given:

D	E	F
t-Test: Two-sample Assuming Equal Variances
	Sample 1	Sample 2
Mean	28.667	28.286
Variance	5.067	2.238
Observations	6	7
Pooled Variance	3.524
Hypothesized Mean Difference	0
df	11
t Stat	0.365
P(T<=t) one-tail	0.361
t Critical one-tail	1.796
P(T<=t) two-tail	0.722
t Critical two-tail	2.201

Calculation:

From the table, the pooled estimate s2 of the population varianceis 3.524

Conclusion: Thus, the pooled estimate s2 of the population variance is 3.524 .

(d)

To determine

To state: conclusions about the difference in the two populations by using part b answer

(d)

Expert Solution

Answer to Problem 10.24E

From part b we can see that the test statistics value is not lie in the critical region we fail to reject the null hypothesis that the population means are equal.

Explanation of Solution

Calculation:

t-critical value for the two-tail is 2.201

From part b we can see that the test statistics value is not lie in the critical region we fail to reject the null hypothesis that the population means are equal.

Conclusion: therefore, from part b we can see that the test statistics value is not lie in the critical region we fail to reject the null hypothesis that the population means are equal.

(e)

To determine

To find: a 95% confidence interval for the difference in the population and explain whether the interval confirm the conclusion in part d

(e)

Expert Solution

Answer to Problem 10.24E

The 95% confidence interval for the difference in the population mean is

(−1.890,2.698) . From the interval we can see that the interval contains 0 value, so the interval conforms the conclusion in part d.

Explanation of Solution

Given:

D	E	F
t-Test: Two-sample Assuming Equal Variances
	Sample 1	Sample 2
Mean	28.667	28.286
Variance	5.067	2.238
Observations	6	7
Pooled Variance	3.524
Hypothesized Mean Difference	0
df	11
t Stat	0.365
P(T<=t) one-tail	0.361
t Critical one-tail	1.796
P(T<=t) two-tail	0.722
t Critical two-tail	2.201

Calculation:

The 95% confidence interval for the difference in the population mean is

((x1¯−x2¯)±tαs2( 1 n 1 + 1 n 2 ))

From the above output by substituting the value, we get

((x1¯−x2¯)±tαs2( 1 n 1 + 1 n 2 ))

((28.667−28.268)±2.2013.524( 1 6 + 1 7 ))

(−1.890,2.698)

From the interval we can see that the interval contains 0 value so the interval conforms the conclusion in part d.

Conclusion: Therefore, 95% confidence interval for the difference in the population mean is

(−1.890,2.698) . From the interval we can see that the interval contains 0 value, so the interval conforms the conclusion in part d.

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Answer 4

Question

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

Chapter 10.4, Problem 10.24E

(a)

To determine

To explain:whether the two sample variances are reasonable to use common population variance

(a)

Expert Solution

Answer to Problem 10.24E

The obtained value is less than 3; therefore it is reasonable to use common population variance.

Explanation of Solution

Given:

D	E	F
t-Test: Two-sample Assuming Equal Variances
	Sample 1	Sample 2
Mean	28.667	28.286
Variance	5.067	2.238
Observations	6	7
Pooled Variance	3.524
Hypothesized Mean Difference	0
df	11
t Stat	0.365
P(T<=t) one-tail	0.361
t Critical one-tail	1.796
P(T<=t) two-tail	0.722
t Critical two-tail	2.201

Calculation:

To test the common population variance is reasonable or not, we have the test statistics asLarger s2smaller s2>3

From the output by substituting the values we get

5.0672.238=2.264

This value is less than 3 so it is reasonable to use common population variance.

Conclusion: The obtained value is less than 3, therefore it is reasonable to use common population variance.

(b)

To determine

To find: the observed value of the test statistic andthe P-value associated with the test

(b)

Expert Solution

Answer to Problem 10.24E

The observed value for the test statistics is tSTAT=0.365 . The P-value associated with the test statistics for two tailed test is P−value=0.722 .

Explanation of Solution

Given:

D	E	F
t-Test: Two-sample Assuming Equal Variances
	Sample 1	Sample 2
Mean	28.667	28.286
Variance	5.067	2.238
Observations	6	7
Pooled Variance	3.524
Hypothesized Mean Difference	0
df	11
t Stat	0.365
P(T<=t) one-tail	0.361
t Critical one-tail	1.796
P(T<=t) two-tail	0.722
t Critical two-tail	2.201

Calculation:

From the above output the observed value for the test statistics is tSTAT=0.365 .

The P-value associated with the test statistics for two tailed test is P−value=0.722 .

The observed value for the test statistics is tSTAT=0.365 . The P-value associated with the test statistics for two tailed test is P−value=0.722 .

Conclusion:

Therefore, observed value for the test statistics is tSTAT=0.365 and the P-value associated with the test statistics for two tailed test is P−value=0.722 .

(c)

To determine

To find: the pooled estimate s2 of the population variance

(c)

Expert Solution

Answer to Problem 10.24E

The pooled estimate s2 of the population variance is 3.524 .

Explanation of Solution

Given:

D	E	F
t-Test: Two-sample Assuming Equal Variances
	Sample 1	Sample 2
Mean	28.667	28.286
Variance	5.067	2.238
Observations	6	7
Pooled Variance	3.524
Hypothesized Mean Difference	0
df	11
t Stat	0.365
P(T<=t) one-tail	0.361
t Critical one-tail	1.796
P(T<=t) two-tail	0.722
t Critical two-tail	2.201

Calculation:

From the table, the pooled estimate s2 of the population varianceis 3.524

Conclusion: Thus, the pooled estimate s2 of the population variance is 3.524 .

(d)

To determine

To state: conclusions about the difference in the two populations by using part b answer

(d)

Expert Solution

Answer to Problem 10.24E

From part b we can see that the test statistics value is not lie in the critical region we fail to reject the null hypothesis that the population means are equal.

Explanation of Solution

Calculation:

t-critical value for the two-tail is 2.201

From part b we can see that the test statistics value is not lie in the critical region we fail to reject the null hypothesis that the population means are equal.

Conclusion: therefore, from part b we can see that the test statistics value is not lie in the critical region we fail to reject the null hypothesis that the population means are equal.

(e)

To determine

To find: a 95% confidence interval for the difference in the population and explain whether the interval confirm the conclusion in part d

(e)

Expert Solution

Answer to Problem 10.24E

The 95% confidence interval for the difference in the population mean is

(−1.890,2.698) . From the interval we can see that the interval contains 0 value, so the interval conforms the conclusion in part d.

Explanation of Solution

Given:

D	E	F
t-Test: Two-sample Assuming Equal Variances
	Sample 1	Sample 2
Mean	28.667	28.286
Variance	5.067	2.238
Observations	6	7
Pooled Variance	3.524
Hypothesized Mean Difference	0
df	11
t Stat	0.365
P(T<=t) one-tail	0.361
t Critical one-tail	1.796
P(T<=t) two-tail	0.722
t Critical two-tail	2.201

Calculation:

The 95% confidence interval for the difference in the population mean is

((x1¯−x2¯)±tαs2( 1 n 1 + 1 n 2 ))

From the above output by substituting the value, we get

((x1¯−x2¯)±tαs2( 1 n 1 + 1 n 2 ))

((28.667−28.268)±2.2013.524( 1 6 + 1 7 ))

(−1.890,2.698)

From the interval we can see that the interval contains 0 value so the interval conforms the conclusion in part d.

Conclusion: Therefore, 95% confidence interval for the difference in the population mean is

(−1.890,2.698) . From the interval we can see that the interval contains 0 value, so the interval conforms the conclusion in part d.

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Answer 5

Chapter 10.4, Problem 10.24E

(a)

To determine

To explain:whether the two sample variances are reasonable to use common population variance

(a)

Expert Solution

Answer to Problem 10.24E

The obtained value is less than 3; therefore it is reasonable to use common population variance.

Explanation of Solution

Given:

D	E	F
t-Test: Two-sample Assuming Equal Variances
	Sample 1	Sample 2
Mean	28.667	28.286
Variance	5.067	2.238
Observations	6	7
Pooled Variance	3.524
Hypothesized Mean Difference	0
df	11
t Stat	0.365
P(T<=t) one-tail	0.361
t Critical one-tail	1.796
P(T<=t) two-tail	0.722
t Critical two-tail	2.201

Calculation:

To test the common population variance is reasonable or not, we have the test statistics asLarger s2smaller s2>3

From the output by substituting the values we get

5.0672.238=2.264

This value is less than 3 so it is reasonable to use common population variance.

Conclusion: The obtained value is less than 3, therefore it is reasonable to use common population variance.

(b)

To determine

To find: the observed value of the test statistic andthe P-value associated with the test

(b)

Expert Solution

Answer to Problem 10.24E

The observed value for the test statistics is tSTAT=0.365 . The P-value associated with the test statistics for two tailed test is P−value=0.722 .

Explanation of Solution

Given:

D	E	F
t-Test: Two-sample Assuming Equal Variances
	Sample 1	Sample 2
Mean	28.667	28.286
Variance	5.067	2.238
Observations	6	7
Pooled Variance	3.524
Hypothesized Mean Difference	0
df	11
t Stat	0.365
P(T<=t) one-tail	0.361
t Critical one-tail	1.796
P(T<=t) two-tail	0.722
t Critical two-tail	2.201

Calculation:

From the above output the observed value for the test statistics is tSTAT=0.365 .

The P-value associated with the test statistics for two tailed test is P−value=0.722 .

The observed value for the test statistics is tSTAT=0.365 . The P-value associated with the test statistics for two tailed test is P−value=0.722 .

Conclusion:

Therefore, observed value for the test statistics is tSTAT=0.365 and the P-value associated with the test statistics for two tailed test is P−value=0.722 .

(c)

To determine

To find: the pooled estimate s2 of the population variance

(c)

Expert Solution

Answer to Problem 10.24E

The pooled estimate s2 of the population variance is 3.524 .

Explanation of Solution

Given:

D	E	F
t-Test: Two-sample Assuming Equal Variances
	Sample 1	Sample 2
Mean	28.667	28.286
Variance	5.067	2.238
Observations	6	7
Pooled Variance	3.524
Hypothesized Mean Difference	0
df	11
t Stat	0.365
P(T<=t) one-tail	0.361
t Critical one-tail	1.796
P(T<=t) two-tail	0.722
t Critical two-tail	2.201

Calculation:

From the table, the pooled estimate s2 of the population varianceis 3.524

Conclusion: Thus, the pooled estimate s2 of the population variance is 3.524 .

(d)

To determine

To state: conclusions about the difference in the two populations by using part b answer

(d)

Expert Solution

Answer to Problem 10.24E

From part b we can see that the test statistics value is not lie in the critical region we fail to reject the null hypothesis that the population means are equal.

Explanation of Solution

Calculation:

t-critical value for the two-tail is 2.201

From part b we can see that the test statistics value is not lie in the critical region we fail to reject the null hypothesis that the population means are equal.

Conclusion: therefore, from part b we can see that the test statistics value is not lie in the critical region we fail to reject the null hypothesis that the population means are equal.

(e)

To determine

To find: a 95% confidence interval for the difference in the population and explain whether the interval confirm the conclusion in part d

(e)

Expert Solution

Answer to Problem 10.24E

The 95% confidence interval for the difference in the population mean is

(−1.890,2.698) . From the interval we can see that the interval contains 0 value, so the interval conforms the conclusion in part d.

Explanation of Solution

Given:

D	E	F
t-Test: Two-sample Assuming Equal Variances
	Sample 1	Sample 2
Mean	28.667	28.286
Variance	5.067	2.238
Observations	6	7
Pooled Variance	3.524
Hypothesized Mean Difference	0
df	11
t Stat	0.365
P(T<=t) one-tail	0.361
t Critical one-tail	1.796
P(T<=t) two-tail	0.722
t Critical two-tail	2.201

Calculation:

The 95% confidence interval for the difference in the population mean is

((x1¯−x2¯)±tαs2( 1 n 1 + 1 n 2 ))

From the above output by substituting the value, we get

((x1¯−x2¯)±tαs2( 1 n 1 + 1 n 2 ))

((28.667−28.268)±2.2013.524( 1 6 + 1 7 ))

(−1.890,2.698)

From the interval we can see that the interval contains 0 value so the interval conforms the conclusion in part d.

Conclusion: Therefore, 95% confidence interval for the difference in the population mean is

(−1.890,2.698) . From the interval we can see that the interval contains 0 value, so the interval conforms the conclusion in part d.

Answer 6

Chapter 10.4, Problem 10.24E

(a)

To determine

To explain:whether the two sample variances are reasonable to use common population variance

(a)

Expert Solution

Answer to Problem 10.24E

The obtained value is less than 3; therefore it is reasonable to use common population variance.

Explanation of Solution

Given:

D	E	F
t-Test: Two-sample Assuming Equal Variances
	Sample 1	Sample 2
Mean	28.667	28.286
Variance	5.067	2.238
Observations	6	7
Pooled Variance	3.524
Hypothesized Mean Difference	0
df	11
t Stat	0.365
P(T<=t) one-tail	0.361
t Critical one-tail	1.796
P(T<=t) two-tail	0.722
t Critical two-tail	2.201

Calculation:

To test the common population variance is reasonable or not, we have the test statistics asLarger s2smaller s2>3

From the output by substituting the values we get

5.0672.238=2.264

This value is less than 3 so it is reasonable to use common population variance.

Conclusion: The obtained value is less than 3, therefore it is reasonable to use common population variance.

Answer 7

Chapter 10.4, Problem 10.24E

(a)

To determine

To explain:whether the two sample variances are reasonable to use common population variance

Answer 8

(a)

Expert Solution

Answer to Problem 10.24E

The obtained value is less than 3; therefore it is reasonable to use common population variance.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

D	E	F
t-Test: Two-sample Assuming Equal Variances
	Sample 1	Sample 2
Mean	28.667	28.286
Variance	5.067	2.238
Observations	6	7
Pooled Variance	3.524
Hypothesized Mean Difference	0
df	11
t Stat	0.365
P(T<=t) one-tail	0.361
t Critical one-tail	1.796
P(T<=t) two-tail	0.722
t Critical two-tail	2.201

Calculation:

To test the common population variance is reasonable or not, we have the test statistics asLarger s2smaller s2>3

From the output by substituting the values we get

5.0672.238=2.264

This value is less than 3 so it is reasonable to use common population variance.

Conclusion: The obtained value is less than 3, therefore it is reasonable to use common population variance.

Answer 13

(b)

To determine

To find: the observed value of the test statistic andthe P-value associated with the test

(b)

Expert Solution

Answer to Problem 10.24E

The observed value for the test statistics is tSTAT=0.365 . The P-value associated with the test statistics for two tailed test is P−value=0.722 .

Explanation of Solution

Given:

D	E	F
t-Test: Two-sample Assuming Equal Variances
	Sample 1	Sample 2
Mean	28.667	28.286
Variance	5.067	2.238
Observations	6	7
Pooled Variance	3.524
Hypothesized Mean Difference	0
df	11
t Stat	0.365
P(T<=t) one-tail	0.361
t Critical one-tail	1.796
P(T<=t) two-tail	0.722
t Critical two-tail	2.201

Calculation:

From the above output the observed value for the test statistics is tSTAT=0.365 .

The P-value associated with the test statistics for two tailed test is P−value=0.722 .

The observed value for the test statistics is tSTAT=0.365 . The P-value associated with the test statistics for two tailed test is P−value=0.722 .

Conclusion:

Therefore, observed value for the test statistics is tSTAT=0.365 and the P-value associated with the test statistics for two tailed test is P−value=0.722 .

Answer 14

(b)

To determine

To find: the observed value of the test statistic andthe P-value associated with the test

Answer 15

(b)

Expert Solution

Answer to Problem 10.24E

The observed value for the test statistics is tSTAT=0.365 . The P-value associated with the test statistics for two tailed test is P−value=0.722 .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given:

D	E	F
t-Test: Two-sample Assuming Equal Variances
	Sample 1	Sample 2
Mean	28.667	28.286
Variance	5.067	2.238
Observations	6	7
Pooled Variance	3.524
Hypothesized Mean Difference	0
df	11
t Stat	0.365
P(T<=t) one-tail	0.361
t Critical one-tail	1.796
P(T<=t) two-tail	0.722
t Critical two-tail	2.201

Calculation:

From the above output the observed value for the test statistics is tSTAT=0.365 .

The P-value associated with the test statistics for two tailed test is P−value=0.722 .

The observed value for the test statistics is tSTAT=0.365 . The P-value associated with the test statistics for two tailed test is P−value=0.722 .

Conclusion:

Therefore, observed value for the test statistics is tSTAT=0.365 and the P-value associated with the test statistics for two tailed test is P−value=0.722 .

Answer 20

(c)

To determine

To find: the pooled estimate s2 of the population variance

(c)

Expert Solution

Answer to Problem 10.24E

The pooled estimate s2 of the population variance is 3.524 .

Explanation of Solution

Given:

D	E	F
t-Test: Two-sample Assuming Equal Variances
	Sample 1	Sample 2
Mean	28.667	28.286
Variance	5.067	2.238
Observations	6	7
Pooled Variance	3.524
Hypothesized Mean Difference	0
df	11
t Stat	0.365
P(T<=t) one-tail	0.361
t Critical one-tail	1.796
P(T<=t) two-tail	0.722
t Critical two-tail	2.201

Calculation:

From the table, the pooled estimate s2 of the population varianceis 3.524

Conclusion: Thus, the pooled estimate s2 of the population variance is 3.524 .

Answer 21

(c)

To determine

To find: the pooled estimate s2 of the population variance

Answer 22

(c)

Expert Solution

Answer to Problem 10.24E

The pooled estimate s2 of the population variance is 3.524 .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given:

D	E	F
t-Test: Two-sample Assuming Equal Variances
	Sample 1	Sample 2
Mean	28.667	28.286
Variance	5.067	2.238
Observations	6	7
Pooled Variance	3.524
Hypothesized Mean Difference	0
df	11
t Stat	0.365
P(T<=t) one-tail	0.361
t Critical one-tail	1.796
P(T<=t) two-tail	0.722
t Critical two-tail	2.201

Calculation:

From the table, the pooled estimate s2 of the population varianceis 3.524

Conclusion: Thus, the pooled estimate s2 of the population variance is 3.524 .

Answer 27

(d)

To determine

To state: conclusions about the difference in the two populations by using part b answer

(d)

Expert Solution

Answer to Problem 10.24E

From part b we can see that the test statistics value is not lie in the critical region we fail to reject the null hypothesis that the population means are equal.

Explanation of Solution

Calculation:

t-critical value for the two-tail is 2.201

From part b we can see that the test statistics value is not lie in the critical region we fail to reject the null hypothesis that the population means are equal.

Conclusion: therefore, from part b we can see that the test statistics value is not lie in the critical region we fail to reject the null hypothesis that the population means are equal.

Answer 28

(d)

To determine

To state: conclusions about the difference in the two populations by using part b answer

Answer 29

(d)

Expert Solution

Answer to Problem 10.24E

From part b we can see that the test statistics value is not lie in the critical region we fail to reject the null hypothesis that the population means are equal.

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Calculation:

t-critical value for the two-tail is 2.201

From part b we can see that the test statistics value is not lie in the critical region we fail to reject the null hypothesis that the population means are equal.

Conclusion: therefore, from part b we can see that the test statistics value is not lie in the critical region we fail to reject the null hypothesis that the population means are equal.

Answer 34

(e)

To determine

To find: a 95% confidence interval for the difference in the population and explain whether the interval confirm the conclusion in part d

(e)

Expert Solution

Answer to Problem 10.24E

The 95% confidence interval for the difference in the population mean is

(−1.890,2.698) . From the interval we can see that the interval contains 0 value, so the interval conforms the conclusion in part d.

Explanation of Solution

Given:

D	E	F
t-Test: Two-sample Assuming Equal Variances
	Sample 1	Sample 2
Mean	28.667	28.286
Variance	5.067	2.238
Observations	6	7
Pooled Variance	3.524
Hypothesized Mean Difference	0
df	11
t Stat	0.365
P(T<=t) one-tail	0.361
t Critical one-tail	1.796
P(T<=t) two-tail	0.722
t Critical two-tail	2.201

Calculation:

The 95% confidence interval for the difference in the population mean is

((x1¯−x2¯)±tαs2( 1 n 1 + 1 n 2 ))

From the above output by substituting the value, we get

((x1¯−x2¯)±tαs2( 1 n 1 + 1 n 2 ))

((28.667−28.268)±2.2013.524( 1 6 + 1 7 ))

(−1.890,2.698)

From the interval we can see that the interval contains 0 value so the interval conforms the conclusion in part d.

Conclusion: Therefore, 95% confidence interval for the difference in the population mean is

(−1.890,2.698) . From the interval we can see that the interval contains 0 value, so the interval conforms the conclusion in part d.

Answer 35

(e)

To determine

To find: a 95% confidence interval for the difference in the population and explain whether the interval confirm the conclusion in part d

Answer 36

(e)

Expert Solution

Answer to Problem 10.24E

The 95% confidence interval for the difference in the population mean is

(−1.890,2.698) . From the interval we can see that the interval contains 0 value, so the interval conforms the conclusion in part d.

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

Given:

D	E	F
t-Test: Two-sample Assuming Equal Variances
	Sample 1	Sample 2
Mean	28.667	28.286
Variance	5.067	2.238
Observations	6	7
Pooled Variance	3.524
Hypothesized Mean Difference	0
df	11
t Stat	0.365
P(T<=t) one-tail	0.361
t Critical one-tail	1.796
P(T<=t) two-tail	0.722
t Critical two-tail	2.201

Calculation:

The 95% confidence interval for the difference in the population mean is

((x1¯−x2¯)±tαs2( 1 n 1 + 1 n 2 ))

From the above output by substituting the value, we get

((x1¯−x2¯)±tαs2( 1 n 1 + 1 n 2 ))

((28.667−28.268)±2.2013.524( 1 6 + 1 7 ))

(−1.890,2.698)

From the interval we can see that the interval contains 0 value so the interval conforms the conclusion in part d.

Conclusion: Therefore, 95% confidence interval for the difference in the population mean is

(−1.890,2.698) . From the interval we can see that the interval contains 0 value, so the interval conforms the conclusion in part d.

Answer 41

Ch. 10.3 - Prob. 10.1E Ch. 10.3 - Prob. 10.2E Ch. 10.3 - Prob. 10.3E Ch. 10.3 - Prob. 10.4E Ch. 10.3 - Prob. 10.5E Ch. 10.3 - Prob. 10.6E Ch. 10.3 - Dissolved O2 Content Industrial wastes andsewage...Ch. 10.3 - Prob. 10.8E Ch. 10.3 - Prob. 10.10E Ch. 10.3 - Prob. 10.11E

whether the two sample variances are reasonable to use common population variance

Videos

To explain:whether the two sample variances are reasonable to use common population variance

Answer to Problem 10.24E

Explanation of Solution

To find: the observed value of the test statistic andthe P-value associated with the test

Answer to Problem 10.24E

Explanation of Solution

To find: the pooled estimate s2 of the population variance

Answer to Problem 10.24E

Explanation of Solution

To state: conclusions about the difference in the two populations by using part b answer

Answer to Problem 10.24E

Explanation of Solution

To find: a 95% confidence interval for the difference in the population and explain whether the interval confirm the conclusion in part d

Answer to Problem 10.24E

Explanation of Solution

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