Question 1

Using the table for the chi-square distributions,find the lower 10% point when d.f.= 17.

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Question 2

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The variance of the prices of all college textbooks was 81.30 last year.A recent sample of 20 college textbooks produced a variance of their prices equal to 164.649.The null hypothesis is that the current variance of the prices of all college textbooks is equal to 81.30 and the alternative hypothesis is that the current variance of the prices of all college textbooks is greater than 81.30.Assume that the prices of college textbooks are normally distributed.The significance level is 1%.
-The value of the chi-square test statistic is,rounded to three decimal places:

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Question 3

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A random sample of 29 workers selected from a large company produced a variance of their weekly earnings equal to 8124.Assume that the population is normally distributed.
-Based on this sample,the 95% confidence interval for the standard deviation of the weekly earnings of all of this company's employees,rounded to the nearest cent,is:

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Question 4

Find the probability of  when d.f.= 9. 
A)0.005
B)0.1
C)0.01
D)0.025

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Question 5

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A researcher claims that the variance of the heights of all female college basketball players is 4.80 square inches.A random sample of 28 female college basketball players produced a variance of their heights equal to 11.122 square inches.The null hypothesis is that the variance of the heights of all female college basketball players is equal to 4.80 square inches and the alternative hypothesis is that the variance of the heights of all female college basketball players is not equal to 4.80.Assume that the population is normally distributed.The significance level is 5%.
-The critical values of chi-square are:
A)13.844 and 41.923
B)14.573 and 43.195
C)15.308 and 44.461
D)12.879 and 40.113

Accepted Answer

The degrees of freedom for the chi-square test is n-1, which is 27. At a 5% significance level for a two-tailed test, the critical chi-square values for 27 degrees of freedom are approximately 14.573 and 43.195.

Question 6

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A random sample of 20 elements selected from a population produced a variance equal to 11.072.Assume that the population is normally distributed.
-Based on this sample,the 90% confidence interval for the population standard deviation,rounded to three decimal places,is:

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Question 7

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A random sample of 29 workers selected from a large company produced a variance of their weekly earnings equal to 8124.Assume that the population is normally distributed.
-Based on this sample,the 95% confidence interval for the variance of the weekly earnings of all of this company's employees,rounded to three decimal places,is:

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Question 8

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A random sample of 28 elements selected from a population produced a variance equal to 16.682.The null hypothesis is that the population variance is greater than or equal to 18.70 and the alternative hypothesis is that the population variance is less than 18.70.Assume that the population is normally distributed.The significance level is 2.5%.
-The value of the chi-square test statistic is,rounded to three decimal places:

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Question 9

Find the probability of  when d.f.= 24.

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Question 10

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A random sample of 28 elements selected from a population produced a variance equal to 16.682.The null hypothesis is that the population variance is greater than or equal to 18.70 and the alternative hypothesis is that the population variance is less than 18.70.Assume that the population is normally distributed.The significance level is 2.5%.
-The critical value of chi-square is:
A)14.573
B)43.195
C)41.923
D)13.844

Accepted Answer

The critical value for a chi-square test with 27 degrees of freedom (n-1) at a 2.5% significance level for a left-tailed test is approximately 14.573.

Question 11

Jean,an expert on dental hygiene,hypothesizes that the dental hygiene of students at the Westwood Elementary School is similar to that of students at Easton Elementary.In order to test her hypothesis,she collects information about children at each school from dentists in the area,recording the number of cavities that each child has had in the past year.Her findings appear in the following table.
  The null hypothesis is that the proportions of students with no cavities,one cavity,two cavities,and three or more cavities are relatively the same between schools.What is the value of the test statistic,rounded to three decimal places,that Jean will use?

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Question 12

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A random sample of 25 elements selected from a population produced a variance equal to 4.959.The null hypothesis is that the population variance is less than or equal to 3.90 and the alternative hypothesis is that the population variance is greater than 3.90.Assume that the population is normally distributed.The significance level is 1%.
-The value of the chi-square test statistic is,rounded to three decimal places:

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Question 13

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A random sample of 25 elements selected from a population produced a variance equal to 4.959.The null hypothesis is that the population variance is less than or equal to 3.90 and the alternative hypothesis is that the population variance is greater than 3.90.Assume that the population is normally distributed.The significance level is 1%.
-The critical value of chi-square is:
A)42.980
B)45.559
C)41.638
D)44.181

Accepted Answer

The critical value of chi-square for a significance level of 1% and 24 degrees of freedom is 42.980.

Question 14

Use the following to answer questions :
A researcher claims that the variance of the heights of all female college basketball players is 4.80 square inches.A random sample of 28 female college basketball players produced a variance of their heights equal to 11.122 square inches.The null hypothesis is that the variance of the heights of all female college basketball players is equal to 4.80 square inches and the alternative hypothesis is that the variance of the heights of all female college basketball players is not equal to 4.80.Assume that the population is normally distributed.The significance level is 5%.
-The value of the chi-square test statistic is,rounded to three decimal places:

Accepted Answer

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Question 15

Use the following to answer questions :
The variance of the prices of all college textbooks was 81.30 last year.A recent sample of 20 college textbooks produced a variance of their prices equal to 164.649.The null hypothesis is that the current variance of the prices of all college textbooks is equal to 81.30 and the alternative hypothesis is that the current variance of the prices of all college textbooks is greater than 81.30.Assume that the prices of college textbooks are normally distributed.The significance level is 1%.
-The critical value of chi-square is:
A)36.191
B)38.582
C)39.997
D)37.566

Accepted Answer

The critical value of chi-square for a significance level of 1% and 19 degrees of freedom is 36.191.

Question 16

Find the probability of  when d.f.= 19.

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Question 17

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A random sample of 20 elements selected from a population produced a variance equal to 9.606.The null hypothesis is that the population variance is equal to 10.05 and the alternative hypothesis is that the population variance is not equal to 10.05.Assume that the population is normally distributed.The significance level is 5%.
-The critical values of chi-square are:
A)10.117 and 30.144
B)8.231 and 31.526
C)7.633 and 36.191
D)8.907 and 32.852

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Question 18

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A random sample of 20 elements selected from a population produced a variance equal to 9.606.The null hypothesis is that the population variance is equal to 10.05 and the alternative hypothesis is that the population variance is not equal to 10.05.Assume that the population is normally distributed.The significance level is 5%.
-The value of the chi-square test statistic is,rounded to three decimal places:

Accepted Answer

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Quiz 11: Chi-Square Tests

Using the table for the chi-square distributions,find the lower 10% point when d.f.= 17.

Find the probability of when d.f.= 9.

Find the probability of when d.f.= 24.

Find the probability of when d.f.= 19.