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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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Essentials Of Statistics

4th Edition
HEALEY + 1 other
Publisher: Cengage Learning,
ISBN: 9781305093836

Solutions

Chapter
Section
BuyFindarrow_forward

Essentials Of Statistics

4th Edition
HEALEY + 1 other
Publisher: Cengage Learning,
ISBN: 9781305093836
Chapter 10, Problem 10.1P
Textbook Problem
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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

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=Rowmarginal×ColumnmarginalN

Where N is the total of frequencies.

The chi square statistic is given by,

χ2(obtained)=(fofe)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=Rowmarginal×ColumnmarginalN

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 fofe (fofe)2 (fofe)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 fofe is obtained as,

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

fofe=2022.5=2.5

Squaring the above obtained result,

(fofe)2=(2.5)2=6.25

Divide the above obtained result by fe.

(fofe)2fe=6.2522.5=0.28

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

The chi square value is given as,

χ2(obtained)=(fofe)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.

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=Rowmarginal×ColumnmarginalN

Where N is the total of frequencies.

The chi square statistic is given by,

χ2(obtained)=(fofe)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=Rowmarginal×ColumnmarginalN……(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 fofe (fofe)2 (fofe)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 fofe is obtained as,

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

fofe=1010=0

Squaring the above obtained result,

(fofe)2=(0)2=0

Divide the above obtained result by fe.

(fofe)2fe=0

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

The chi square value is given as,

χ2(obtained)=(fofe)2fe=0+0+0+0=0.0

Thus, the chi square value is 0.0.

Conclusion:

The chi square value is 0.0.

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=Rowmarginal×ColumnmarginalN

Where N is the total of frequencies.

The chi square statistic is given by,

χ2(obtained)=(fofe)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=Rowmarginal×ColumnmarginalN……(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 fofe (fofe)2 (fofe)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 fofe is obtained as,

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

fofe=2522=3

Squaring the above obtained result,

(fofe)2=(3)2=9

Divide the above obtained result by fe.

(fofe)2fe=922=0.41

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

The chi square value is given as,

χ2(obtained)=(fofe)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.

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=Rowmarginal×ColumnmarginalN

Where N is the total of frequencies.

The chi square statistic is given by,

χ2(obtained)=(fofe)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=Rowmarginal×ColumnmarginalN……(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 fofe (fofe)2 (fofe)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 fofe is obtained as,

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

fofe=2022.75=2.75

Squaring the above obtained result,

(fofe)2=(2.75)2=7.56

Divide the above obtained result by fe.

(fofe)2fe=7.5622.75=0.33

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

The chi square value is given as,

χ2(obtained)=(fofe)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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