In 2007, the United States had 1.5 million homeschooled students, according to the U.S. National Center for Education Statistics. In Table 11.56 you can see that parents decide to homeschool their children for different reasons, and some reasons are ranked by parents as more important than others. According to the survey results shown in the table, is the distribution of applicable reasons the same as the distribution of the most important reason? Provide your assessment at the

5% significance level. Did you expect the result you obtained?

Short Answer

Expert verified

There is sufficient evidence to insure that the distribution of applicable reasons is different as the distribution of the most important reason.

Step by step solution

01

Given Information

Given that the survey results shown in the table is the distribution of applicable reasons the same as the distribution of the most important reason

02

Null Hypothesis

The null hypothesis is given below:

H0 : The distribution of applicable reasons is the same as the distribution of the most important reason.

03

Alternative Hypothesis

Against the alternative hypothesis as given below:

Ha:The distribution of applicable reasons is different from the distribution of the most important reason.

04

The Degrees of Freedom  

The degrees of freedom can be calculated by the formula given below:

df=(number of columns-1)

Therefore.

df=(number of columns-1)

=2-1

=1

05

Distribution Identification

From above calculation, it is clear that the distribution for the test is χ12.

06

Calculation of Expected Frequencies

All calculations can be done in excel worksheet. Hence, the expected (E) values table is shown below:

07

The Test Statistic

The test statistic of the independence test is given below

Test staristic=(<1)(O-E)2E

To calculate (O-E)2Eapply the formula

=(B4-B16)2/B16in cell B27 and drag the same formula up to cell C32. After that, take the total of columns total and rows total. The table of the test statistic is shown below:

Hence the test statistic is234

08

p-value

The p-value can be calculated in excel by using CHIDIST ( ) formula as shown below:

The p-value is 0.000

09

Chi-square sketch

10

Decision, reason and conclusion

i. Alpha:0.05

ii: Decision: Reject the null hypothesisH0

iii. Reason for decision: Because p-value <α

iv. Conclusion: There is sufficient evidence to insure that the distribution of applicable reasons is different as the distribution of the most important reason.

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Most popular questions from this chapter

Suppose an airline claims that its flights are consistently on time with an average delay of at most 15 minutes. It claims that the average delay is so consistent that the variance is no more than 150 minutes. Doubting the consistency part of the claim, a disgruntled traveler calculates the delays for his next 25 flights. The average delay for those 25 flights is 22 minutes with a standard deviation of 15 minutes.

How did you know to test the variance instead of the mean?

Decide whether the following statements are true or false:

As the number of degrees of freedom increases, the graph of the chi-square distribution looks more and more symmetrical.

Table 11.42contains information from a survey among 499participants classified according to their age groups. The second column shows the percentage of obese people per age class among the study participants. The last column comes from a different study at the national level that shows the corresponding percentages of obese people in the same age classes in the USA. Perform a hypothesis test at the 5% significance level to determine whether the survey participants are a representative sample of the USA obese population.

Age Class (Years) Obese Expected (percentage) Obese Observed (Frequencies)
20-30
22.4
122
31-40
18.6
104
41-50
12.8
78
51-60
10.4
64
61-70
35.8
168

Table11.42

Use a solution sheet to solve the hypothesis test problem. Go to Appendix E for the chi-square solution sheet. Round expected frequency to two decimal places.

A manager of a sports club keeps information concerning the main sport in which members participate and their ages. To test whether there is a relationship between the age of a member and his or her choice of sport, 643members of the sports club are randomly selected. Conduct a test of independence.

Sport18-2526-3031-4041andover
racquetball42583046
tennis58763865
swimming72606533

Suppose an airline claims that its flights are consistently on time with an average delay of at most 15minutes. It claims that the average delay is so consistent that the variance is no more than 150minutes. Doubting the consistency part of the claim, a disgruntled traveler calculates the delays for his next 25flights. The average delay for those 25flights is 22minutes with a standard deviation of 15minutes.

If an additional test were done on the claim of the average delay, which distribution would you use?

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