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Chapter 10: Problem 56

Suppose that, in the past, \(40 \%\) of all adults favored capital punishment. Do we have reason to believe that the proportion of adults favoring capital punishment today has increased if, in a random sample of 15 adults, 8 favor capital punishment? Use a 0.05 level of significance.

Short Answer

Expert verified
There is not enough evidence at the 0.05 significance level to support the claim that the proportion of adults favoring capital punishment has increased.

Step by step solution

01

State the hypotheses

The first step in hypothesis testing is to set up the null and alternative hypotheses. For this problem, the null hypothesis \(H0\) is that the population proportion \(p\) is 0.40, i.e. \(H0: p = 0.40\). The alternative hypothesis \(H1\) is that the population proportion \(p\) has increased, i.e. \(H1: p > 0.40\).
02

Compute the test statistic

The next step is to compute the test statistic which follows a normal distribution. The formula for the test statistic is \((\hat{p} - p0) / sqrt(p0 * (1 - p0) / n)\), where \(\hat{p}\) is the sample proportion, p0 is the proportion as per null hypothesis, and n is the sample size. Computing the sample proportion, we have \(\hat{p} = 8 / 15 = 0.53 \). The test statistic becomes \((0.53 - 0.40) / sqrt((0.40 * 0.60) / 15) = 0.90\).
03

Find the p-value

The next step is to find the p-value. It's the probability that a test statistic at least as extreme as the one observed would be obtained under the null hypothesis. Because the alternative hypothesis is that the proportion has increased, this is a one-tailed test. The p-value is found by looking up the value of the test statistic in the tables of the standard normal distribution. The p-value associated with a z-score of 0.90 is 0.1841.
04

Make the Decision

If the p-value is less than the significance level, the null hypothesis is rejected. Here, the p-value (0.1841) is more than the given significance level (0.05), therefore, 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
The null hypothesis is the pillar of hypothesis testing. It's a statement that assumes there is no effect or no difference. In our scenario, the null hypothesis ( H0 ) asserts that the proportion of adults who favor capital punishment has remained the same as in the past, at 40%. When you begin hypothesis testing, the null hypothesis is your starting point. It is what you are aiming to test or disprove.

In mathematical terms, it's presented as:
  • H0: p = 0.40
where

Alternative Hypothesis
The alternative hypothesis ( H1 or Ha ) is the counterclaim to the null hypothesis. In simple terms, if the null hypothesis suggests there's no change or effect, the alternative hypothesis suggests the opposite. In our problem, the alternative hypothesis suggests that the proportion of adults who favor capital punishment has increased.

It opposes the null hypothesis:
  • H1: p > 0.40
This claim is what we are seeking evidence for in our hypothesis test. The alternative hypothesis is crucial because it highlights the direction of the effect or change that researchers aim to detect. In this case, since we're interested to see if there's an increase, we've used a one-tailed test.
Significance Level
The significance level, denoted as \( \alpha \), is a threshold set by the researcher to judge the strength of the evidence. It's the probability of rejecting the null hypothesis when it is actually true, also known as the risk of a Type I error. A commonly used significance level is 0.05, which reflects a 5% risk of making this error.

In our exercise, the significance level is set at 0.05. This means we would only reject the null hypothesis if the probability of observing our sample data, or something more extreme, is less than 5% under the assumptions of the null hypothesis. This threshold ensures that we have a high enough level of evidence before claiming a real effect or difference.
P-value
The p-value provides a measure of the evidence against the null hypothesis presented by the sample data. It represents the probability of obtaining a test statistic at least as extreme as the one we have calculated, assuming the null hypothesis is true. It helps us decide whether to reject the null hypothesis.

In our example, the computed p-value is 0.1841 . This value indicates the likelihood of getting our observed sample proportion or more extreme, given that the true proportion is still 40% . Since our p-value (0.1841) is greater than our significance level (0.05) , we do not have enough evidence to reject the null hypothesis.

The p-value serves as a critical quantitative summary that assists in making informed conclusions from hypothesis tests.

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Most popular questions from this chapter

\( \quad \wedge\) machine is supposed to mix peanuts, hazelnuts, cashews, and pecans in the ratio 5: 2: 2: 1 . A can containing 500 of these mixed nuts was found to have 269 peanuts, 112 hazelnuts, 74 cashews, and 45 pecans. At the 0.05 level of significance, test the hypothesis that the machine is mixing the nuts in the ratio \(5: 2 ; 2: 1\).

A UCLA researcher claims that the average life span of mice can be extended by as much as 8 months when the calories in their food are reduced by approximately \(40 \%\) from the time they are weaned. The restricted diets are enriched to normal levels by vitamins and protein. Suppose that a random sample of 10 mice are fed a normal diet and live an average life span of 32.1 months with a standard deviation of 3.2 months. while a random sample of 15 mice are fod the restricted diet and live an average life span of 37.6 months with a standard deviation of 2.8 months. Test the hypothesis at the 0.05 level of significance that the average life span of mice on this restricted diet is increased by 8 months against the alternative that the increase is less than 8 months. Assume the distributions of life spans for the regular and restricted diets are approximately normal with equal variances.

A fuel oil company claims that one-fifth of the homes in a certain city are heated by oil. Do we have reason to believe that fewer than \(1 / 5\) are heated by oil if, in a random sample of 1000 homes in this city, it is found that 136 are heated by oil? Use a P-value in your conclusion.

10.2 A sociologist is concerned about the effectiveness of a training course designed to get more drivers to use seat, belts in automobiles. (a) What hypothesis is she testing if she commits a type I error by erroneously concluding that the training course is ineffective? b) What hypothesis is she testing if she commits a type II error by erroneously concluding that the training course is effective?

In Exercise 1.19 on page \(28,\) test the goodness of fit between the observed class frequencies and the corresponding expected frequencies of a normal distribution with \(p=1.8\) and \(a=0.4,\) using a 0.01 level of significance.
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