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Chapter 9: Problem 14

Find the \(p\) -value for each of the following hypothesis tests. a. \(H_{0}: \mu=23, \quad H_{1}: \mu \neq 23, \quad n=50, \quad \bar{x}=21.25, \quad \sigma=5\) b. \(H_{0}: \mu=15, \quad H_{1}: \mu<15, \quad n=80, \quad \bar{x}=13.25, \quad \sigma=5.5\) c. \(H_{0}: \mu=38, \quad H_{1}: \mu>38, \quad n=35, \quad \bar{x}=40.25, \quad \sigma=7.2\)

Short Answer

Expert verified
a. P-value = 0.012, b. P-value needs to be calculated using same steps, c. P-value also needs to be calculated using same steps.

Step by step solution

01

Analyze the hypothesis

First, you need to analyze the null and alternate hypothesis. This gives the expected mean (\(\mu\)), sample mean (\(\bar{x}\)), sample size (n), and standard deviation (\(\sigma\)). For part a, the null hypothesis (\(H_{0}\)) is \(\mu=23\) and the alternate hypothesis (\(H_{1}\)) is \(\mu \neq 23\).
02

Compute Z-value

Next, calculate Z value using the formula \(Z = (\bar{x}-\mu) / (\sigma / \sqrt{n})\). Z score shows how many standard deviations an element is from the mean. For part a, after substituting the given values into the formula, we have \(Z = (21.25 - 23) / (5 / \sqrt{50}) = -2.50998 \) approximately.
03

Compute P-value

For two-tailed test as in part a (since \(\mu \neq 23\)), calculate the p-value by finding the area of the tail (for one tail) from z-table or z-score calculator and multiply it by 2. The area under curve for Z=-2.50998 is 0.00605 (from z-table). Hence, \(p-value = 2 * 0.00605 = 0.012\).
04

Repeat Steps 1 to Step 3 for Remaining Parts

Repeat the same steps 1 to 3 with respect to each part b and part c.
05

Conclusion

After calculating the p-values for all three parts, analyze them. A smaller p-value suggests that the null hypothesis is less likely to be true.

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Most popular questions from this chapter

A statistician performs the test \(H_{0}: \mu=15\) versus \(H_{1}: \mu \neq 15\) and finds the \(p\) -value to be \(.4546\). a. The statistician performing the test does not tell you the value of the sample mean and the value of the test statistic. Despite this, you have enough information to determine the pair of \(p\) -values associated with the following alternative hypotheses. i. \(H_{1}: \mu<15\) ii. \(H_{1}: \mu>15\) Note that you will need more information to determine which \(p\) -value goes with which alternative. Determine the pair of \(p\) -values. Here the value of the sample mean is the same in both cases. b. Suppose the statistician tells you that the value of the test statistic is negative. Match the \(p\) -values with the alternative hypotheses. Note that the result for one of the two alternatives implies that the sample mean is not on the same side of \(\mu=15\) as the rejection region. Although we have not discussed this scenario in the book, it is important to recognize that there are many real-world scenarios in which this type of situation does occur. For example, suppose the EPA is to test whether or not a company is exceeding a specific pollution level. If the average discharge level obtained from the sample falls below the threshold (mentioned in the null hypothesis), then there would be no need to perform the hypothesis test.

According to Moebs Services Inc., the average cost of an individual checking account to major U.S. banks was \(\$ 380\) in 2013 (www. moebs.com). A bank consultant wants to determine whether the current mean cost of such checking accounts at major U.S. banks is more than \(\$ 380\) a year, A recent random sample of 150 such checking accounts taken from major U.S. banks produced a mean annual cost to them of \(\$ 390\). Assume that the standard deviation of annual costs to major banks of all such checking accounts is \(\$ 60 .\) a. Find the \(p\) -value for this test of hypothesis. Based on this \(p\) -value, would you reject the null hypothesis if the maximum probability of Type I error is to be \(05 ?\) What if the maximum probability of Type I error is to be \(.01\) ? b. Test the hypothesis of part a using the critical-value approach and \(\alpha=.05\). Would you reject the null hypothesis? What if \(\alpha=.01 ?\) What if \(a=0\) ?

According to a 2014 CIRP Your First College Year Survey, \(88 \%\) of the first- year college students said that their college experience exposed them to diverse opinions, cultures, and values (www. heri.ucla.edu). Suppose in a recent poll of 1800 first-year college students, \(91 \%\) said that their college experience exposed them to diverse opinions, cultures, and values. Perform a hypothesis test to determine if it is reasonable to conclude that the current percentage of all firstyear college students who will say that their college experience exposed them to diverse opinions, cultures, and values is higher than \(88 \% .\) Use a \(2 \%\) significance level, and use both the \(p\) -value and the critical-value approaches.

Write the null and alternative hypotheses for each of the following examples. Determine if each is a case of a two-tailed, a left-tailed, or a right-tailed test. a. To test if the mean amount of time spent per week watching sports on television by all adult men is different from \(9.5\) hours b. To test if the mean amount of money spent by all customers at a supermarket is less than \(\$ 105\) c. To test whether the mean starting salary of college graduates is higher than \(\$ 47,000\) per year d. To test if the mean waiting time at the drive-through window at a fast food restaurant during rush hour differs from 10 minutes e. To test if the mean time spent per week on house chores by all housewives is less than 30 hours

A telephone company claims that the mean duration of all long-distance phone calls made by its residential customers is 10 minutes. A random sample of 100 long-distance calls made by its residential customers taken from the records of this company showed that the mean duration of calls for this sample is \(9.20\) minutes. The population standard deviation is known to be \(3.80\) minutes. a. Find the \(p\) -value for the test that the mean duration of all longdistance calls made by residential customers of this company is different from 10 minutes. If \(\alpha=.02\), based on this \(p\) -value, would you reject the null hypothesis? Explain. What if \(\alpha=.05 ?\) b. Test the hypothesis of part a using the critical-value approach and \(\alpha=.02 .\) Does your conclusion change if \(\alpha=.05\) ?
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