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Chapter 8: Problem 13

In a previous study conducted several years ago, a man owned on average 15 dress shirts. The standard deviation of the population is \(3 .\) A researcher wishes to see if that average has changed. He selected a random sample of 42 men and found that the average number of dress shirts that they owned was \(13.8 .\) At \(\alpha=0.05,\) is there enough evidence to support the claim that the average has changed?

Short Answer

Expert verified
Reject the null hypothesis; the average number of shirts has changed.

Step by step solution

01

Set Up the Hypotheses

Start by setting up null and alternative hypotheses. The null hypothesis (\( H_0 \)) states that the average number of shirts has not changed, so the mean \( \mu = 15 \). The alternative hypothesis (\( H_a \)) states that the average number of shirts has changed, so \( \mu eq 15 \).
02

Identify the Information Given

From the problem, we know the standard deviation of the population \( \sigma = 3 \), the sample size \( n = 42 \), the sample mean \( \bar{x} = 13.8 \), and the significance level \( \alpha = 0.05 \).
03

Calculate the Standard Error

The standard error (SE) is calculated using the formula: \( SE = \frac{\sigma}{\sqrt{n}} \). Substituting the values, \( SE = \frac{3}{\sqrt{42}} \approx 0.463 \).
04

Compute the Test Statistic

Calculate the test statistic using the formula: \( z = \frac{\bar{x} - \mu}{SE} \). Substituting the known values, \( z = \frac{13.8 - 15}{0.463} \approx -2.594 \).
05

Determine the Critical Z-Values

For a two-tailed test at \( \alpha = 0.05 \), the critical z-values are \( z = \pm 1.96 \). Any test statistic beyond these values will lead to the rejection of the null hypothesis.
06

Make the Decision

Since the calculated test statistic \( -2.594 \) is less than the lower critical value \( -1.96 \), we reject the null hypothesis.
07

Conclusion

Conclude that there is enough statistical evidence at the 0.05 significance level to support the claim that the average number of dress shirts has changed.

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Key Concepts

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

Understanding the Null Hypothesis
The null hypothesis, often symbolized as \( H_0 \), is the foundation of hypothesis testing. It represents a statement of no effect or no difference and acts as a default position. In our exercise, the null hypothesis is that the average number of dress shirts owned by men has not changed over the years. This implies the population mean, \( \mu \), remains at 15 shirts. The null hypothesis serves several important functions:
  • It provides a baseline against which we can compare actual data.
  • It simplifies our analysis by offering a clear statement to test.
  • It is presumed true until statistical evidence suggests otherwise.
In practical terms, the null hypothesis sets the stage for test statistics and helps determine whether any observed differences are statistically significant.
Grasping the Alternative Hypothesis
The alternative hypothesis, denoted as \( H_a \) or \( H_1 \), opposes the null hypothesis. It suggests that there is a change, effect, or difference. In the context of our exercise, the alternative hypothesis proposes that the average number of dress shirts owned by men is no longer 15. Instead, it assumes \( \mu eq 15 \), indicating a change has taken place.This hypothesis is crucial because:
  • It directs the research focus and objectives.
  • It determines the type of test (one-tailed or two-tailed).
  • It helps in assessing whether the observed data can actually refute the null hypothesis.
In hypothesis testing, we gather data to provide evidence in favor of the alternative hypothesis, proving it with statistical significance.
Decoding the Significance Level
The significance level, represented as \( \alpha \), is a threshold we set to determine how strongly the data must support \( H_a \) before rejecting \( H_0 \). In our exercise, the significance level is \( \alpha = 0.05 \), signifying a 5% risk of concluding that a difference exists when there is none. Here鈥檚 why significance level matters:
  • It balances the risk of Type I error (false positive), where we wrongly reject \( H_0 \).
  • The choice of \( \alpha \) often reflects the level of confidence researchers need in their findings.
  • Lower \( \alpha \) levels require stronger evidence to reject the null hypothesis, which means the findings are more robust.
When a test statistic falls beyond the critical value associated with \( \alpha \), we consider the results statistically significant and reject \( H_0 \), lending credibility to the alternative hypothesis.

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Most popular questions from this chapter

A motorist claims that the South Boro Police issue an average of 60 speeding tickets per day. These data show the number of speeding tickets issued each day for a randomly selected period of 30 days. Assume \(\sigma\) is \(13.42 .\) Is there enough evidence to reject the motorist's claim at \(\alpha=0.05 ?\) Use the \(P\) -value method. \(\begin{array}{llllllll}72 & 45 & 36 & 68 & 69 & 71 & 57 & 60 \\ 83 & 26 & 60 & 72 & 58 & 87 & 48 & 59 \\ 60 & 56 & 64 & 68 & 42 & 57 & 57 & \\ 58 & 63 & 49 & 73 & 75 & 42 & 63 & \end{array}\)

The average expenditure per student (based on average daily attendance) for a certain school year was \(\$ 10,337\) with a population standard deviation of \(\$ 1560 .\) A survey for the next school year of 150 randomly selected students resulted in a sample mean of \(\$ 10,798\). Do these results indicate that the average expenditure has changed? Choose your own level of significance.

Assume that the variables are normally or approximately normally distributed. Use the traditional method of hypothesis testing unless otherwise specified. A random sample of heights (in feet) of active volcanoes in North America, outside of Alaska, is shown. Is there sufficient evidence that the standard deviation in heights of volcanoes outside Alaska is less than the standard deviation in heights of Alaskan volcanoes, which is 2385.9 feet? Use \(\alpha=0.05 .\) \(\begin{array}{llll}10,777 & 8159 & 11,240 & 10,456 \\ 14,163 & 8363 & & \end{array}\)

Perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Find the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Assume that the population is approximately normally distributed. A survey of 15 large U.S. cities finds that the average commute time one way is 25.4 minutes. A chamber of commerce executive feels that the commute in his city is less and wants to publicize this. He randomly selects 25 commuters and finds the average is 22.1 minutes with a standard deviation of 5.3 minutes. At \(\alpha=0.10\), is he correct?

Find the critical value (or values) for the \(t\) test for each. a. \(n=15, \alpha=0.05,\) right-tailed b. \(n=23, \alpha=0.005,\) left-tailed c. \(n=28, \alpha=0.01,\) two-tailed d. \(n=17, \alpha=0.02,\) two-tailed
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