# Chi Square

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(Using  = 0.01, the appropriate decision for the treatment effects is ________

For a one-way ANOVA, the Between Sum of Squares (SSB) is ________

Using  = 0.05, the appropriate decision is _____________.

4. Suppose the mean squares for treatment in a one-way ANOVA are 24.4 and the mean squares for error are 9.8. There were four treatments and 7 subjects received each treatment (for a total of 28). The calculated value of F is _______.

A chi-square goodness of fit test is to be performed. The degrees of freedom are 12, and alpha is 0.10. The table chi-square value that defines the rejection region is _______.

For a chi-square goodness of fit test, the calculated chi-square value is 6.74. The table chi-square value is 9.488. The appropriate decision for this is _______.

7. A variable contains five categories. It is expected that data are uniformly distributed across these five categories. To test this, a sample of observed data is gathered on this variable resulting in frequencies of 26, 10, 17, 21, 26. Using  = .01, the critical chi square value is _______.

A variable contains five categories. It is expected that data are uniformly distributed across these five categories. To test this, a sample of observed data is gathered on this variable resulting in frequencies of 26, 10, 17, 21, 26. Using  = .01, the appropriate decision is _______.

Using  = .10, the critical chi square value for this goodness of fit test is ________

A contingency table is to be used to test for independence. There are 3 rows and 3 columns in the table. How many degrees of freedom are there for this problem?)

1.

Pate's Pharmacy, Inc. operates a regional chain of 120 pharmacies. Each pharmacy's floor plan includes a greeting card department which is relatively isolated. Sandra Royo, Marketing Manager, feels that the lighting level in the greeting card department may affect sales in that department. She divides the 120 pharmacies into 3 groups (urban, suburban, and rural) and randomly assigns six pharmacies to each lighting level (soft, medium, and bright). Analysis of Sandra's data produced the following ANOVA table.

Source of Variation SS df MS F

Treatment 50 2 25 0.230769

Block 6016.667 2 3008.333 27.76923

Error 433.3333 4 108.3333

Total 6500 8

Using  = 0.01, the appropriate decision for the treatment effects is ________.

do not reject the null hypothesis  1   2   3

do not reject the null hypothesis  1 =  2 = 

reject the null hypothesis  1   2   3

reject the null hypothesis  1   2   3

2.

Data from a completely randomized design are shown in the following table.

Treatment Level

1 2 3

27 26 27

26 22 29

23 21 27

24 23 26

For a one-way ANOVA, the Between Sum of Squares (SSB) is

36.17

28.75

64.92

18.03

3.

BigShots, Inc. is a specialty e-tailer that operates 87 catalog Web sites on the Internet. Kevin Conn, Sales Director, feels that the style (color scheme, graphics, fonts, etc.) of a Web site may affect its sales. He chooses three levels of design style (neon, old world, and sophisticated) and randomly assigns six catalog Web sites to each design style. Analysis of Kevin's data yielded the following ANOVA table.

Source of Variation SS df MS F

Between Groups 384.3333 2 192.1667

Within Groups 1359.667 15 90.64444

Total 1744 17

Using  = 0.05, the appropriate decision is _____________.

do not reject the null hypothesis  1 =  2 = 3

do not reject the null hypothesis  1   2  3

reject the null hypothesis  1   2   3

reject the null hypothesis  1   2   3

4. Suppose the mean squares for treatment in a one-way ANOVA are 24.4 and the mean squares for error are 9.8. There were four treatments and 7 subjects received each treatment (for a total of 28). The calculated value of F is _______.

9.8

34.2

2.49

14.6

5.

A chi-square goodness of fit test is to be performed. The degrees of freedom are 12, and alpha is 0.10. The table chi-square value that defines the rejection region is _______.

26.217

6.304

18.549

17.275

6.

For a chi-square goodness of fit test, the calculated chi-square value is 6.74. The table chi-square value is 9.488. The appropriate decision for this is _______.

reject the null hypothesis

accept the null hypothesis

accept the alternative hypothesis

impossible to determine from this information

7. A variable contains five categories. It is expected that data are uniformly distributed across these five categories. To test this, a sample of observed data is gathered on this variable resulting in frequencies of 26, 10, 17, 21, 26. Using  = .01, the critical chi square value is _______. (Points: 10)

13.277

15.086

7.779

11.070

8.

A variable contains five categories. It is expected that data are uniformly distributed across these five categories. To test this, a sample of observed data is gathered on this variable resulting in frequencies of 26, 10, 17, 21, 26. Using  = .01, the appropriate decision is _______.

accept the null hypothesis that the observed distribution is uniform

accept the null hypothesis that the observed distribution is not uniform

reject the null hypothesis that the observed distribution is uniform

reject the null hypothesis that the observed distribution is not uniform

9.

A researcher believes that a variable is Poisson distributed across six categories. To test this, a random sample of observations is made for the variable resulting in the following data:

Category 0 1 2 3 4 5

Observed 47 56 39 22 18 10

Using  = .10, the critical chi square value for this goodness of fit test is _______.

1.064

13.277

9.236

7.779

10.

A contingency table is to be used to test for independence. There are 3 rows and 3 columns in the table. How many degrees of freedom are there for this problem?

6

5

4

3

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##### Solution Summary

(Using  = 0.01, the appropriate decision for the treatment effects is ________

For a one-way ANOVA, the Between Sum of Squares (SSB) is ________

Using  = 0.05, the appropriate decision is _____________.

4. Suppose the mean squares for treatment in a one-way ANOVA are 24.4 and the mean squares for error are 9.8. There were four treatments and 7 subjects received each treatment (for a total of 28). The calculated value of F is _______.

A chi-square goodness of fit test is to be performed. The degrees of freedom are 12, and alpha is 0.10. The table chi-square value that defines the rejection region is _______.

For a chi-square goodness of fit test, the calculated chi-square value is 6.74. The table chi-square value is 9.488. The appropriate decision for this is _______.

7. A variable contains five categories. It is expected that data are uniformly distributed across these five categories. To test this, a sample of observed data is gathered on this variable resulting in frequencies of 26, 10, 17, 21, 26. Using  = .01, the critical chi square value is _______.

A variable contains five categories. It is expected that data are uniformly distributed across these five categories. To test this, a sample of observed data is gathered on this variable resulting in frequencies of 26, 10, 17, 21, 26. Using  = .01, the appropriate decision is _______.

Using  = .10, the critical chi square value for this goodness of fit test is ________

A contingency table is to be used to test for independence. There are 3 rows and 3 columns in the table. How many degrees of freedom are there for this problem?)

###### Education

- BSc , Wuhan Univ. China
- MA, Shandong Univ.

###### Recent Feedback

- "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
- "excellent work"
- "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
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- "Thank you very much for your valuable time and assistance!"

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