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Chapter 10: Problem 18

NCAA rules require the circumference of a softball to be \(12 \pm 0.125\) inches. Suppose that the NCAA also requires that the standard deviation of the softball circumferences not exceed 0.05 inch. A representative from the NCAA believes the manufacturer does not meet this requirement. She collects a random sample of 20 softballs from the production line and finds that \(s=0.09\) inch. Is there enough evidence to support the representative's belief at the \(\alpha=0.05\) level of significance?

Short Answer

Expert verified
The test statistic exceeds the critical value; therefore, we reject the null hypothesis and conclude that there is enough evidence to support the representative's belief.

Step by step solution

01

Title - Identify the hypothesis

We need to determine whether the standard deviation requirement is met. The null hypothesis (H鈧�) is that the standard deviation of the softballs, \(\sigma\), is less than or equal to 0.05 inch. The alternative hypothesis (H鈧�) is that the standard deviation is greater than 0.05 inch. Formally: \(\ H_0: \sigma \leq 0.05\) and \(\ H_1: \sigma > 0.05\).
02

- Choose the appropriate test and calculate test statistic

Use the Chi-Square test for variance. The test statistic for the Chi-Square distribution is given by \(\ \chi^2 = \frac{(n - 1) \cdot s^2}{\sigma_0^2}\), where \(\ n \) is the sample size, \(\ s \) is the sample standard deviation, and \(\ \sigma_0 \) is the population standard deviation under the null hypothesis. Plug in the values: \(\ \chi^2 = \frac{(20 - 1) \cdot (0.09)^2}{(0.05)^2} \).
03

- Perform the calculation

Perform the calculation: \(\ \chi^2 = \frac{19 \cdot 0.0081}{0.0025} = \frac{0.1539}{0.0025} = 61.56\).
04

- Determine the critical value

We need the critical value for the Chi-Square distribution with \(\ n - 1 = 19\) degrees of freedom at \(\ \alpha = 0.05 \). From the Chi-Square table, the critical value is approximately 30.14 for \(\ d.f. = 19 \).
05

- Compare the test statistic to the critical value

Compare the test statistic \(\ \chi^2 = 61.56\) to the critical value 30.14. Since 61.56 > 30.14, we reject the null hypothesis.
06

- Conclusion

There is enough evidence at the \(\ \alpha = 0.05 \) level of significance to support the representative's belief that the manufacturer does not meet the standard deviation requirement.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chi-Square Test
The Chi-Square test is used to determine if there is a significant difference between the observed data and the expected data under the null hypothesis. It is often applied in testing the variance or standard deviation of a sample.
In this exercise, we use the Chi-Square test to check if the standard deviation of softball circumferences exceeds the NCAA requirement.

The test statistic is calculated using the formula: \[ \chi^2 = \frac{(n - 1) \times s^2}{\text{null standard deviation}^2} \]
Here, \( n \) is the sample size, and \( s \) is the sample standard deviation.
We collect a sample of 20 softballs, calculate the sample standard deviation as 0.09, and compare it to the null hypothesis standard deviation of 0.05.

In this case, our test statistic value \( \chi^2 = 61.56 \). We compare this with the critical value from the Chi-Square distribution table with 19 degrees of freedom (since \( n - 1 = 20 - 1 \)). If the test statistic is higher than the critical value, we reject the null hypothesis.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. It is a crucial concept in statistics for understanding how spread out the data points are around the mean.

For instance, if the standard deviation is low, it means the data points are close to the mean, indicating consistency. A high standard deviation means data points are spread out over a wider range, indicating variability.
In this exercise, the NCAA sets a requirement that the standard deviation of softball circumferences should not exceed 0.05 inches. The sample collected showed a standard deviation of 0.09 inches. This large value compared to the requirement suggests variability outside the limit, prompting a hypothesis test.

A hypothesis test like the Chi-Square test helps determine if this observed standard deviation significantly exceeds the stipulated limit, helping us make an informed decision.
Significance Level
The significance level, symbolized by \( \alpha \), is the probability of rejecting the null hypothesis when it is actually true. This is also known as the Type I error rate.

Common significance levels used in hypothesis testing are 0.01, 0.05, and 0.10. In this exercise, we use a significance level of 0.05, which means we are willing to accept a 5% chance of incorrectly rejecting the null hypothesis.

To conclude our test, we compare the test statistic with the critical Chi-Square value corresponding to the significance level of 0.05 and 19 degrees of freedom. If our test statistic (61.56) is greater than the critical value (30.14), it indicates that there is a significant difference.
This higher test statistic leads us to reject the null hypothesis, concluding that there is enough evidence to support the NCAA representative's belief that the manufacturer does not meet the standard deviation requirement.

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Most popular questions from this chapter

In Problems \(9-14,\) the null and alternative hypotheses are given. Determine whether the hypothesis test is lefi-tailed, right-tailed, or two-tailed. What parameter is being tested? \(H_{0}: \sigma=4.2\) \(H_{1}: \sigma \neq 4.2\)

Professors Honey Kirk and Diane Lerma of Palo Alto College developed a "learning community curriculum that blended the developmental mathematics and the reading curriculum with a structured emphasis on study skills." In a typical developmental mathematics course at Palo Alto College, \(50 \%\) of the students complete the course with a letter grade of \(\mathrm{A}, \mathrm{B},\) or \(\mathrm{C} .\) In the experimental course, of the 16 students enrolled, 11 completed the course with a letter grade of \(\mathrm{A}, \mathrm{B},\) or \(\mathrm{C} .\) Do you believe the experimental course was effective at the \(\alpha=0.05\) level of significance? (a) State the appropriate null and alternative hypotheses. (b) Verify that the normal model may not be used to estimate the P-value (c) Explain why this is a binomial experiment. (d) Determine the \(P\) -value using the binomial probability distribution. State your conclusion to the hypothesis test. (e) Suppose the course is taught with 48 students and 33 complete the course with a letter grade of \(\mathrm{A}, \mathrm{B},\) or \(\mathrm{C}\). Verify the normal model may now be used to estimate the \(P\) -value. (f) Use the normal model to obtain and interpret the \(P\) -value. State your conclusion to the hypothesis test. (g) Explain the role that sample size plays in the ability to reject statements in the null hypothesis.

Based on historical birthing records, the proportion of males born worldwide is \(0.51 .\) In other words, the commonly held belief that boys are just as likely as girls is false. Systematic lupus erythematosus (SLE), or lupus for short, is a disease in which one's immune system attacks healthy cells and tissue by mistake. It is well known that lupus tends to exist more in females than in males Researchers wondered, however, if families with a child who had lupus had a lower ratio of males to females than the general population. If this were true, it would suggest that something happens during conception that causes males to be conceived at a lower rate when the SLE gene is present. To determine if this hypothesis is true, the researchers obtained records of families with a child who had SLE A total of 23 males and 79 females were found to have SLE. The 23 males with SL \(E\) had \(a\) total of 23 male siblings and 22 female siblings The 79 females with SLE had a total of 69 male siblings and 80 female siblings Source L.N. Moorthy, M.G.E. Peterson, K.B. Onel, and T.J.A. Lehman. "Do Children with Lupus Have Fewer Male Siblings" Laprus 2008 \(17: 128-131,2008\) (a) Explain why this is an observational study. (b) Is the study retrospective or prospective? Why? (c) There are a total of \(23+69=92\) male siblings in the study How many female siblings are in the study? (d) Draw a relative frequency bar graph of gender of the siblings. (e) Find a point estimate for the proportion of male siblings in families where one of the children has SLE (f) Does the sample evidence suggest that the proportion of male siblings in families where one of the children has SLE is less than 0.51 , the accepted proportion of males born in the general population? Use the \(\alpha=0.05\) level of significance. (g) Construct a \(95 \%\) confidence interval for the proportion of male siblings in a family where one of the children has SLE.

In Problems \(9-14,\) the null and alternative hypotheses are given. Determine whether the hypothesis test is lefi-tailed, right-tailed, or two-tailed. What parameter is being tested? \(H_{0}: \sigma=7.8\) \(H_{1}: \sigma \neq 7.8\)

In Problems \(9-14,\) the null and alternative hypotheses are given. Determine whether the hypothesis test is lefi-tailed, right-tailed, or two-tailed. What parameter is being tested? \(H_{0}: p=0.2\) \(H_{1}: p<0.2\)
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