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Chapter 8: Problem 41

A plan for an executive traveler's club has been developed by an airline on the premise that \(5 \%\) of its current customers would qualify for membership. A random sample of 500 customers yielded 40 who would qualify. a. Using this data, test at level .01 the null hypothesis that the company's premise is correct against the alternative that it is not correct. b. What is the probability that when the test of part (a) is used, the company's premise will be judged correct when in fact \(10 \%\) of all current customers qualify?

Short Answer

Expert verified
a) Reject the null hypothesis; b) Probability of Type II error is negligible.

Step by step solution

01

Define Hypotheses

Formally state the null hypothesis \( H_0 \) and the alternative hypothesis \( H_1 \). Define \( p \) as the proportion of current customers who qualify for membership. Thus, \( H_0: p = 0.05 \) and \( H_1: p eq 0.05 \).
02

Calculate Sample Proportion

In the sample of 500 customers, 40 qualify for the membership. Thus, the sample proportion \( \hat{p} \) is \( \hat{p} = \frac{40}{500} = 0.08 \).
03

Determine Standard Deviation

Use the formula for the standard deviation of the sampling distribution of a proportion, \( \sigma_{\hat{p}} = \sqrt{ \frac{p(1-p)}{n} } \), where \( n = 500 \) and \( p = 0.05 \). Thus, \( \sigma_{\hat{p}} = \sqrt{ \frac{0.05 \times 0.95}{500} } = 0.0097 \).
04

Calculate Z-Score

Determine the Z-score using the formula \( Z = \frac{\hat{p} - p}{\sigma_{\hat{p}}} = \frac{0.08 - 0.05}{0.0097} = 3.09 \).
05

Determine Critical Value

For a two-tailed test at the \( \alpha = 0.01 \) significance level, the critical Z-values are \( \pm 2.576 \).
06

Make Decision (a)

Since the calculated Z-score of 3.09 exceeds the critical value of 2.576, we reject the null hypothesis \( H_0 \). This suggests that there is significant evidence at the 0.01 level to conclude that the premise is incorrect.
07

Calculate Type II Error Probability (b)

If the true proportion \( p = 0.10 \), calculate \( \beta \), the probability of Type II error. Using the same standard deviation, determine the new Z-score comparing this true proportion: \( Z = \frac{0.10 - 0.05}{0.0097} = 5.15 \). Determine the probability from the standard normal table: for Z-score corresponding to \( 2.576 \) , the lower bound is 0.005 (lowest tail). For \( Z = 5.15 \), the upper bound is practically negligible as it lies far in the tail, solidifying that \( \beta \) is exceedingly small.

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Key Concepts

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

Null Hypothesis
In hypothesis testing, the null hypothesis, often denoted as \( H_0 \), is a fundamental concept. It represents a statement of no effect or no difference. Essentially, it is the hypothesis that there is no change from the status quo, or that the observed results are due to random chance.

In our example, the airline's premise is that 5% of its customers qualify for a traveler's club. The null hypothesis posits that the true proportion \( p \) of qualifying customers is indeed 0.05. Formally, this is expressed as \( H_0: p = 0.05 \).

The null hypothesis serves as a starting point for statistical testing. It is usually tested against an alternative hypothesis. By applying statistical methods, we can challenge or fail to reject the null hypothesis based on sample data.
Alternative Hypothesis
The alternative hypothesis, denoted as \( H_1 \), stands opposite to the null hypothesis. It suggests that there exists a significant effect or a distinct difference.

When conducting the test in our scenario, the alternative hypothesis proposes that the proportion \( p \) of current customers who qualify for membership is different from 0.05. This is expressed as \( H_1: p eq 0.05 \).

The alternative hypothesis is what researchers typically aim to support. With the calculated Z-score of 3.09 in our example, which exceeds the critical value of 2.576, the evidence suggests that we should reject the null hypothesis in favor of the alternative. The data indicates that the proportion of qualifying customers is significantly different from the airline's initial claim.
Type II Error
Type II error, also known as a false negative, occurs when the null hypothesis is not rejected despite being false. It represents a missed opportunity to identify a true effect.

In our exercise, after rejecting the null hypothesis, we also calculated the probability of committing a Type II error (denoted as \( \beta \)) given that in reality, 10% of customers qualify for the club. A Type II error would mean concluding that no significant change from 5% exists when, in fact, 10% qualify.

Calculate the new Z-score assuming \( p = 0.10 \), giving \( Z = 5.15 \). The probability of \( Z \) falling within the bounds at a higher threshold than our calculated 3.09 is very low, indicating an exceedingly small chance of committing a Type II error in this scenario. This low probability assures us that if there is indeed a 10% qualification rate, it's unlikely we'd mistakenly uphold the original claim.

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Most popular questions from this chapter

To obtain information on the corrosion-resistance properties of a certain type of steel conduit, 45 specimens are buried in soil for a 2-year period. The maximum penetration (in mils) for each specimen is then measured, yielding a sample average penetration of \(\bar{x}=52.7\) and a sample standard deviation of \(s=4.8\). The conduits were manufactured with the specification that true average penetration be at most 50 mils. They will be used unless it can be demonstrated conclusively that the specification has not been met. What would you conclude?

Pairs of \(P\)-values and significance levels, \(\alpha\), are given. For each pair, state whether the observed \(P\)-value would lead to rejection of \(H_{0}\) at the given significance level. a. \(P\)-value \(=.084, \alpha=.05\) b. \(P\)-value \(=.003, \alpha=.001\) c. \(P\)-value \(=.498, \alpha=.05\) d. \(P\)-value \(-.084, \alpha-.10\) e. \(P\)-value \(=.039, \alpha=.01\) f. \(P\)-value \(=.218, \alpha=.10\)

The times of first sprinkler activation for a series of tests with fire prevention sprinkler systems using an aqueous film-forming foam were (in sec) \(\begin{array}{lllllllllllll}27 & 41 & 22 & 27 & 23 & 35 & 30 & 33 & 24 & 27 & 28 & 22 & 24\end{array}\) (see "Use of AFFF in Sprinkler Systems," Fire Technology, 1976: 5). The system has been designed so that true average activation time is at most \(25 \mathrm{sec}\) under such conditions. Does the data strongly contradict the validity of this design specification? Test the relevant hypotheses at significance level \(.05\) using the \(P\)-value approach.

For the following pairs of assertions, indicate which do not comply with our rules for setting up hypotheses and why (the subscripts 1 and 2 differentiate between quantities for two different populations or samples): a. \(H_{0}: \mu=100, H_{\mathrm{a}}: \mu>100\) b. \(H_{0}: \sigma=20, H_{\mathrm{a}}: \sigma \leq 20\) c. \(H_{0}: p \neq .25, H_{\mathrm{a}}: p=.25\) d. \(H_{0}: \mu_{1}-\mu_{2}=25, H_{\mathrm{a}}: \mu_{1}-\mu_{2}>100\) e. \(H_{0}: S_{1}^{2}=S_{2}^{2}, H_{\mathrm{a}}: S_{1}^{2} \neq S_{2}^{2}\) f. \(H_{0}: \mu=120, H_{\mathrm{a}}: \mu=150\) g. \(H_{0}: \sigma_{1} / \sigma_{2}=1, H_{\mathrm{a}}: \sigma_{1} / \sigma_{2} \neq 1\) h. \(H_{0}: p_{1}-p_{2}=-.1, H_{\mathrm{a}}: p_{1}-p_{2}<-.1\)

After a period of apprenticeship, an organization gives an exam that must be passed to be eligible for membership. Let \(p=P\) (randomly chosen apprentice passes). The organization wishes an exam that most but not all should be able to pass, so it decides that \(p=.90\) is desirable. For a particular exam, the relevant hypotheses are \(H_{0}: p=.90\) versus the alternative \(H_{\mathrm{a}}: p \neq 90\). Suppose ten people take the exam, and let \(X=\) the number who pass. a. Does the lower-tailed region \(\\{0,1, \ldots, 5\\}\) specify a level \(.01\) test? b. Show that even though \(H_{\mathrm{a}}\) is two-sided, no two-tailed test is a level \(.01\) test. c. Sketch a graph of \(\beta\left(p^{\prime}\right)\) as a function of \(p^{\prime}\) for this test. Is this desirable?
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