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Chapter 11: Problem 19

In October 1947, the Gallup organization surveyed 1100 adult Americans and asked, 鈥淎re you a total abstainer from, or do you on occasion consume, alcoholic beverages?鈥� Of the 1100 adults surveyed, 407 indicated that they were total abstainers. In a recent survey, the same question was asked of 1100 adult Americans and 333 indicated that they were total abstainers. Has the proportion of adult Americans who totally abstain from alcohol changed? Use the a = 0.05 level of significance.

Short Answer

Expert verified
The proportion of adult Americans who totally abstain from alcohol has changed.

Step by step solution

01

- Define the null and alternative hypotheses

Let \(p_1\) be the proportion of adult Americans who were total abstainers in 1947, and \(p_2\) be the proportion in the recent survey. The null hypothesis \(H_0\) states that there is no change in the proportions: \(H_0: p_1 = p_2\). The alternative hypothesis \(H_1\) states that there is a change in the proportions: \(H_1: p_1 eq p_2\).
02

- Calculate sample proportions

The sample proportion for 1947 (\(\hat{p_1}\)) is calculated as follows: \(\hat{p_1} = \frac{407}{1100} \approx 0.37\). The sample proportion for the recent survey (\(\hat{p_2}\)) is calculated as follows: \(\hat{p_2} = \frac{333}{1100} \approx 0.30\).
03

- Calculate the pooled sample proportion

The pooled sample proportion \(\hat{p}\) combines both sample proportions and is calculated as: \(\hat{p} = \frac{407 + 333}{1100 + 1100} = \frac{740}{2200} \approx 0.336\).
04

- Calculate the standard error

The standard error (SE) of the difference between two proportions is calculated as: \[ SE = \sqrt{ \hat{p} (1 - \hat{p}) \left( \frac{1}{n_1} + \frac{1}{n_2} \right) } = \sqrt{ 0.336 (1 - 0.336) \left( \frac{1}{1100} + \frac{1}{1100} \right) } \approx 0.022\].
05

- Calculate the test statistic

The test statistic (z) is calculated as: \[ z = \frac{ (\hat{p_1} - \hat{p_2}) }{ SE } = \frac{ 0.37 - 0.30 }{ 0.022 } \approx 3.18\].
06

- Determine the critical value

For a two-tailed test with \(\alpha = 0.05\), the critical value for z is \(\pm 1.96\).
07

- Compare test statistic and critical value

Since \(|3.18| > 1.96\), we reject the null hypothesis \(H_0\).
08

- Conclusion

Since the null hypothesis is rejected, there is sufficient evidence to conclude that the proportion of adult Americans who totally abstain from alcohol has changed.

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Key Concepts

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

proportion
In statistics, a proportion is a number between 0 and 1 that represents a part of a whole. It tells us how much of the sample has a particular characteristic. For example, in our exercise, the proportion of total abstainers in 1947 is calculated as follows: \(\frac{407}{1100} \approx 0.37\). This means about 37% of the surveyed adults in 1947 were total abstainers. Similarly, in the recent survey, the proportion is \(\frac{333}{1100} \approx 0.30\), meaning 30% of surveyed adults are abstainers. Proportions help us understand and compare different aspects of data more easily.
significance level
The significance level, denoted as \( \alpha \) , is the threshold we set to decide when to reject the null hypothesis. It represents the probability of rejecting the null hypothesis when it is actually true. In our exercise, we use a significance level of \( \alpha = 0.05 \), which means we are willing to accept a 5% risk of making a Type I error. This allows us to determine the critical value for our test. For a two-tailed test with \( \alpha = 0.05 \), the critical value for the z-score is \( \pm 1.96 \). If our test statistic exceeds this value, we reject the null hypothesis.
test statistic
The test statistic measures how far our sample statistic is from the null hypothesis value, under the assumption that the null hypothesis is true. It allows us to determine if the observed results are statistically significant. In our exercise, the test statistic (z) is calculated using the formula: \[ z = \frac{ ( \hat{p_1} - \hat{p_2} ) }{ SE } = \frac{ 0.37 - 0.30 }{ 0.022 } \approx 3.18 \]. This formula helps us standardize the difference between two sample proportions by the standard error (SE) of the difference. By comparing the test statistic to the critical value, we decide whether or not to reject the null hypothesis.
null hypothesis
The null hypothesis (\( H_0 \)) is a statement that there is no effect or no difference, and it serves as the default assumption in hypothesis testing. It represents the idea that any observed effect or difference is due to randomness or sampling error. For our exercise, the null hypothesis is: \[ H_0: p_1 = p_2 \]. This means we assume that the proportion of adult Americans who totally abstain from alcohol has not changed from 1947 to the recent survey. We test this hypothesis against the alternative hypothesis and decide whether to reject \( H_0 \) based on the test results.
alternative hypothesis
The alternative hypothesis (\( H_1 \)) is a statement that there is an effect or a difference. It represents what we aim to support through our hypothesis test. In the exercise, the alternative hypothesis is: \[ H_1: p_1 \e p_2 \]. This indicates that there is a change in the proportion of adult Americans who totally abstain from alcohol from 1947 to the recent survey. By conducting the hypothesis test, we determine if there is enough evidence to support the alternative hypothesis. If our test statistic falls beyond the critical value, we reject the null hypothesis, supporting the conclusion that the proportions have indeed changed.

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Most popular questions from this chapter

A Secchi disk is an 8 -inch-diameter weighted disk that is painted black and white and attached to a rope. The disk is lowered into water and the depth (in inches) at which it is no longer visible is recorded. The measurement is an indication of water clarity. An environmental biologist interested in determining whether the water clarity of the lake at Joliet Junior College is improving takes measurements at the same location on eight dates during the course of a year and repeats the measurements on the same dates five years later. She obtains the following results: $$ \begin{array}{lcccccccc} \text { Observation } & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} & \mathbf{5} & \mathbf{6} & \mathbf{7} & \mathbf{8} \\ \text { Date } & \mathbf{5 / 1 1} & \mathbf{6 / 7} & \mathbf{6 / 2 4} & \mathbf{7 / 8} & \mathbf{7 / 2 7} & \mathbf{8 / 3 1} & 9 / 30 & \mathbf{1 0 / 1 2} \\ \hline \begin{array}{l} \text { Initial } \\ \text { depth, } X_{i} \end{array} & 38 & 58 & 65 & 74 & 56 & 36 & 56 & 52 \\ \hline \begin{array}{l} \text { Depth five } \\ \text { years later, } Y_{i} \end{array} & 52 & 60 & 72 & 72 & 54 & 48 & 58 & 60 \\ \hline \end{array} $$ (a) Why is it important to take the measurements on the same date? (b) Does the evidence suggest that the clarity of the lake is improving at the \(\alpha=0.05\) level of significance? Note: A normal probability plot and boxplot of the data indicate that the differences are approximately normally distributed with no outliers. (c) Draw a boxplot of the differenced data. Does this visual evidence support the results obtained in part (b)?

Perform the appropriate hypothesis test. A random sample of size \(n=13\) obtained from a population that is normally distributed results in a sample mean of 45.3 and sample standard deviation of \(12.4 .\) An independent sample of size \(n=18\) obtained from a population that is normally distributed results in a sample mean of 52.1 and sample standard deviation of 14.7 . Does this constitute sufficient evidence to conclude that the population means differ at the \(\alpha=0.05\) level of significance?

Clifford Adelman, a researcher with the Department of Education, followed a cohort of students who graduated from high school in \(1992,\) monitoring the progress the students made toward completing a bachelor's degree. One aspect of his research was to compare students who first attended a community college to those who immediately attended and remained at a four-year institution. The sample standard deviation time to complete a bachelor's degree of the 268 students who transferred to a four-year school after attending community college was \(1.162 .\) The sample standard deviation time to complete a bachelor's degree of the 1145 students who immediately attended and remained at a four-year institution was \(1.015 .\) Assuming the time to earn a bachelor's degree is normally distributed, does the evidence suggest the standard deviation time to earn a bachelor's degree is different between the two groups? Use the \(\alpha=0.05\) level of significance.

To illustrate the effects of driving under the influence (DUI) of alcohol, a police officer brought a DUI simulator to a local high school. Student reaction time in an emergency was measured with unimpaired vision and also while wearing a pair of special goggles to simulate the effects of alcohol on vision. For a random sample of nine teenagers, the time (in seconds) required to bring the vehicle to a stop from a speed of 60 miles per hour was recorded. $$ \begin{array}{lccccccccc} \text { Subject } & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} & \mathbf{5} & \mathbf{6} & \mathbf{7} & \mathbf{8} & \mathbf{9} \\ \hline \text { Normal, } X_{i} & 4.47 & 4.24 & 4.58 & 4.65 & 4.31 & 4.80 & 4.55 & 5.00 & 4.79 \\ \hline \text { Impaired, } Y_{i} & 5.77 & 5.67 & 5.51 & 5.32 & 5.83 & 5.49 & 5.23 & 5.61 & 5.63 \\ \hline \end{array} $$ (a) Whether the student had unimpaired vision or wore goggles first was randomly selected. Why is this a good idea in designing the experiment? (b) Use a \(95 \%\) confidence interval to test if there is a difference in braking time with impaired vision and normal vision where the differences are computed as "impaired minus normal." Note: A normal probability plot and boxplot of the data indicate that the differences are approximately normally distributed with no outliers.

For each study, explain which statistical procedure (estimating a single proportion; estimating a single mean; hypothesis test for a single proportion; hypothesis test for a single mean; hypothesis test or estimation of two proportions, hypothesis test or estimation of two means, dependent or independent) would most likely be used for the research objective given. Assume all model requirements for conducting the appropriate procedure have been satisfied. Do adult males who take a single aspirin daily experience a lower rate of heart attacks than adult males who do not take aspirin daily?
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