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Chapter 9: Problem 16

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
The p-values need to be calculated and interpreted for each part independently. The calculated z-scores need to be looked up in the standard normal probability table to find the corresponding p-values. The interpretation of the p-value determines whether to reject or fail to reject the null hypothesis.

Step by step solution

01

Calculate z-score for part (a)

This is a two-tailed test because \(H_1: \mu \neq 23\). The z-score formula is \(Z = \frac{{\bar{x}-\mu}}{{\sigma / \sqrt{n}}}\). Substituting the given values we would get: \(Z = \frac{{21.25-23}}{{5 / \sqrt{50}}}\).
02

Find p-value for part (a)

Refer to the z-table to find the probability corresponding to the calculated z-score. As it's a two-tailed test, double the calculated probability to get the p-value.
03

Calculate z-score for part (b)

This is a left-tailed test because \(H_1: \mu < 15\). Using the z-score formula, substituting the given values we get: \(Z = \frac{{13.25-15}}{{5.5 / \sqrt{80}}}\).
04

Find p-value for part (b)

This is a left-tailed test, so the p-value is the probability corresponding to the calculated z-score in the z-table.
05

Calculate z-score for part (c)

This is a right-tailed test because \(H_1: \mu > 38\). Using the z-score formula, substituting the given values we get: \(Z = \frac{{40.25-38}}{{7.2 / \sqrt{35}}}\).
06

Find p-value for part (c)

This is a right-tailed test, so the p-value is \(1 - \) the probability corresponding to the calculated z-score in the z-table.
07

Interpret the p-value

A small p-value (typically 鈮� 0.05) indicates strong evidence against the null hypothesis, so you reject the null hypothesis. A large p-value (> 0.05) indicates weak evidence against the null hypothesis, so you fail to reject the null hypothesis.

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Most popular questions from this chapter

The president of a university claims that the mean time spent partying by all students at this university is not more than 7 hours per week. A random sample of 40 students taken from this university showed that they spent an average of \(9.50\) hours partying the previous week with a standard deviation of \(2.3\) hours. Test at a \(2.5 \%\) significance level whether the president's claim is true. Explain your conclusion in words.

For each of the following significance levels, what is the probability of making a Type I error? \(\begin{array}{lll}\text { a. } \alpha=.025 & \text { b. } \alpha=.05 & \text { c. } \alpha=.01\end{array}\)

According to a book published in \(2011,45 \%\) of the undergraduate students in the United States show almost no gain in learning in their first 2 years of college (Richard Arum et al., Academically Adrift, University of Chicago Press, Chicago, 2011 ). A recent sample of 1500 undergraduate students showed that this percentage is \(38 \%\). Can you reject the null hypothesis at a \(1 \%\) significance level in favor of the alternative that the percentage of undergraduate students in the United States who show almost no gain in learning in their first 2 years of college is currently lower than \(45 \%\) ? Use both the \(p\) -value and the critical-value approaches.

Explain which of the following is a two-tailed test, a left-tailed test, or a right-tailed test. a. \(H_{0}: \mu=12, \quad H_{1}: \mu<12 \quad\) b. \(H_{0}: \mu \leq 85, \quad H_{1}: \mu>85\) c. \(H_{0}: \mu=33, \quad H_{1}: \mu \neq 33\) Show the rejection and nonrejection regions for each of these cases by drawing a sampling distribution curve for the sample mean, assuming that it is normally distributed.

A consumer advocacy group suspects that a local supermarket's 10 -ounce packages of cheddar cheese actually weigh less than 10 ounces. The group took a random sample of 20 such packages and found that the mean weight for the sample was \(9.955\) ounces. The population follows a normal distribution with the population standard deviation of \(.15\) ounce. a. Find the \(p\) -value for the test of hypothesis with the alternative hypothesis that the mean weight of all such packages is less than 10 ounces. Will you reject the null hypothesis at \(\alpha=.01\) ? b. Test the hypothesis of part a using the critical-value approach and \(\alpha=.01\).
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