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Isabella, an accomplished Bay to Breakers runner, claims that the standard deviation for her time to run the 7.5mile race is at most three minutes. To test her claim, Rupinder looks up five of her race times. They are 55minutes, 61minutes, 58minutes, 63 minutes, and 57 minutes.

Short Answer

Expert verified

Decision: Do not reject the null hypothesis H0

Conclusion: There is insufficient evidence to ensure that the standard deviation of 7.5mile race is greater than three minutes.

Step by step solution

01

Given Information

The null hypothesis is shown below:

H0:σ2(3)2

That is, the standard deviation of 7.5mile race is at most three minutes.

Against the alternative hypothesis as shown below:

Ha:σ2>(15)2

That is, the standard deviation of 7.5mile race is greater than three minutes.

02

Explanation

The degrees of freedom can be calculated as shown below:

n-1=5-1

=4

From the above calculation, it is clear that the distribution for the test is χ42.

Calculate the mean and standard deviation as shown below:

03

Calculation

The test statistics can be calculated by using the formula shown below:

Teststatistic=(n-1)s2σ2

Here, nis sample size, s2is sample variance and σ2is population variance. Therefore, the calculation is shown below:

Test statistic=(n-1)s2σ2

=(5-1)(3.19)2(3)2

=4×10.17619

=4.52

Hence, the test statistic is 4.52.

04

Graph

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

Hence the p- value is 0.340.

The Chi-Square sketch is given below:

05

Final Answer

Alpha: 0.05

Decision: Do not reject the null hypothesisH0

Reason for decision: Because p-value >α

Conclusion: There is insufficient evidence to ensure that the standard deviation of 7.5mile race is greater than three minutes.

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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 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.

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