# Chi square test - statistics

1. As an alternative to flextime, many companies allow employees to do some of their

work at home. Individuals in a random sample of 300 workers were classified

according to salary and number of workdays per week spent at home. The results of

are given in the table. Is there sufficient evidence to indicate that family income and

work-at-home status are independent? Use 5% level of significance level.

Workdays at Home per Week

Less than 1 hour At least one, but not all All at home Total

Under 25 38 16 14 68

25 to 49.999 54 26 12 92

50 to 74.999 35 22 9 66

75 33 29 12 74

Total 160 93 47 300

-H0: The two classifications are?

-H1: The two classifications are?

-Degree of Freedom?

-Test Statistics?

-x2=

-Critical Value=

Therefore, we can conclude that income and work-at-home status are _____________classifications.

2. An educational testing service, in developing a standardized test, would like the test

to have a standard deviation of at least 10. The present form of the test has produced

a standard deviation of s = 8.9 based on n = 30 test scores. Should the present form

of the test be revised based on these sample data? What p-value would you report?

-H0:σ 2 ≥

-H1: σ 2 ≥

-Test Statistic=

-x2=

-Critical value is ?

-Therefore, we _____________ reject the null hypothesis.

3. A quick technique for determining the concentration of a chemical solution has been

proposed to replace the standard technique, which takes much longer. In testing a

standardized solution, 30 determinations using the new technique produced a standard

deviation of s = 7.3 parts per million.

a. Does it appear that the new technique is less sensitive (has larger variability) than

the standard technique whose standard deviation is s = 5 parts per million?

b. Estimate the true standard deviation for the new technique with a 95% confidence

interval.

Solution:

-H0: σ ≤5

-H1: σ ≤5

- Test statistic

-X2=

Critical value=

-Thus, we ____________ reject the null hypothesis

-__________ <

-σ <

4. Fifty fifth-grade students from each of four city schools were given a standardized

fifth-grade reading test. After grading, each student was rated as satisfactory or not

satisfactory in reading ability, with the following results:

School

1 2 3 4

Not satisfactory 7 10 13 6

Is there sufficient evidence to indicate that the percentage of fifth-grade students with

an unsatisfactory reading ability varies from school to school? Use 1% level of

significance level.

-H0: p1=p2=p3=p4=p

-H1

-At least one proportion differs from at least one another. Test statistic?

-Degree of freedom=

-Therefore, we _____________ reject the null hypothesis that reading ability for the fifth grades as measured by the test does not vary from school to school.

5. An experiment to explore the pain thresholds to electrical shock for males and

females resulted in the following data summary:

Males Females

n 10 13

x 15.1 12.6

s2 11.3 26.9

Do these data supply sufficient evidence to indicate a significant difference in

variability of thresholds for these two groups at the 10% level of significance? What

p-value would you report?

-H0:n 10 13

x 15.1 12.6

s 2 11.3 26.9

Do these data supply sufficient evidence to indicate a significant difference in

variability of thresholds for these two groups at the 10% level of significance?

-H0: σ21= σ22

-H1:σ21 ≠ σ22

-Where population 1 is the population of thresholds for females.

Test statistics: F=____________

-Critical Value is _____________

-Therefore, we _____________ reject the null hypothesis and we ______________conclude that the two groups exhibit the same basic variation.

7. A company specializing in kitchen products offers a refrigerator in five different

models. A random sample of n = 250 sales has produced the following data:

Model 1 2 3 4 5

Number sold 62 48 56 39 45

Test the hypothesis that there is no model preference at the a = .05 level of

significance.

-H0:p1=p2=p3=p4=p5=

-H1

At least one of these equalities is incorrect.

-Degree of freedom=

-Test Statistic=

-Critical value-=

-Therefore, we ______________reject the null hypothesis.

8. On the basis of the following data, is there a significant relationship between levels of

income and political party affiliation at the a = .05 level of significance?

Party Affiliation

Income

Low Average High

Republican 33 85 27

Democrat 19 71 56

Other 22 25 13

#### Solution Preview

1. As an alternative to flextime, many companies allow employees to do some of their

work at home. Individuals in a random sample of 300 workers were classified

according to salary and number of workdays per week spent at home. The results of

are given in the table. Is there sufficient evidence to indicate that family income and

work-at-home status are independent? Use 5% level of significance level.

Workdays at Home per Week

Less than 1 hour At least one, but not all All at home Total

Under 25 38 16 14 68

25 to 49.999 54 26 12 92

50 to 74.999 35 22 9 66

75 33 29 12 74

Total 160 93 47 300

-H0: The two classifications are independent from each other.

-H1: The two classifications are dependent on each other

-Degree of Freedom=(4-1)*(3-1)=6

Expected frequency for each category:

160*68/300=36.27, 93*68/300=21.08, 47*68/30=10.65

160*92/300=49.07, 93*92/300=28.52, 47*92/300=14.41;

160*66/300=35.2, 93*66/300=20.46, 47*66/300=10.34

160*74/300=39.47, 93*74/300=22.94, 47*74/300=11.59

Test statistics=(38-36.27)^2/36.27+(16-21.08)^2/21.08+(14-10.65)^2/10.65+(54-49.07)^2/49.07+(26-28.52)^2/28.52+(14.41-12)^2/14.41+(35.2-35)^2/35.2+(22-20.46)^2/20.46+(9-10.34)^2/10.34+(33-39.47)^2/39.47+(22.94-29)^2/22.94+(11.59-12)^2/11.59=6.448

-Test Statistics?

-x2=6.448

-Critical Value=12.592

Therefore, we can conclude that income and work-at-home status are ____independent_________classifications. (since 6.448<12.592, we fail to reject Ho)

2. An educational testing service, in developing a standardized test, would like the test

to have a standard deviation of at least 10. The ...

#### Solution Summary

The solution provides examples how chi square test is used to test variance, association and others.