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Chapter 9: Problem 27

Based on information from Harper's Index, \(r_{1}=\) 37 people out of a random sample of \(n_{1}=100\) adult Americans who did not attend college believe in extraterrestrials. However, out of a random sample of \(n_{2}=100\) adult Americans who did attend college, \(r_{2}=47\) claim that they believe in extraterrestrials. Does this indicate that the proportion of people who attended college and who believe in extraterrestrials is higher than the proportion who did not attend college? Use \(\alpha=0.01\).

Short Answer

Expert verified
There is not enough evidence to indicate a higher proportion for college attendees.

Step by step solution

01

Identify Variables and Hypotheses

We are comparing two proportions: those who did not attend college and those who did. Let \( p_1 \) be the proportion of non-college attendees who believe in extraterrestrials, and \( p_2 \) be the proportion of college attendees who believe. The null hypothesis \( H_0 \) is that \( p_1 = p_2 \), and the alternative hypothesis \( H_a \) is that \( p_1 < p_2 \).
02

Calculate Sample Proportions

Calculate the sample proportions for both groups: \( \hat{p}_1 = \frac{r_1}{n_1} = \frac{37}{100} = 0.37 \) and \( \hat{p}_2 = \frac{r_2}{n_2} = \frac{47}{100} = 0.47 \).
03

Calculate Pooled Proportion

The pooled proportion \( \hat{p} \) is used to estimate the common proportion given the null hypothesis: \[ \hat{p} = \frac{r_1 + r_2}{n_1 + n_2} = \frac{37 + 47}{100 + 100} = 0.42 \]
04

Calculate Standard Error

The standard error for the difference in proportions is calculated as: \[ SE = \sqrt{ \hat{p} (1 - \hat{p}) \left( \frac{1}{n_1} + \frac{1}{n_2} \right) } = \sqrt{ 0.42 \times 0.58 \times \left( \frac{1}{100} + \frac{1}{100} \right) } \approx 0.069 \]
05

Calculate Z-score

The Z-score is calculated using the formula: \[ Z = \frac{\hat{p}_2 - \hat{p}_1}{SE} = \frac{0.47 - 0.37}{0.069} \approx 1.45 \]
06

Determine Critical Value and Decision

At \( \alpha = 0.01 \), the critical value for a one-tailed test is approximately 2.33 from the Z-table. Since the calculated Z-score (1.45) is less than 2.33, we do not reject the null hypothesis.

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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 represented as \( H_0 \), is a statement of no effect or no difference. It serves as a baseline or default position that we aim to challenge through our test. In our exercise, the null hypothesis states that the proportion of non-college attendees who believe in extraterrestrials is the same as that of college attendees. Mathematically, this can be expressed as \( p_1 = p_2 \).
  • If the test results support the null hypothesis, it suggests that any observed difference in the data could be due to random chance.
  • Failing to reject the null hypothesis does not prove that it is true; it merely indicates insufficient evidence to support the alternative hypothesis.
The null hypothesis is crucial because it provides a statement to test against. By determining whether evidence allows us to reject this assumption, we gain insights into whether any true difference exists between the groups studied.
Alternative Hypothesis
Standing in opposition to the null is the alternative hypothesis, denoted as \( H_a \). This hypothesis proposes that there is indeed a difference or effect, specifically tailored to the context of the research question. In our scenario, the alternative hypothesis asserts that the proportion of college attendees who believe in extraterrestrials (2) is greater than those who never attended college.
  • It is expressed mathematically as \( p_1 < p_2 \), suggesting a higher belief rate among college attendees.
  • The alternative hypothesis forms the basis of a directional test. Therefore, we use a one-tailed test approach to check if this predicted direction holds true.
Validated acceptance of the alternative hypothesis implies that the differences observed are statistically significant and unlikely the result of random sampling variability.
Proportions
Proportions are a type of statistic that represents a part of a total relationship, often expressed as a percentage or a fraction. In hypothesis testing involving categorical data, proportions help to convert raw counts into interpretable measures. For example:
  • The non-college group has a proportion of \( 7_1 = \frac{37}{100} = 0.37 \).
  • Meanwhile, the college group yields a proportion of \( 7_2 = \frac{47}{100} = 0.47 \).
By comparing these proportions, we aim to deduce if there is a tangible difference between the groups. Furthermore, the pooled proportion \( \hat{p} = 0.42 \) serves as an estimate for a common proportion under the null hypothesis, facilitating further calculations such as the standard error. Understanding these foundational elements allows for accurate statistical assessments of group differences.
Z-score
A Z-score is a statistical measure that indicates the number of standard deviations a data point is from the mean of the data set. In hypothesis testing, the Z-score is utilized to determine the position of a sample mean (or proportion) concerning the null hypothesis. Here鈥檚 how it鈥檚 applied in our problem:
  • Derived as \( Z = \frac{p_2 - p_1}{SE} \), where \( SE \) is the standard error of the difference in proportions.
  • Our calculated Z-score of approximately 1.45 expresses how many standard errors the observed difference of 0.10 (0.47 - 0.37) is from zero, assuming the null hypothesis is true.
Statisticians compare this Z-score against critical values from the Z-distribution table to decide whether to support or refute the null hypothesis. With an alpha level of 0.01 and a one-tailed test, the critical value is 2.33. Since 1.45 does not exceed 2.33, the null hypothesis remains intact, hinting that the observed difference may not be statistically significant.

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Most popular questions from this chapter

Again suppose you are the auditor for a very large corporation. The revenue file contains millions of numbers in a large computer data bank (see Problem 5). You draw a random sample of \(n=228\) numbers from this file and \(r=92\) have a first nonzero digit of \(1 .\) Let \(p\) represent the population proportion of all numbers in the computer file that have a leading digit of 1 . i. Test the claim that \(p\) is more than \(0.301\). Use \(\alpha=0.01\). ii. If \(p\) is in fact larger than \(0.301\), it would seem there are too many numbers in the file with leading 1's. Could this indicate that the books have been "cooked" by artificially lowering numbers in the file? Comment from the point of view of the Internal Revenue Service. Comment from the perspective of the Federal Bureau of Investigation as it looks for "profit skimming" by unscrupulous employees. iii. Comment on the following statement: If we reject the null hypothesis at level of significance \(\alpha\), we have not proved \(H_{0}\) to be false. We can say that the probability is \(\alpha\) that we made a mistake in rejecting \(H_{0} .\) Based on the outcome of the test, would you recommend further investigation before accusing the company of fraud?

In the following data pairs, A represents the cost of living index for housing and \(B\) represents the cost of living index for groceries. The data are paired by metropolitan areas in the United States. A random sample of 36 metropolitan areas gave the following information. (Reference: Statistical Abstract of the United States, 121 st edition.) \(\begin{array}{l|lllllllll} \hline A: & 132 & 109 & 128 & 122 & 100 & 96 & 100 & 131 & 97 \\ \hline B: & 125 & 118 & 139 & 104 & 103 & 107 & 109 & 117 & 105 \\ \hline A: & 120 & 115 & 98 & 111 & 93 & 97 & 111 & 110 & 92 \\ \hline B: & 110 & 109 & 105 & 109 & 104 & 102 & 100 & 106 & 103 \\ \hline \end{array}\) \(\begin{array}{l|rrrrrrrrr} \hline A: & 85 & 109 & 123 & 115 & 107 & 96 & 108 & 104 & 128 \\ \hline B: & 98 & 102 & 100 & 95 & 93 & 98 & 93 & 90 & 108 \\ \hline \\ \hline A: & 121 & 85 & 91 & 115 & 114 & 86 & 115 & 90 & 113 \\ \hline B: & 102 & 96 & 92 & 108 & 117 & 109 & 107 & 100 & 95 \\ \hline \end{array}\) i. Let \(d\) be the random variable \(d=A-B\). Use a calculator to verify that \(\bar{d} \approx 2.472\) and \(s_{d} \approx 12.124 .\) ii. Do the data indicate that the U.S. population mean cost of living index for housing is higher than that for groceries in these areas? Use \(\alpha=0.05\).

Please provide the following information. (a) What is the level of significance? State the null and alternate hypotheses. Will you use a left-tailed, right-tailed, or two-tailed test? (b) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution. What is the value of the sample test statistic? (c) Find (or estimate) the \(P\) -value. Sketch the sampling distribution and show the area corresponding to the \(P\) -value. (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level \(\alpha\) ? (e) State your conclusion in the context of the application. Bill Alther is a zoologist who studies Anna's hummingbird (Calypte anna). (Reference: Hummingbirds, K. Long, W. Alther.) Suppose that in a remote part of the Grand Canyon, a random sample of six of these birds was caught, weighed, and released. The weights (in grams) were \(\begin{array}{llllll}3.7 & 2.9 & 3.8 & 4.2 & 4.8 & 3.1\end{array}\) The sample mean is \(\bar{x}=3.75\) grams. Let \(x\) be a random variable representing weights of Anna's hummingbirds in this part of the Grand Canyon. We assume that \(x\) has a normal distribution and \(\sigma=0.70\) gram. It is known that for the population of all Anna's hummingbirds, the mean weight is \(\mu=4.55\) grams. Do the data indicate that the mean weight of these birds in this part of the Grand Canyon is less than \(4.55\) grams? Use \(\alpha=0.01\).

Consider a hypothesis test of difference of means for two independent populations \(x_{1}\) and \(x_{2}\). What are two ways of expressing the null hypothesis?

The western United States has a number of four-lane interstate highways that cut through long tracts of wilderness. To prevent car accidents with wild animals, the highways are bordered on both sides with 12 -foot-high woven wire fences. Although the fences prevent accidents, they also disturb the winter migration pattern of many animals. To compensate for this disturbance, the highways have frequent wilderness underpasses designed for exclusive use by deer, elk, and other animals. In Colorado, there is a large group of deer that spend their summer months in a region on one side of a highway and survive the winter months in a lower region on the other side. To determine if the highway has disturbed deer migration to the winter feeding area, the following data were gathered on a random sample of 10 wilderness districts in the winter feeding area. Row \(B\) represents the average January deer count for a 5 -year period before the highway was built, and row \(A\) represents the average January deer count for a 5 -year period after the highway was built. The highway department claims that the January population has not changed. Test this claim against the claim that the January population has dropped. Use a \(5 \%\) level of significance. Units used in the table are hundreds of deer. \(\begin{array}{l|cccccccccc} \hline \text { Wilderness District } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\\ \hline \text { B: Before highway } & 10.3 & 7.2 & 12.9 & 5.8 & 17.4 & 9.9 & 20.5 & 16.2 & 18.9 & 11.6 \\ \hline \text { A: After highway } & 9.1 & 8.4 & 10.0 & 4.1 & 4.0 & 7.1 & 15.2 & 8.3 & 12.2 & 7.3 \\ \hline \end{array}\)
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