The following four 2 × 2 tables, presented without labels or titles, provide simplified opportunities to practice your computational skills. Calculate chi square for each. HINTS: • Calculate the expected frequencies for each cell with Formula 10.2. Double check to make sure you are using the correct row and column marginals for each cell. • It may he helpful to record the expected frequencies in table format—see Tables 10.4 and 10.8. • Use a computational table to organize the calculation for Formula 10.1—see Tables 10.5 and 10.9. • Follow the step-by-step instructions. • Double-check to make sure you are using the correct values for each cell. a. 20 25 45 c. 25 15 40 25 45 20 45 45 90 30 55 30 45 60 100 b. 10 15 25 d. 20 45 65 20 30 30 45 50 75 15 35 20 65 35 100

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Answer 1

Textbook Question

Chapter 10, Problem 10.1P

The following four $2 × 2$ tables, presented without labels or titles, provide simplified opportunities to practice your computational skills. Calculate chi square for each.

HINTS:

• Calculate the expected frequencies for each cell with Formula 10.2. Double check to make sure you are using the correct row and column marginals for each cell.

• It may he helpful to record the expected frequencies in table format—see Tables 10.4 and 10.8.

• Use a computational table to organize the calculation for Formula 10.1—see Tables 10.5 and 10.9.

• Follow the step-by-step instructions.

• Double-check to make sure you are using the correct values for each cell.

a.	$20$	$25$	$45$	c.	$25$	$15$	$40$
	$25 45$	$20 45$	$45 90$		$30 55$	$30 45$	$60 100$
b.	$10$	$15$	$25$	d.	$20$	$45$	$65$
	$20 30$	$30 45$	$50 75$		$15 35$	$20 65$	$35 100$

Expert Solution

To determine

(a)

To find:

The chi square value for the given information.

Answer to Problem 10.1P

Solution:

The chi square value is 1.12.

Explanation of Solution

Given:

The given table of information is,

a.	$20$	$25$	$45$
	$25$	$20$	$45$
	$45$	$45$	$90$

Description:

The confidence interval is an interval estimate from the statistics of the observed data that might contain the true value of the unknown population parameter.

Formula used:

For a chi square, the expected frequency $fe$ is given as,

$fe=Row marginal×Column marginalN$

Where N is the total of frequencies.

The chi square statistic is given by,

$χ2(obtained)=∑(fo−fe)2fe$

Where $fo$ is the observed frequency,

And $fe$ is the expected frequency.

Calculation:

From the given information,

The row marginal is 45, the column marginal is 45 and N is 90.

The observed frequency is given as,

$fe=Row marginal×Column marginalN$

Substitute 45 for row marginal, 45 for column marginal and 90 for N in the above formula.

$fe=45×4590=22.5$

Consider the following table,

$6.25$	$fo$	$fe$	$fo−fe$	$(fo−fe)2$	$(fo−fe)2/fe$
	20	22.5	$−2.5$	$6.25$	0.28
	25	22.5	$2.5$	$6.25$	0.28
	25	22.5	$2.5$	$6.25$	0.28
	20	22.5	$−2.5$	$6.25$	0.28
Total	90	90	0		$χ2=1.12$

The value $fo−fe$ is obtained as,

Substitute 20 for $fo$ and 22.5 for $fe$ in the above formula.

$fo−fe=20−22.5=−2.5$

Squaring the above obtained result,

$(fo−fe)2=(−2.5)2=6.25$

Divide the above obtained result by $fe$ .

$(fo−fe)2fe=6.2522.5=0.28$

Proceed in a similar manner to obtain rest of the values of $(fo−fe)2/fe$ and refer above table for the rest of the values of $(fo−fe)2/fe$ .

The chi square value is given as,

$χ2(obtained)=∑(fo−fe)2fe=0.28+0.28+0.28+0.28=1.12$

Thus, the chi square value is 1.12.

Conclusion:

The chi square value is 1.12.

Expert Solution

To determine

(b)

To find:

The chi square value for the given information.

Answer to Problem 10.1P

Solution:

The chi square value is 0.0.

Explanation of Solution

Given:

The given table of information is,

b.	10	15	25
	20	30	50
	30	45	75

Description:

The confidence interval is an interval estimate from the statistics of the observed data that might contain the true value of the unknown population parameter.

Formula used:

For a chi square, the expected frequency $fe$ is given as,

$fe=Row marginal×Column marginalN$

Where N is the total of frequencies.

The chi square statistic is given by,

$χ2(obtained)=∑(fo−fe)2fe$

Where $fo$ is the observed frequency,

And $fe$ is the expected frequency.

Calculation:

From the given information,

The observed frequency is given as,

$fe=Row marginal×Column marginalN$ …… $(1)$

Substitute 25 for row marginal, 30 for column marginal and 75 for N in equation $(1)$ .

$f1=25×3075=10$

Substitute 25 for row marginal, 45 for column marginal and 75 for N in equation $(1)$ .

$f1=25×4575=15$

Substitute 50 for row marginal, 30 for column marginal and 75 for N in equation $(1)$ .

$f1=50×3075=20$

Substitute 50 for row marginal, 45 for column marginal and 75 for N in equation $(1)$ .

$f1=50×4575=30$

Consider the following table,

$6.25$	$fo$	$fe$	$fo−fe$	$(fo−fe)2$	$(fo−fe)2/fe$
	10	10	0	0	0
	15	15	0	0	0
	20	20	0	0	0
	30	30	0	0	0
Total	75	75	0	0	$χ2=0.0$

The value $fo−fe$ is obtained as,

Substitute 10 for $fo$ and 10 for $fe$ in the above formula.

$fo−fe=10−10=0$

Squaring the above obtained result,

$(fo−fe)2=(0)2=0$

Divide the above obtained result by $fe$ .

$(fo−fe)2fe=0$

Proceed in a similar manner to obtain rest of the values of $(fo−fe)2/fe$ and refer above table for the rest of the values of $(fo−fe)2/fe$ .

The chi square value is given as,

$χ2(obtained)=∑(fo−fe)2fe=0+0+0+0=0.0$

Thus, the chi square value is 0.0.

Conclusion:

The chi square value is 0.0.

Expert Solution

To determine

(c)

To find:

The chi square value for the given information.

Answer to Problem 10.1P

Solution:

The chi square value is 1.51.

Explanation of Solution

Given:

The given table of information is,

c.	25	15	40
	30	30	60
	55	45	100

Description:

The confidence interval is an interval estimate from the statistics of the observed data that might contain the true value of the unknown population parameter.

Formula used:

For a chi square, the expected frequency $fe$ is given as,

$fe=Row marginal×Column marginalN$

Where N is the total of frequencies.

The chi square statistic is given by,

$χ2(obtained)=∑(fo−fe)2fe$

Where $fo$ is the observed frequency,

And $fe$ is the expected frequency.

Calculation:

From the given information,

The observed frequency is given as,

$fe=Row marginal×Column marginalN$ …… $(1)$

Substitute 40 for row marginal, 55 for column marginal and 100 for N in equation $(1)$ .

$f1=40×55100=22$

Substitute 40 for row marginal, 45 for column marginal and 100 for N in equation $(1)$ .

$f1=40×45100=18$

Substitute 60 for row marginal, 55 for column marginal and 100 for N in equation $(1)$ .

$f1=60×55100=33$

Substitute 60 for row marginal, 45 for column marginal and 100 for N in equation $(1)$ .

$f1=60×45100=27$

Consider the following table,

$6.25$	$fo$	$fe$	$fo−fe$	$(fo−fe)2$	$(fo−fe)2/fe$
	25	22	3	9	0.41
	15	18	$−3$	9	0.50
	30	33	$−3$	9	0.27
	30	27	3	9	0.33
Total	100	100	0		$χ2=1.51$

The value $fo−fe$ is obtained as,

Substitute 25 for $fo$ and 22 for $fe$ in the above formula.

$fo−fe=25−22=3$

Squaring the above obtained result,

$(fo−fe)2=(3)2=9$

Divide the above obtained result by $fe$ .

$(fo−fe)2fe=922=0.41$

Proceed in a similar manner to obtain rest of the values of $(fo−fe)2/fe$ and refer above table for the rest of the values of $(fo−fe)2/fe$ .

The chi square value is given as,

$χ2(obtained)=∑(fo−fe)2fe=0.41+0.50+0.27+0.33=1.51$

Thus, the chi square value is 1.51.

Conclusion:

The chi square value is 1.51.

Expert Solution

To determine

(d)

To find:

The chi square value for the given information.

Answer to Problem 10.1P

Solution:

The chi square value is 1.43.

Explanation of Solution

Given:

The given table of information is,

d.	20	45	65
	15	20	35
	35	65	100

Description:

The confidence interval is an interval estimate from the statistics of the observed data that might contain the true value of the unknown population parameter.

Formula used:

For a chi square, the expected frequency $fe$ is given as,

$fe=Row marginal×Column marginalN$

Where N is the total of frequencies.

The chi square statistic is given by,

$χ2(obtained)=∑(fo−fe)2fe$

Where $fo$ is the observed frequency,

And $fe$ is the expected frequency.

Calculation:

From the given information,

The observed frequency is given as,

$fe=Row marginal×Column marginalN$ …… $(1)$

Substitute 65 for row marginal, 35 for column marginal and 100 for N in equation $(1)$ .

$f1=65×35100=22.75$

Substitute 65 for row marginal, 65 for column marginal and 100 for N in equation $(1)$ .

$f1=65×65100=42.25$

Substitute 35 for row marginal, 35 for column marginal and 100 for N in equation $(1)$ .

$f1=35×35100=12.25$

Substitute 35 for row marginal, 65 for column marginal and 100 for N in equation $(1)$ .

$f1=35×65100=22.75$

Consider the following table,

$6.25$	$fo$	$fe$	$fo−fe$	$(fo−fe)2$	$(fo−fe)2/fe$
	20	22.75	-2.75	7.5625	0.332417582
	45	42.25	2.75	7.5625	0.178994083
	15	12.25	2.75	7.5625	0.617346939
	20	22.75	$−2.75$	7.5625	0.332417582
Total	100	100	0		$χ2=1.46$

The value $fo−fe$ is obtained as,

Substitute 20 for $fo$ and 22.75 for $fe$ in the above formula.

$fo−fe=20−22.75=−2.75$

Squaring the above obtained result,

$(fo−fe)2=(−2.75)2=7.56$

Divide the above obtained result by $fe$ .

$(fo−fe)2fe=7.5622.75=0.33$

Proceed in a similar manner to obtain rest of the values of $(fo−fe)2/fe$ and refer above table for the rest of the values of $(fo−fe)2/fe$ .

The chi square value is given as,

$χ2(obtained)=∑(fo−fe)2fe=0.33+0.18+0.62+0.33=1.46$

Thus, the chi square value is 1.46.

Conclusion:

The chi square value is 1.46.

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Answer 2

Textbook Question

Chapter 10, Problem 10.1P

The following four $2 × 2$ tables, presented without labels or titles, provide simplified opportunities to practice your computational skills. Calculate chi square for each.

HINTS:

• Calculate the expected frequencies for each cell with Formula 10.2. Double check to make sure you are using the correct row and column marginals for each cell.

• It may he helpful to record the expected frequencies in table format—see Tables 10.4 and 10.8.

• Use a computational table to organize the calculation for Formula 10.1—see Tables 10.5 and 10.9.

• Follow the step-by-step instructions.

• Double-check to make sure you are using the correct values for each cell.

a.	$20$	$25$	$45$	c.	$25$	$15$	$40$
	$25 45$	$20 45$	$45 90$		$30 55$	$30 45$	$60 100$
b.	$10$	$15$	$25$	d.	$20$	$45$	$65$
	$20 30$	$30 45$	$50 75$		$15 35$	$20 65$	$35 100$

Expert Solution

To determine

(a)

To find:

The chi square value for the given information.

Answer to Problem 10.1P

Solution:

The chi square value is 1.12.

Explanation of Solution

Given:

The given table of information is,

a.	$20$	$25$	$45$
	$25$	$20$	$45$
	$45$	$45$	$90$

Description:

The confidence interval is an interval estimate from the statistics of the observed data that might contain the true value of the unknown population parameter.

Formula used:

For a chi square, the expected frequency $fe$ is given as,

$fe=Row marginal×Column marginalN$

Where N is the total of frequencies.

The chi square statistic is given by,

$χ2(obtained)=∑(fo−fe)2fe$

Where $fo$ is the observed frequency,

And $fe$ is the expected frequency.

Calculation:

From the given information,

The row marginal is 45, the column marginal is 45 and N is 90.

The observed frequency is given as,

$fe=Row marginal×Column marginalN$

Substitute 45 for row marginal, 45 for column marginal and 90 for N in the above formula.

$fe=45×4590=22.5$

Consider the following table,

$6.25$	$fo$	$fe$	$fo−fe$	$(fo−fe)2$	$(fo−fe)2/fe$
	20	22.5	$−2.5$	$6.25$	0.28
	25	22.5	$2.5$	$6.25$	0.28
	25	22.5	$2.5$	$6.25$	0.28
	20	22.5	$−2.5$	$6.25$	0.28
Total	90	90	0		$χ2=1.12$

The value $fo−fe$ is obtained as,

Substitute 20 for $fo$ and 22.5 for $fe$ in the above formula.

$fo−fe=20−22.5=−2.5$

Squaring the above obtained result,

$(fo−fe)2=(−2.5)2=6.25$

Divide the above obtained result by $fe$ .

$(fo−fe)2fe=6.2522.5=0.28$

Proceed in a similar manner to obtain rest of the values of $(fo−fe)2/fe$ and refer above table for the rest of the values of $(fo−fe)2/fe$ .

The chi square value is given as,

$χ2(obtained)=∑(fo−fe)2fe=0.28+0.28+0.28+0.28=1.12$

Thus, the chi square value is 1.12.

Conclusion:

The chi square value is 1.12.

Expert Solution

To determine

(b)

To find:

The chi square value for the given information.

Answer to Problem 10.1P

Solution:

The chi square value is 0.0.

Explanation of Solution

Given:

The given table of information is,

b.	10	15	25
	20	30	50
	30	45	75

Description:

The confidence interval is an interval estimate from the statistics of the observed data that might contain the true value of the unknown population parameter.

Formula used:

For a chi square, the expected frequency $fe$ is given as,

$fe=Row marginal×Column marginalN$

Where N is the total of frequencies.

The chi square statistic is given by,

$χ2(obtained)=∑(fo−fe)2fe$

Where $fo$ is the observed frequency,

And $fe$ is the expected frequency.

Calculation:

From the given information,

The observed frequency is given as,

$fe=Row marginal×Column marginalN$ …… $(1)$

Substitute 25 for row marginal, 30 for column marginal and 75 for N in equation $(1)$ .

$f1=25×3075=10$

Substitute 25 for row marginal, 45 for column marginal and 75 for N in equation $(1)$ .

$f1=25×4575=15$

Substitute 50 for row marginal, 30 for column marginal and 75 for N in equation $(1)$ .

$f1=50×3075=20$

Substitute 50 for row marginal, 45 for column marginal and 75 for N in equation $(1)$ .

$f1=50×4575=30$

Consider the following table,

$6.25$	$fo$	$fe$	$fo−fe$	$(fo−fe)2$	$(fo−fe)2/fe$
	10	10	0	0	0
	15	15	0	0	0
	20	20	0	0	0
	30	30	0	0	0
Total	75	75	0	0	$χ2=0.0$

The value $fo−fe$ is obtained as,

Substitute 10 for $fo$ and 10 for $fe$ in the above formula.

$fo−fe=10−10=0$

Squaring the above obtained result,

$(fo−fe)2=(0)2=0$

Divide the above obtained result by $fe$ .

$(fo−fe)2fe=0$

Proceed in a similar manner to obtain rest of the values of $(fo−fe)2/fe$ and refer above table for the rest of the values of $(fo−fe)2/fe$ .

The chi square value is given as,

$χ2(obtained)=∑(fo−fe)2fe=0+0+0+0=0.0$

Thus, the chi square value is 0.0.

Conclusion:

The chi square value is 0.0.

Expert Solution

To determine

(c)

To find:

The chi square value for the given information.

Answer to Problem 10.1P

Solution:

The chi square value is 1.51.

Explanation of Solution

Given:

The given table of information is,

c.	25	15	40
	30	30	60
	55	45	100

Description:

The confidence interval is an interval estimate from the statistics of the observed data that might contain the true value of the unknown population parameter.

Formula used:

For a chi square, the expected frequency $fe$ is given as,

$fe=Row marginal×Column marginalN$

Where N is the total of frequencies.

The chi square statistic is given by,

$χ2(obtained)=∑(fo−fe)2fe$

Where $fo$ is the observed frequency,

And $fe$ is the expected frequency.

Calculation:

From the given information,

The observed frequency is given as,

$fe=Row marginal×Column marginalN$ …… $(1)$

Substitute 40 for row marginal, 55 for column marginal and 100 for N in equation $(1)$ .

$f1=40×55100=22$

Substitute 40 for row marginal, 45 for column marginal and 100 for N in equation $(1)$ .

$f1=40×45100=18$

Substitute 60 for row marginal, 55 for column marginal and 100 for N in equation $(1)$ .

$f1=60×55100=33$

Substitute 60 for row marginal, 45 for column marginal and 100 for N in equation $(1)$ .

$f1=60×45100=27$

Consider the following table,

$6.25$	$fo$	$fe$	$fo−fe$	$(fo−fe)2$	$(fo−fe)2/fe$
	25	22	3	9	0.41
	15	18	$−3$	9	0.50
	30	33	$−3$	9	0.27
	30	27	3	9	0.33
Total	100	100	0		$χ2=1.51$

The value $fo−fe$ is obtained as,

Substitute 25 for $fo$ and 22 for $fe$ in the above formula.

$fo−fe=25−22=3$

Squaring the above obtained result,

$(fo−fe)2=(3)2=9$

Divide the above obtained result by $fe$ .

$(fo−fe)2fe=922=0.41$

Proceed in a similar manner to obtain rest of the values of $(fo−fe)2/fe$ and refer above table for the rest of the values of $(fo−fe)2/fe$ .

The chi square value is given as,

$χ2(obtained)=∑(fo−fe)2fe=0.41+0.50+0.27+0.33=1.51$

Thus, the chi square value is 1.51.

Conclusion:

The chi square value is 1.51.

Expert Solution

To determine

(d)

To find:

The chi square value for the given information.

Answer to Problem 10.1P

Solution:

The chi square value is 1.43.

Explanation of Solution

Given:

The given table of information is,

d.	20	45	65
	15	20	35
	35	65	100

Description:

The confidence interval is an interval estimate from the statistics of the observed data that might contain the true value of the unknown population parameter.

Formula used:

For a chi square, the expected frequency $fe$ is given as,

$fe=Row marginal×Column marginalN$

Where N is the total of frequencies.

The chi square statistic is given by,

$χ2(obtained)=∑(fo−fe)2fe$

Where $fo$ is the observed frequency,

And $fe$ is the expected frequency.

Calculation:

From the given information,

The observed frequency is given as,

$fe=Row marginal×Column marginalN$ …… $(1)$

Substitute 65 for row marginal, 35 for column marginal and 100 for N in equation $(1)$ .

$f1=65×35100=22.75$

Substitute 65 for row marginal, 65 for column marginal and 100 for N in equation $(1)$ .

$f1=65×65100=42.25$

Substitute 35 for row marginal, 35 for column marginal and 100 for N in equation $(1)$ .

$f1=35×35100=12.25$

Substitute 35 for row marginal, 65 for column marginal and 100 for N in equation $(1)$ .

$f1=35×65100=22.75$

Consider the following table,

$6.25$	$fo$	$fe$	$fo−fe$	$(fo−fe)2$	$(fo−fe)2/fe$
	20	22.75	-2.75	7.5625	0.332417582
	45	42.25	2.75	7.5625	0.178994083
	15	12.25	2.75	7.5625	0.617346939
	20	22.75	$−2.75$	7.5625	0.332417582
Total	100	100	0		$χ2=1.46$

The value $fo−fe$ is obtained as,

Substitute 20 for $fo$ and 22.75 for $fe$ in the above formula.

$fo−fe=20−22.75=−2.75$

Squaring the above obtained result,

$(fo−fe)2=(−2.75)2=7.56$

Divide the above obtained result by $fe$ .

$(fo−fe)2fe=7.5622.75=0.33$

Proceed in a similar manner to obtain rest of the values of $(fo−fe)2/fe$ and refer above table for the rest of the values of $(fo−fe)2/fe$ .

The chi square value is given as,

$χ2(obtained)=∑(fo−fe)2fe=0.33+0.18+0.62+0.33=1.46$

Thus, the chi square value is 1.46.

Conclusion:

The chi square value is 1.46.

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Answer 3

Textbook Question

Chapter 10, Problem 10.1P

The following four $2 × 2$ tables, presented without labels or titles, provide simplified opportunities to practice your computational skills. Calculate chi square for each.

HINTS:

• Calculate the expected frequencies for each cell with Formula 10.2. Double check to make sure you are using the correct row and column marginals for each cell.

• It may he helpful to record the expected frequencies in table format—see Tables 10.4 and 10.8.

• Use a computational table to organize the calculation for Formula 10.1—see Tables 10.5 and 10.9.

• Follow the step-by-step instructions.

• Double-check to make sure you are using the correct values for each cell.

a.	$20$	$25$	$45$	c.	$25$	$15$	$40$
	$25 45$	$20 45$	$45 90$		$30 55$	$30 45$	$60 100$
b.	$10$	$15$	$25$	d.	$20$	$45$	$65$
	$20 30$	$30 45$	$50 75$		$15 35$	$20 65$	$35 100$

Expert Solution

To determine

(a)

To find:

The chi square value for the given information.

Answer to Problem 10.1P

Solution:

The chi square value is 1.12.

Explanation of Solution

Given:

The given table of information is,

a.	$20$	$25$	$45$
	$25$	$20$	$45$
	$45$	$45$	$90$

Description:

The confidence interval is an interval estimate from the statistics of the observed data that might contain the true value of the unknown population parameter.

Formula used:

For a chi square, the expected frequency $fe$ is given as,

$fe=Row marginal×Column marginalN$

Where N is the total of frequencies.

The chi square statistic is given by,

$χ2(obtained)=∑(fo−fe)2fe$

Where $fo$ is the observed frequency,

And $fe$ is the expected frequency.

Calculation:

From the given information,

The row marginal is 45, the column marginal is 45 and N is 90.

The observed frequency is given as,

$fe=Row marginal×Column marginalN$

Substitute 45 for row marginal, 45 for column marginal and 90 for N in the above formula.

$fe=45×4590=22.5$

Consider the following table,

$6.25$	$fo$	$fe$	$fo−fe$	$(fo−fe)2$	$(fo−fe)2/fe$
	20	22.5	$−2.5$	$6.25$	0.28
	25	22.5	$2.5$	$6.25$	0.28
	25	22.5	$2.5$	$6.25$	0.28
	20	22.5	$−2.5$	$6.25$	0.28
Total	90	90	0		$χ2=1.12$

The value $fo−fe$ is obtained as,

Substitute 20 for $fo$ and 22.5 for $fe$ in the above formula.

$fo−fe=20−22.5=−2.5$

Squaring the above obtained result,

$(fo−fe)2=(−2.5)2=6.25$

Divide the above obtained result by $fe$ .

$(fo−fe)2fe=6.2522.5=0.28$

Proceed in a similar manner to obtain rest of the values of $(fo−fe)2/fe$ and refer above table for the rest of the values of $(fo−fe)2/fe$ .

The chi square value is given as,

$χ2(obtained)=∑(fo−fe)2fe=0.28+0.28+0.28+0.28=1.12$

Thus, the chi square value is 1.12.

Conclusion:

The chi square value is 1.12.

Expert Solution

To determine

(b)

To find:

The chi square value for the given information.

Answer to Problem 10.1P

Solution:

The chi square value is 0.0.

Explanation of Solution

Given:

The given table of information is,

b.	10	15	25
	20	30	50
	30	45	75

Description:

The confidence interval is an interval estimate from the statistics of the observed data that might contain the true value of the unknown population parameter.

Formula used:

For a chi square, the expected frequency $fe$ is given as,

$fe=Row marginal×Column marginalN$

Where N is the total of frequencies.

The chi square statistic is given by,

$χ2(obtained)=∑(fo−fe)2fe$

Where $fo$ is the observed frequency,

And $fe$ is the expected frequency.

Calculation:

From the given information,

The observed frequency is given as,

$fe=Row marginal×Column marginalN$ …… $(1)$

Substitute 25 for row marginal, 30 for column marginal and 75 for N in equation $(1)$ .

$f1=25×3075=10$

Substitute 25 for row marginal, 45 for column marginal and 75 for N in equation $(1)$ .

$f1=25×4575=15$

Substitute 50 for row marginal, 30 for column marginal and 75 for N in equation $(1)$ .

$f1=50×3075=20$

Substitute 50 for row marginal, 45 for column marginal and 75 for N in equation $(1)$ .

$f1=50×4575=30$

Consider the following table,

$6.25$	$fo$	$fe$	$fo−fe$	$(fo−fe)2$	$(fo−fe)2/fe$
	10	10	0	0	0
	15	15	0	0	0
	20	20	0	0	0
	30	30	0	0	0
Total	75	75	0	0	$χ2=0.0$

The value $fo−fe$ is obtained as,

Substitute 10 for $fo$ and 10 for $fe$ in the above formula.

$fo−fe=10−10=0$

Squaring the above obtained result,

$(fo−fe)2=(0)2=0$

Divide the above obtained result by $fe$ .

$(fo−fe)2fe=0$

Proceed in a similar manner to obtain rest of the values of $(fo−fe)2/fe$ and refer above table for the rest of the values of $(fo−fe)2/fe$ .

The chi square value is given as,

$χ2(obtained)=∑(fo−fe)2fe=0+0+0+0=0.0$

Thus, the chi square value is 0.0.

Conclusion:

The chi square value is 0.0.

Expert Solution

To determine

(c)

To find:

The chi square value for the given information.

Answer to Problem 10.1P

Solution:

The chi square value is 1.51.

Explanation of Solution

Given:

The given table of information is,

c.	25	15	40
	30	30	60
	55	45	100

Description:

The confidence interval is an interval estimate from the statistics of the observed data that might contain the true value of the unknown population parameter.

Formula used:

For a chi square, the expected frequency $fe$ is given as,

$fe=Row marginal×Column marginalN$

Where N is the total of frequencies.

The chi square statistic is given by,

$χ2(obtained)=∑(fo−fe)2fe$

Where $fo$ is the observed frequency,

And $fe$ is the expected frequency.

Calculation:

From the given information,

The observed frequency is given as,

$fe=Row marginal×Column marginalN$ …… $(1)$

Substitute 40 for row marginal, 55 for column marginal and 100 for N in equation $(1)$ .

$f1=40×55100=22$

Substitute 40 for row marginal, 45 for column marginal and 100 for N in equation $(1)$ .

$f1=40×45100=18$

Substitute 60 for row marginal, 55 for column marginal and 100 for N in equation $(1)$ .

$f1=60×55100=33$

Substitute 60 for row marginal, 45 for column marginal and 100 for N in equation $(1)$ .

$f1=60×45100=27$

Consider the following table,

$6.25$	$fo$	$fe$	$fo−fe$	$(fo−fe)2$	$(fo−fe)2/fe$
	25	22	3	9	0.41
	15	18	$−3$	9	0.50
	30	33	$−3$	9	0.27
	30	27	3	9	0.33
Total	100	100	0		$χ2=1.51$

The value $fo−fe$ is obtained as,

Substitute 25 for $fo$ and 22 for $fe$ in the above formula.

$fo−fe=25−22=3$

Squaring the above obtained result,

$(fo−fe)2=(3)2=9$

Divide the above obtained result by $fe$ .

$(fo−fe)2fe=922=0.41$

Proceed in a similar manner to obtain rest of the values of $(fo−fe)2/fe$ and refer above table for the rest of the values of $(fo−fe)2/fe$ .

The chi square value is given as,

$χ2(obtained)=∑(fo−fe)2fe=0.41+0.50+0.27+0.33=1.51$

Thus, the chi square value is 1.51.

Conclusion:

The chi square value is 1.51.

Expert Solution

To determine

(d)

To find:

The chi square value for the given information.

Answer to Problem 10.1P

Solution:

The chi square value is 1.43.

Explanation of Solution

Given:

The given table of information is,

d.	20	45	65
	15	20	35
	35	65	100

Description:

The confidence interval is an interval estimate from the statistics of the observed data that might contain the true value of the unknown population parameter.

Formula used:

For a chi square, the expected frequency $fe$ is given as,

$fe=Row marginal×Column marginalN$

Where N is the total of frequencies.

The chi square statistic is given by,

$χ2(obtained)=∑(fo−fe)2fe$

Where $fo$ is the observed frequency,

And $fe$ is the expected frequency.

Calculation:

From the given information,

The observed frequency is given as,

$fe=Row marginal×Column marginalN$ …… $(1)$

Substitute 65 for row marginal, 35 for column marginal and 100 for N in equation $(1)$ .

$f1=65×35100=22.75$

Substitute 65 for row marginal, 65 for column marginal and 100 for N in equation $(1)$ .

$f1=65×65100=42.25$

Substitute 35 for row marginal, 35 for column marginal and 100 for N in equation $(1)$ .

$f1=35×35100=12.25$

Substitute 35 for row marginal, 65 for column marginal and 100 for N in equation $(1)$ .

$f1=35×65100=22.75$

Consider the following table,

$6.25$	$fo$	$fe$	$fo−fe$	$(fo−fe)2$	$(fo−fe)2/fe$
	20	22.75	-2.75	7.5625	0.332417582
	45	42.25	2.75	7.5625	0.178994083
	15	12.25	2.75	7.5625	0.617346939
	20	22.75	$−2.75$	7.5625	0.332417582
Total	100	100	0		$χ2=1.46$

The value $fo−fe$ is obtained as,

Substitute 20 for $fo$ and 22.75 for $fe$ in the above formula.

$fo−fe=20−22.75=−2.75$

Squaring the above obtained result,

$(fo−fe)2=(−2.75)2=7.56$

Divide the above obtained result by $fe$ .

$(fo−fe)2fe=7.5622.75=0.33$

Proceed in a similar manner to obtain rest of the values of $(fo−fe)2/fe$ and refer above table for the rest of the values of $(fo−fe)2/fe$ .

The chi square value is given as,

$χ2(obtained)=∑(fo−fe)2fe=0.33+0.18+0.62+0.33=1.46$

Thus, the chi square value is 1.46.

Conclusion:

The chi square value is 1.46.

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Answer 4

Textbook Question

Chapter 10, Problem 10.1P

The following four $2 × 2$ tables, presented without labels or titles, provide simplified opportunities to practice your computational skills. Calculate chi square for each.

HINTS:

• Calculate the expected frequencies for each cell with Formula 10.2. Double check to make sure you are using the correct row and column marginals for each cell.

• It may he helpful to record the expected frequencies in table format—see Tables 10.4 and 10.8.

• Use a computational table to organize the calculation for Formula 10.1—see Tables 10.5 and 10.9.

• Follow the step-by-step instructions.

• Double-check to make sure you are using the correct values for each cell.

a.	$20$	$25$	$45$	c.	$25$	$15$	$40$
	$25 45$	$20 45$	$45 90$		$30 55$	$30 45$	$60 100$
b.	$10$	$15$	$25$	d.	$20$	$45$	$65$
	$20 30$	$30 45$	$50 75$		$15 35$	$20 65$	$35 100$

Expert Solution

To determine

(a)

To find:

The chi square value for the given information.

Answer to Problem 10.1P

Solution:

The chi square value is 1.12.

Explanation of Solution

Given:

The given table of information is,

a.	$20$	$25$	$45$
	$25$	$20$	$45$
	$45$	$45$	$90$

Description:

The confidence interval is an interval estimate from the statistics of the observed data that might contain the true value of the unknown population parameter.

Formula used:

For a chi square, the expected frequency $fe$ is given as,

$fe=Row marginal×Column marginalN$

Where N is the total of frequencies.

The chi square statistic is given by,

$χ2(obtained)=∑(fo−fe)2fe$

Where $fo$ is the observed frequency,

And $fe$ is the expected frequency.

Calculation:

From the given information,

The row marginal is 45, the column marginal is 45 and N is 90.

The observed frequency is given as,

$fe=Row marginal×Column marginalN$

Substitute 45 for row marginal, 45 for column marginal and 90 for N in the above formula.

$fe=45×4590=22.5$

Consider the following table,

$6.25$	$fo$	$fe$	$fo−fe$	$(fo−fe)2$	$(fo−fe)2/fe$
	20	22.5	$−2.5$	$6.25$	0.28
	25	22.5	$2.5$	$6.25$	0.28
	25	22.5	$2.5$	$6.25$	0.28
	20	22.5	$−2.5$	$6.25$	0.28
Total	90	90	0		$χ2=1.12$

The value $fo−fe$ is obtained as,

Substitute 20 for $fo$ and 22.5 for $fe$ in the above formula.

$fo−fe=20−22.5=−2.5$

Squaring the above obtained result,

$(fo−fe)2=(−2.5)2=6.25$

Divide the above obtained result by $fe$ .

$(fo−fe)2fe=6.2522.5=0.28$

Proceed in a similar manner to obtain rest of the values of $(fo−fe)2/fe$ and refer above table for the rest of the values of $(fo−fe)2/fe$ .

The chi square value is given as,

$χ2(obtained)=∑(fo−fe)2fe=0.28+0.28+0.28+0.28=1.12$

Thus, the chi square value is 1.12.

Conclusion:

The chi square value is 1.12.

Expert Solution

To determine

(b)

To find:

The chi square value for the given information.

Answer to Problem 10.1P

Solution:

The chi square value is 0.0.

Explanation of Solution

Given:

The given table of information is,

b.	10	15	25
	20	30	50
	30	45	75

Description:

The confidence interval is an interval estimate from the statistics of the observed data that might contain the true value of the unknown population parameter.

Formula used:

For a chi square, the expected frequency $fe$ is given as,

$fe=Row marginal×Column marginalN$

Where N is the total of frequencies.

The chi square statistic is given by,

$χ2(obtained)=∑(fo−fe)2fe$

Where $fo$ is the observed frequency,

And $fe$ is the expected frequency.

Calculation:

From the given information,

The observed frequency is given as,

$fe=Row marginal×Column marginalN$ …… $(1)$

Substitute 25 for row marginal, 30 for column marginal and 75 for N in equation $(1)$ .

$f1=25×3075=10$

Substitute 25 for row marginal, 45 for column marginal and 75 for N in equation $(1)$ .

$f1=25×4575=15$

Substitute 50 for row marginal, 30 for column marginal and 75 for N in equation $(1)$ .

$f1=50×3075=20$

Substitute 50 for row marginal, 45 for column marginal and 75 for N in equation $(1)$ .

$f1=50×4575=30$

Consider the following table,

$6.25$	$fo$	$fe$	$fo−fe$	$(fo−fe)2$	$(fo−fe)2/fe$
	10	10	0	0	0
	15	15	0	0	0
	20	20	0	0	0
	30	30	0	0	0
Total	75	75	0	0	$χ2=0.0$

The value $fo−fe$ is obtained as,

Substitute 10 for $fo$ and 10 for $fe$ in the above formula.

$fo−fe=10−10=0$

Squaring the above obtained result,

$(fo−fe)2=(0)2=0$

Divide the above obtained result by $fe$ .

$(fo−fe)2fe=0$

Proceed in a similar manner to obtain rest of the values of $(fo−fe)2/fe$ and refer above table for the rest of the values of $(fo−fe)2/fe$ .

The chi square value is given as,

$χ2(obtained)=∑(fo−fe)2fe=0+0+0+0=0.0$

Thus, the chi square value is 0.0.

Conclusion:

The chi square value is 0.0.

Expert Solution

To determine

(c)

To find:

The chi square value for the given information.

Answer to Problem 10.1P

Solution:

The chi square value is 1.51.

Explanation of Solution

Given:

The given table of information is,

c.	25	15	40
	30	30	60
	55	45	100

Description:

The confidence interval is an interval estimate from the statistics of the observed data that might contain the true value of the unknown population parameter.

Formula used:

For a chi square, the expected frequency $fe$ is given as,

$fe=Row marginal×Column marginalN$

Where N is the total of frequencies.

The chi square statistic is given by,

$χ2(obtained)=∑(fo−fe)2fe$

Where $fo$ is the observed frequency,

And $fe$ is the expected frequency.

Calculation:

From the given information,

The observed frequency is given as,

$fe=Row marginal×Column marginalN$ …… $(1)$

Substitute 40 for row marginal, 55 for column marginal and 100 for N in equation $(1)$ .

$f1=40×55100=22$

Substitute 40 for row marginal, 45 for column marginal and 100 for N in equation $(1)$ .

$f1=40×45100=18$

Substitute 60 for row marginal, 55 for column marginal and 100 for N in equation $(1)$ .

$f1=60×55100=33$

Substitute 60 for row marginal, 45 for column marginal and 100 for N in equation $(1)$ .

$f1=60×45100=27$

Consider the following table,

$6.25$	$fo$	$fe$	$fo−fe$	$(fo−fe)2$	$(fo−fe)2/fe$
	25	22	3	9	0.41
	15	18	$−3$	9	0.50
	30	33	$−3$	9	0.27
	30	27	3	9	0.33
Total	100	100	0		$χ2=1.51$

The value $fo−fe$ is obtained as,

Substitute 25 for $fo$ and 22 for $fe$ in the above formula.

$fo−fe=25−22=3$

Squaring the above obtained result,

$(fo−fe)2=(3)2=9$

Divide the above obtained result by $fe$ .

$(fo−fe)2fe=922=0.41$

Proceed in a similar manner to obtain rest of the values of $(fo−fe)2/fe$ and refer above table for the rest of the values of $(fo−fe)2/fe$ .

The chi square value is given as,

$χ2(obtained)=∑(fo−fe)2fe=0.41+0.50+0.27+0.33=1.51$

Thus, the chi square value is 1.51.

Conclusion:

The chi square value is 1.51.

Expert Solution

To determine

(d)

To find:

The chi square value for the given information.

Answer to Problem 10.1P

Solution:

The chi square value is 1.43.

Explanation of Solution

Given:

The given table of information is,

d.	20	45	65
	15	20	35
	35	65	100

Description:

The confidence interval is an interval estimate from the statistics of the observed data that might contain the true value of the unknown population parameter.

Formula used:

For a chi square, the expected frequency $fe$ is given as,

$fe=Row marginal×Column marginalN$

Where N is the total of frequencies.

The chi square statistic is given by,

$χ2(obtained)=∑(fo−fe)2fe$

Where $fo$ is the observed frequency,

And $fe$ is the expected frequency.

Calculation:

From the given information,

The observed frequency is given as,

$fe=Row marginal×Column marginalN$ …… $(1)$

Substitute 65 for row marginal, 35 for column marginal and 100 for N in equation $(1)$ .

$f1=65×35100=22.75$

Substitute 65 for row marginal, 65 for column marginal and 100 for N in equation $(1)$ .

$f1=65×65100=42.25$

Substitute 35 for row marginal, 35 for column marginal and 100 for N in equation $(1)$ .

$f1=35×35100=12.25$

Substitute 35 for row marginal, 65 for column marginal and 100 for N in equation $(1)$ .

$f1=35×65100=22.75$

Consider the following table,

$6.25$	$fo$	$fe$	$fo−fe$	$(fo−fe)2$	$(fo−fe)2/fe$
	20	22.75	-2.75	7.5625	0.332417582
	45	42.25	2.75	7.5625	0.178994083
	15	12.25	2.75	7.5625	0.617346939
	20	22.75	$−2.75$	7.5625	0.332417582
Total	100	100	0		$χ2=1.46$

The value $fo−fe$ is obtained as,

Substitute 20 for $fo$ and 22.75 for $fe$ in the above formula.

$fo−fe=20−22.75=−2.75$

Squaring the above obtained result,

$(fo−fe)2=(−2.75)2=7.56$

Divide the above obtained result by $fe$ .

$(fo−fe)2fe=7.5622.75=0.33$

Proceed in a similar manner to obtain rest of the values of $(fo−fe)2/fe$ and refer above table for the rest of the values of $(fo−fe)2/fe$ .

The chi square value is given as,

$χ2(obtained)=∑(fo−fe)2fe=0.33+0.18+0.62+0.33=1.46$

Thus, the chi square value is 1.46.

Conclusion:

The chi square value is 1.46.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution

To determine

(a)

To find:

The chi square value for the given information.

Answer to Problem 10.1P

Solution:

The chi square value is 1.12.

Explanation of Solution

Given:

The given table of information is,

a.	$20$	$25$	$45$
	$25$	$20$	$45$
	$45$	$45$	$90$

Description:

The confidence interval is an interval estimate from the statistics of the observed data that might contain the true value of the unknown population parameter.

Formula used:

For a chi square, the expected frequency $fe$ is given as,

$fe=Row marginal×Column marginalN$

Where N is the total of frequencies.

The chi square statistic is given by,

$χ2(obtained)=∑(fo−fe)2fe$

Where $fo$ is the observed frequency,

And $fe$ is the expected frequency.

Calculation:

From the given information,

The row marginal is 45, the column marginal is 45 and N is 90.

The observed frequency is given as,

$fe=Row marginal×Column marginalN$

Substitute 45 for row marginal, 45 for column marginal and 90 for N in the above formula.

$fe=45×4590=22.5$

Consider the following table,

$6.25$	$fo$	$fe$	$fo−fe$	$(fo−fe)2$	$(fo−fe)2/fe$
	20	22.5	$−2.5$	$6.25$	0.28
	25	22.5	$2.5$	$6.25$	0.28
	25	22.5	$2.5$	$6.25$	0.28
	20	22.5	$−2.5$	$6.25$	0.28
Total	90	90	0		$χ2=1.12$

The value $fo−fe$ is obtained as,

Substitute 20 for $fo$ and 22.5 for $fe$ in the above formula.

$fo−fe=20−22.5=−2.5$

Squaring the above obtained result,

$(fo−fe)2=(−2.5)2=6.25$

Divide the above obtained result by $fe$ .

$(fo−fe)2fe=6.2522.5=0.28$

Proceed in a similar manner to obtain rest of the values of $(fo−fe)2/fe$ and refer above table for the rest of the values of $(fo−fe)2/fe$ .

The chi square value is given as,

$χ2(obtained)=∑(fo−fe)2fe=0.28+0.28+0.28+0.28=1.12$

Thus, the chi square value is 1.12.

Conclusion:

The chi square value is 1.12.

Expert Solution

To determine

(b)

To find:

The chi square value for the given information.

Answer to Problem 10.1P

Solution:

The chi square value is 0.0.

Explanation of Solution

Given:

The given table of information is,

b.	10	15	25
	20	30	50
	30	45	75

Description:

The confidence interval is an interval estimate from the statistics of the observed data that might contain the true value of the unknown population parameter.

Formula used:

For a chi square, the expected frequency $fe$ is given as,

$fe=Row marginal×Column marginalN$

Where N is the total of frequencies.

The chi square statistic is given by,

$χ2(obtained)=∑(fo−fe)2fe$

Where $fo$ is the observed frequency,

And $fe$ is the expected frequency.

Calculation:

From the given information,

The observed frequency is given as,

$fe=Row marginal×Column marginalN$ …… $(1)$

Substitute 25 for row marginal, 30 for column marginal and 75 for N in equation $(1)$ .

$f1=25×3075=10$

Substitute 25 for row marginal, 45 for column marginal and 75 for N in equation $(1)$ .

$f1=25×4575=15$

Substitute 50 for row marginal, 30 for column marginal and 75 for N in equation $(1)$ .

$f1=50×3075=20$

Substitute 50 for row marginal, 45 for column marginal and 75 for N in equation $(1)$ .

$f1=50×4575=30$

Consider the following table,

$6.25$	$fo$	$fe$	$fo−fe$	$(fo−fe)2$	$(fo−fe)2/fe$
	10	10	0	0	0
	15	15	0	0	0
	20	20	0	0	0
	30	30	0	0	0
Total	75	75	0	0	$χ2=0.0$

The value $fo−fe$ is obtained as,

Substitute 10 for $fo$ and 10 for $fe$ in the above formula.

$fo−fe=10−10=0$

Squaring the above obtained result,

$(fo−fe)2=(0)2=0$

Divide the above obtained result by $fe$ .

$(fo−fe)2fe=0$

Proceed in a similar manner to obtain rest of the values of $(fo−fe)2/fe$ and refer above table for the rest of the values of $(fo−fe)2/fe$ .

The chi square value is given as,

$χ2(obtained)=∑(fo−fe)2fe=0+0+0+0=0.0$

Thus, the chi square value is 0.0.

Conclusion:

The chi square value is 0.0.

Expert Solution

To determine

(c)

To find:

The chi square value for the given information.

Answer to Problem 10.1P

Solution:

The chi square value is 1.51.

Explanation of Solution

Given:

The given table of information is,

c.	25	15	40
	30	30	60
	55	45	100

Description:

The confidence interval is an interval estimate from the statistics of the observed data that might contain the true value of the unknown population parameter.

Formula used:

For a chi square, the expected frequency $fe$ is given as,

$fe=Row marginal×Column marginalN$

Where N is the total of frequencies.

The chi square statistic is given by,

$χ2(obtained)=∑(fo−fe)2fe$

Where $fo$ is the observed frequency,

And $fe$ is the expected frequency.

Calculation:

From the given information,

The observed frequency is given as,

$fe=Row marginal×Column marginalN$ …… $(1)$

Substitute 40 for row marginal, 55 for column marginal and 100 for N in equation $(1)$ .

$f1=40×55100=22$

Substitute 40 for row marginal, 45 for column marginal and 100 for N in equation $(1)$ .

$f1=40×45100=18$

Substitute 60 for row marginal, 55 for column marginal and 100 for N in equation $(1)$ .

$f1=60×55100=33$

Substitute 60 for row marginal, 45 for column marginal and 100 for N in equation $(1)$ .

$f1=60×45100=27$

Consider the following table,

$6.25$	$fo$	$fe$	$fo−fe$	$(fo−fe)2$	$(fo−fe)2/fe$
	25	22	3	9	0.41
	15	18	$−3$	9	0.50
	30	33	$−3$	9	0.27
	30	27	3	9	0.33
Total	100	100	0		$χ2=1.51$

The value $fo−fe$ is obtained as,

Substitute 25 for $fo$ and 22 for $fe$ in the above formula.

$fo−fe=25−22=3$

Squaring the above obtained result,

$(fo−fe)2=(3)2=9$

Divide the above obtained result by $fe$ .

$(fo−fe)2fe=922=0.41$

Proceed in a similar manner to obtain rest of the values of $(fo−fe)2/fe$ and refer above table for the rest of the values of $(fo−fe)2/fe$ .

The chi square value is given as,

$χ2(obtained)=∑(fo−fe)2fe=0.41+0.50+0.27+0.33=1.51$

Thus, the chi square value is 1.51.

Conclusion:

The chi square value is 1.51.

Expert Solution

To determine

(d)

To find:

The chi square value for the given information.

Answer to Problem 10.1P

Solution:

The chi square value is 1.43.

Explanation of Solution

Given:

The given table of information is,

d.	20	45	65
	15	20	35
	35	65	100

Description:

The confidence interval is an interval estimate from the statistics of the observed data that might contain the true value of the unknown population parameter.

Formula used:

For a chi square, the expected frequency $fe$ is given as,

$fe=Row marginal×Column marginalN$

Where N is the total of frequencies.

The chi square statistic is given by,

$χ2(obtained)=∑(fo−fe)2fe$

Where $fo$ is the observed frequency,

And $fe$ is the expected frequency.

Calculation:

From the given information,

The observed frequency is given as,

$fe=Row marginal×Column marginalN$ …… $(1)$

Substitute 65 for row marginal, 35 for column marginal and 100 for N in equation $(1)$ .

$f1=65×35100=22.75$

Substitute 65 for row marginal, 65 for column marginal and 100 for N in equation $(1)$ .

$f1=65×65100=42.25$

Substitute 35 for row marginal, 35 for column marginal and 100 for N in equation $(1)$ .

$f1=35×35100=12.25$

Substitute 35 for row marginal, 65 for column marginal and 100 for N in equation $(1)$ .

$f1=35×65100=22.75$

Consider the following table,

$6.25$	$fo$	$fe$	$fo−fe$	$(fo−fe)2$	$(fo−fe)2/fe$
	20	22.75	-2.75	7.5625	0.332417582
	45	42.25	2.75	7.5625	0.178994083
	15	12.25	2.75	7.5625	0.617346939
	20	22.75	$−2.75$	7.5625	0.332417582
Total	100	100	0		$χ2=1.46$

The value $fo−fe$ is obtained as,

Substitute 20 for $fo$ and 22.75 for $fe$ in the above formula.

$fo−fe=20−22.75=−2.75$

Squaring the above obtained result,

$(fo−fe)2=(−2.75)2=7.56$

Divide the above obtained result by $fe$ .

$(fo−fe)2fe=7.5622.75=0.33$

Proceed in a similar manner to obtain rest of the values of $(fo−fe)2/fe$ and refer above table for the rest of the values of $(fo−fe)2/fe$ .

The chi square value is given as,

$χ2(obtained)=∑(fo−fe)2fe=0.33+0.18+0.62+0.33=1.46$

Thus, the chi square value is 1.46.

Conclusion:

The chi square value is 1.46.

Answer 8

Expert Solution

To determine

(a)

To find:

The chi square value for the given information.

Answer to Problem 10.1P

Solution:

The chi square value is 1.12.

Explanation of Solution

Given:

The given table of information is,

a.	$20$	$25$	$45$
	$25$	$20$	$45$
	$45$	$45$	$90$

Description:

The confidence interval is an interval estimate from the statistics of the observed data that might contain the true value of the unknown population parameter.

Formula used:

For a chi square, the expected frequency $fe$ is given as,

$fe=Row marginal×Column marginalN$

Where N is the total of frequencies.

The chi square statistic is given by,

$χ2(obtained)=∑(fo−fe)2fe$

Where $fo$ is the observed frequency,

And $fe$ is the expected frequency.

Calculation:

From the given information,

The row marginal is 45, the column marginal is 45 and N is 90.

The observed frequency is given as,

$fe=Row marginal×Column marginalN$

Substitute 45 for row marginal, 45 for column marginal and 90 for N in the above formula.

$fe=45×4590=22.5$

Consider the following table,

$6.25$	$fo$	$fe$	$fo−fe$	$(fo−fe)2$	$(fo−fe)2/fe$
	20	22.5	$−2.5$	$6.25$	0.28
	25	22.5	$2.5$	$6.25$	0.28
	25	22.5	$2.5$	$6.25$	0.28
	20	22.5	$−2.5$	$6.25$	0.28
Total	90	90	0		$χ2=1.12$

The value $fo−fe$ is obtained as,

Substitute 20 for $fo$ and 22.5 for $fe$ in the above formula.

$fo−fe=20−22.5=−2.5$

Squaring the above obtained result,

$(fo−fe)2=(−2.5)2=6.25$

Divide the above obtained result by $fe$ .

$(fo−fe)2fe=6.2522.5=0.28$

Proceed in a similar manner to obtain rest of the values of $(fo−fe)2/fe$ and refer above table for the rest of the values of $(fo−fe)2/fe$ .

The chi square value is given as,

$χ2(obtained)=∑(fo−fe)2fe=0.28+0.28+0.28+0.28=1.12$

Thus, the chi square value is 1.12.

Conclusion:

The chi square value is 1.12.

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

(a)

To find:

The chi square value for the given information.

Answer 13

Answer to Problem 10.1P

Solution:

The chi square value is 1.12.

Answer 14

Explanation of Solution

Given:

The given table of information is,

a.	$20$	$25$	$45$
	$25$	$20$	$45$
	$45$	$45$	$90$

Description:

The confidence interval is an interval estimate from the statistics of the observed data that might contain the true value of the unknown population parameter.

Formula used:

For a chi square, the expected frequency $fe$ is given as,

$fe=Row marginal×Column marginalN$

Where N is the total of frequencies.

The chi square statistic is given by,

$χ2(obtained)=∑(fo−fe)2fe$

Where $fo$ is the observed frequency,

And $fe$ is the expected frequency.

Calculation:

From the given information,

The row marginal is 45, the column marginal is 45 and N is 90.

The observed frequency is given as,

$fe=Row marginal×Column marginalN$

Substitute 45 for row marginal, 45 for column marginal and 90 for N in the above formula.

$fe=45×4590=22.5$

Consider the following table,

$6.25$	$fo$	$fe$	$fo−fe$	$(fo−fe)2$	$(fo−fe)2/fe$
	20	22.5	$−2.5$	$6.25$	0.28
	25	22.5	$2.5$	$6.25$	0.28
	25	22.5	$2.5$	$6.25$	0.28
	20	22.5	$−2.5$	$6.25$	0.28
Total	90	90	0		$χ2=1.12$

The value $fo−fe$ is obtained as,

Substitute 20 for $fo$ and 22.5 for $fe$ in the above formula.

$fo−fe=20−22.5=−2.5$

Squaring the above obtained result,

$(fo−fe)2=(−2.5)2=6.25$

Divide the above obtained result by $fe$ .

$(fo−fe)2fe=6.2522.5=0.28$

Proceed in a similar manner to obtain rest of the values of $(fo−fe)2/fe$ and refer above table for the rest of the values of $(fo−fe)2/fe$ .

The chi square value is given as,

$χ2(obtained)=∑(fo−fe)2fe=0.28+0.28+0.28+0.28=1.12$

Thus, the chi square value is 1.12.

Conclusion:

The chi square value is 1.12.

Answer 15

Expert Solution

To determine

(b)

To find:

The chi square value for the given information.

Answer to Problem 10.1P

Solution:

The chi square value is 0.0.

Explanation of Solution

Given:

The given table of information is,

b.	10	15	25
	20	30	50
	30	45	75

Description:

The confidence interval is an interval estimate from the statistics of the observed data that might contain the true value of the unknown population parameter.

Formula used:

For a chi square, the expected frequency $fe$ is given as,

$fe=Row marginal×Column marginalN$

Where N is the total of frequencies.

The chi square statistic is given by,

$χ2(obtained)=∑(fo−fe)2fe$

Where $fo$ is the observed frequency,

And $fe$ is the expected frequency.

Calculation:

From the given information,

The observed frequency is given as,

$fe=Row marginal×Column marginalN$ …… $(1)$

Substitute 25 for row marginal, 30 for column marginal and 75 for N in equation $(1)$ .

$f1=25×3075=10$

Substitute 25 for row marginal, 45 for column marginal and 75 for N in equation $(1)$ .

$f1=25×4575=15$

Substitute 50 for row marginal, 30 for column marginal and 75 for N in equation $(1)$ .

$f1=50×3075=20$

Substitute 50 for row marginal, 45 for column marginal and 75 for N in equation $(1)$ .

$f1=50×4575=30$

Consider the following table,

$6.25$	$fo$	$fe$	$fo−fe$	$(fo−fe)2$	$(fo−fe)2/fe$
	10	10	0	0	0
	15	15	0	0	0
	20	20	0	0	0
	30	30	0	0	0
Total	75	75	0	0	$χ2=0.0$

The value $fo−fe$ is obtained as,

Substitute 10 for $fo$ and 10 for $fe$ in the above formula.

$fo−fe=10−10=0$

Squaring the above obtained result,

$(fo−fe)2=(0)2=0$

Divide the above obtained result by $fe$ .

$(fo−fe)2fe=0$

Proceed in a similar manner to obtain rest of the values of $(fo−fe)2/fe$ and refer above table for the rest of the values of $(fo−fe)2/fe$ .

The chi square value is given as,

$χ2(obtained)=∑(fo−fe)2fe=0+0+0+0=0.0$

Thus, the chi square value is 0.0.

Conclusion:

The chi square value is 0.0.

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

(b)

To find:

The chi square value for the given information.

Answer 20

Answer to Problem 10.1P

Solution:

The chi square value is 0.0.

Answer 21

Explanation of Solution

Given:

The given table of information is,

b.	10	15	25
	20	30	50
	30	45	75

Description:

The confidence interval is an interval estimate from the statistics of the observed data that might contain the true value of the unknown population parameter.

Formula used:

For a chi square, the expected frequency $fe$ is given as,

$fe=Row marginal×Column marginalN$

Where N is the total of frequencies.

The chi square statistic is given by,

$χ2(obtained)=∑(fo−fe)2fe$

Where $fo$ is the observed frequency,

And $fe$ is the expected frequency.

Calculation:

From the given information,

The observed frequency is given as,

$fe=Row marginal×Column marginalN$ …… $(1)$

Substitute 25 for row marginal, 30 for column marginal and 75 for N in equation $(1)$ .

$f1=25×3075=10$

Substitute 25 for row marginal, 45 for column marginal and 75 for N in equation $(1)$ .

$f1=25×4575=15$

Substitute 50 for row marginal, 30 for column marginal and 75 for N in equation $(1)$ .

$f1=50×3075=20$

Substitute 50 for row marginal, 45 for column marginal and 75 for N in equation $(1)$ .

$f1=50×4575=30$

Consider the following table,

$6.25$	$fo$	$fe$	$fo−fe$	$(fo−fe)2$	$(fo−fe)2/fe$
	10	10	0	0	0
	15	15	0	0	0
	20	20	0	0	0
	30	30	0	0	0
Total	75	75	0	0	$χ2=0.0$

The value $fo−fe$ is obtained as,

Substitute 10 for $fo$ and 10 for $fe$ in the above formula.

$fo−fe=10−10=0$

Squaring the above obtained result,

$(fo−fe)2=(0)2=0$

Divide the above obtained result by $fe$ .

$(fo−fe)2fe=0$

Proceed in a similar manner to obtain rest of the values of $(fo−fe)2/fe$ and refer above table for the rest of the values of $(fo−fe)2/fe$ .

The chi square value is given as,

$χ2(obtained)=∑(fo−fe)2fe=0+0+0+0=0.0$

Thus, the chi square value is 0.0.

Conclusion:

The chi square value is 0.0.

Answer 22

Expert Solution

To determine

(c)

To find:

The chi square value for the given information.

Answer to Problem 10.1P

Solution:

The chi square value is 1.51.

Explanation of Solution

Given:

The given table of information is,

c.	25	15	40
	30	30	60
	55	45	100

Description:

The confidence interval is an interval estimate from the statistics of the observed data that might contain the true value of the unknown population parameter.

Formula used:

For a chi square, the expected frequency $fe$ is given as,

$fe=Row marginal×Column marginalN$

Where N is the total of frequencies.

The chi square statistic is given by,

$χ2(obtained)=∑(fo−fe)2fe$

Where $fo$ is the observed frequency,

And $fe$ is the expected frequency.

Calculation:

From the given information,

The observed frequency is given as,

$fe=Row marginal×Column marginalN$ …… $(1)$

Substitute 40 for row marginal, 55 for column marginal and 100 for N in equation $(1)$ .

$f1=40×55100=22$

Substitute 40 for row marginal, 45 for column marginal and 100 for N in equation $(1)$ .

$f1=40×45100=18$

Substitute 60 for row marginal, 55 for column marginal and 100 for N in equation $(1)$ .

$f1=60×55100=33$

Substitute 60 for row marginal, 45 for column marginal and 100 for N in equation $(1)$ .

$f1=60×45100=27$

Consider the following table,

$6.25$	$fo$	$fe$	$fo−fe$	$(fo−fe)2$	$(fo−fe)2/fe$
	25	22	3	9	0.41
	15	18	$−3$	9	0.50
	30	33	$−3$	9	0.27
	30	27	3	9	0.33
Total	100	100	0		$χ2=1.51$

The value $fo−fe$ is obtained as,

Substitute 25 for $fo$ and 22 for $fe$ in the above formula.

$fo−fe=25−22=3$

Squaring the above obtained result,

$(fo−fe)2=(3)2=9$

Divide the above obtained result by $fe$ .

$(fo−fe)2fe=922=0.41$

Proceed in a similar manner to obtain rest of the values of $(fo−fe)2/fe$ and refer above table for the rest of the values of $(fo−fe)2/fe$ .

The chi square value is given as,

$χ2(obtained)=∑(fo−fe)2fe=0.41+0.50+0.27+0.33=1.51$

Thus, the chi square value is 1.51.

Conclusion:

The chi square value is 1.51.

Answer 23

Expert Solution

Answer 24

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

(c)

To find:

The chi square value for the given information.

Answer 27

Answer to Problem 10.1P

Solution:

The chi square value is 1.51.

Answer 28

Explanation of Solution

Given:

The given table of information is,

c.	25	15	40
	30	30	60
	55	45	100

Description:

The confidence interval is an interval estimate from the statistics of the observed data that might contain the true value of the unknown population parameter.

Formula used:

For a chi square, the expected frequency $fe$ is given as,

$fe=Row marginal×Column marginalN$

Where N is the total of frequencies.

The chi square statistic is given by,

$χ2(obtained)=∑(fo−fe)2fe$

Where $fo$ is the observed frequency,

And $fe$ is the expected frequency.

Calculation:

From the given information,

The observed frequency is given as,

$fe=Row marginal×Column marginalN$ …… $(1)$

Substitute 40 for row marginal, 55 for column marginal and 100 for N in equation $(1)$ .

$f1=40×55100=22$

Substitute 40 for row marginal, 45 for column marginal and 100 for N in equation $(1)$ .

$f1=40×45100=18$

Substitute 60 for row marginal, 55 for column marginal and 100 for N in equation $(1)$ .

$f1=60×55100=33$

Substitute 60 for row marginal, 45 for column marginal and 100 for N in equation $(1)$ .

$f1=60×45100=27$

Consider the following table,

$6.25$	$fo$	$fe$	$fo−fe$	$(fo−fe)2$	$(fo−fe)2/fe$
	25	22	3	9	0.41
	15	18	$−3$	9	0.50
	30	33	$−3$	9	0.27
	30	27	3	9	0.33
Total	100	100	0		$χ2=1.51$

The value $fo−fe$ is obtained as,

Substitute 25 for $fo$ and 22 for $fe$ in the above formula.

$fo−fe=25−22=3$

Squaring the above obtained result,

$(fo−fe)2=(3)2=9$

Divide the above obtained result by $fe$ .

$(fo−fe)2fe=922=0.41$

Proceed in a similar manner to obtain rest of the values of $(fo−fe)2/fe$ and refer above table for the rest of the values of $(fo−fe)2/fe$ .

The chi square value is given as,

$χ2(obtained)=∑(fo−fe)2fe=0.41+0.50+0.27+0.33=1.51$

Thus, the chi square value is 1.51.

Conclusion:

The chi square value is 1.51.

Answer 29

Expert Solution

To determine

(d)

To find:

The chi square value for the given information.

Answer to Problem 10.1P

Solution:

The chi square value is 1.43.

Explanation of Solution

Given:

The given table of information is,

d.	20	45	65
	15	20	35
	35	65	100

Description:

The confidence interval is an interval estimate from the statistics of the observed data that might contain the true value of the unknown population parameter.

Formula used:

For a chi square, the expected frequency $fe$ is given as,

$fe=Row marginal×Column marginalN$

Where N is the total of frequencies.

The chi square statistic is given by,

$χ2(obtained)=∑(fo−fe)2fe$

Where $fo$ is the observed frequency,

And $fe$ is the expected frequency.

Calculation:

From the given information,

The observed frequency is given as,

$fe=Row marginal×Column marginalN$ …… $(1)$

Substitute 65 for row marginal, 35 for column marginal and 100 for N in equation $(1)$ .

$f1=65×35100=22.75$

Substitute 65 for row marginal, 65 for column marginal and 100 for N in equation $(1)$ .

$f1=65×65100=42.25$

Substitute 35 for row marginal, 35 for column marginal and 100 for N in equation $(1)$ .

$f1=35×35100=12.25$

Substitute 35 for row marginal, 65 for column marginal and 100 for N in equation $(1)$ .

$f1=35×65100=22.75$

Consider the following table,

$6.25$	$fo$	$fe$	$fo−fe$	$(fo−fe)2$	$(fo−fe)2/fe$
	20	22.75	-2.75	7.5625	0.332417582
	45	42.25	2.75	7.5625	0.178994083
	15	12.25	2.75	7.5625	0.617346939
	20	22.75	$−2.75$	7.5625	0.332417582
Total	100	100	0		$χ2=1.46$

The value $fo−fe$ is obtained as,

Substitute 20 for $fo$ and 22.75 for $fe$ in the above formula.

$fo−fe=20−22.75=−2.75$

Squaring the above obtained result,

$(fo−fe)2=(−2.75)2=7.56$

Divide the above obtained result by $fe$ .

$(fo−fe)2fe=7.5622.75=0.33$

Proceed in a similar manner to obtain rest of the values of $(fo−fe)2/fe$ and refer above table for the rest of the values of $(fo−fe)2/fe$ .

The chi square value is given as,

$χ2(obtained)=∑(fo−fe)2fe=0.33+0.18+0.62+0.33=1.46$

Thus, the chi square value is 1.46.

Conclusion:

The chi square value is 1.46.

Answer 30

Expert Solution

Answer 31

Expert Solution

Answer 32

Expert Solution

Answer 33

To determine

(d)

To find:

The chi square value for the given information.

Answer 34

Answer to Problem 10.1P

Solution:

The chi square value is 1.43.

Answer 35

Explanation of Solution

Given:

The given table of information is,

d.	20	45	65
	15	20	35
	35	65	100

Description:

The confidence interval is an interval estimate from the statistics of the observed data that might contain the true value of the unknown population parameter.

Formula used:

For a chi square, the expected frequency $fe$ is given as,

$fe=Row marginal×Column marginalN$

Where N is the total of frequencies.

The chi square statistic is given by,

$χ2(obtained)=∑(fo−fe)2fe$

Where $fo$ is the observed frequency,

And $fe$ is the expected frequency.

Calculation:

From the given information,

The observed frequency is given as,

$fe=Row marginal×Column marginalN$ …… $(1)$

Substitute 65 for row marginal, 35 for column marginal and 100 for N in equation $(1)$ .

$f1=65×35100=22.75$

Substitute 65 for row marginal, 65 for column marginal and 100 for N in equation $(1)$ .

$f1=65×65100=42.25$

Substitute 35 for row marginal, 35 for column marginal and 100 for N in equation $(1)$ .

$f1=35×35100=12.25$

Substitute 35 for row marginal, 65 for column marginal and 100 for N in equation $(1)$ .

$f1=35×65100=22.75$

Consider the following table,

$6.25$	$fo$	$fe$	$fo−fe$	$(fo−fe)2$	$(fo−fe)2/fe$
	20	22.75	-2.75	7.5625	0.332417582
	45	42.25	2.75	7.5625	0.178994083
	15	12.25	2.75	7.5625	0.617346939
	20	22.75	$−2.75$	7.5625	0.332417582
Total	100	100	0		$χ2=1.46$