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Would you favor spending more federal tax money on the arts? This question was asked by a research group on behalf of The National Institute (Reference: Painting by Numbers, J. Wypijewski, University of California Press). Of a random sample of \(n_{1}=93\) politically conservative voters, \(r_{1}=21\) responded yes. Another random sample of \(n_{2}=83\) politically moderate voters showed that \(r_{2}=22\) responded yes. Does this information indicate that the population proportion of conservative voters inclined to spend more federal tax money on funding the arts is less than the proportion of moderate voters so inclined? Use \(\alpha=0.05\).

Short Answer

Expert verified
The data does not indicate that conservatives are less inclined than moderates.

Step by step solution

01

State the Hypotheses

Let's define our null and alternative hypotheses. The null hypothesis is that the true proportion of conservative voters who would favor spending more on the arts is equal to the proportion of moderate voters, i.e., \( H_0: p_1 = p_2 \). The alternative hypothesis is that the proportion of conservative voters is less than that of moderate voters, i.e., \( H_a: p_1 < p_2 \), where \( p_1 \) is the proportion of conservatives, and \( p_2 \) is the proportion of moderates.
02

Calculate Sample Proportions

Calculate the sample proportions for each group. For conservative voters, \( \hat{p}_1 = \frac{r_1}{n_1} = \frac{21}{93} \approx 0.2258 \). For moderate voters, \( \hat{p}_2 = \frac{r_2}{n_2} = \frac{22}{83} \approx 0.2651 \).
03

Compute the Pooled Sample Proportion

The pooled sample proportion is calculated by combining the successes and sample sizes from both groups: \( \hat{p} = \frac{r_1 + r_2}{n_1 + n_2} = \frac{21 + 22}{93 + 83} = \frac{43}{176} \approx 0.2443 \).
04

Calculate the Test Statistic

We will use the formula for the test statistic for two proportions: \[ z = \frac{\hat{p}_1 - \hat{p}_2}{\sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}} \]Substituting the values: \[ z = \frac{0.2258 - 0.2651}{\sqrt{0.2443 \times (1-0.2443) \left(\frac{1}{93} + \frac{1}{83}\right)}} \approx -0.826 \].
05

Determine the p-Value

Using a standard normal distribution table or calculator, find the p-value for the calculated z-value of -0.826. This gives us a p-value of approximately 0.204. This p-value represents the probability of observing a test statistic as extreme as, or more extreme than, the observed statistic, under the null hypothesis.
06

Make a Decision

Compare the p-value to the significance level \( \alpha = 0.05 \). Since 0.204 > 0.05, we fail to reject the null hypothesis. This means there is not enough evidence to conclude that the proportion of conservative voters who favor spending more on the arts is less than that of moderate voters.

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Key Concepts

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

Population Proportion
In hypothesis testing, the population proportion is a key concept that helps us understand what fraction of the population holds a particular characteristic. It's often denoted by the symbol \( p \). When conducting such tests, we are typically interested in comparing the population proportion from different groups or samples.

In our exercise, we look into two different populations: politically conservative voters and politically moderate voters. We are interested in the fraction of each group that favors spending more on the arts. The sample proportion \( \hat{p} \) is derived from the sample data and is an estimate of the true population proportion. In the exercise, \( \hat{p}_1 = 0.2258 \) for conservatives and \( \hat{p}_2 = 0.2651 \) for moderates.

Understanding these proportions allows us to form hypotheses and conduct further statistical analyses to determine if there is a significant difference between them.
Test Statistic
The test statistic is a standardized value that helps us make a decision about our hypotheses. It measures how far our sample data is from the null hypothesis. In our case, the test statistic is derived from the formula used for comparing two proportions.

To compute this, we first calculate the pooled sample proportion, \( \hat{p} \), which combines data from both samples to give a more comprehensive view. We then use this pooled proportion to compute our test statistic \( z \) using the formula:
  • \[ z = \frac{\hat{p}_1 - \hat{p}_2}{\sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}} \]

This z-value signifies how many standard deviations our observed sample difference in proportions is away from the expected difference under the null hypothesis. In the exercise, our resulting test statistic is approximately \( -0.826 \).
P-Value
The p-value is a crucial component of hypothesis testing. It tells us the probability of observing a test statistic as extreme as, or more extreme than, our calculated statistic, given that the null hypothesis is true. Essentially, it's a measure of the strength of the evidence against the null hypothesis.

In the exercise, using a standard normal distribution table, we find the p-value corresponding to our test statistic. With a z-value of approximately \( -0.826 \), the p-value comes out to be about 0.204. The larger the p-value, the stronger the evidence supporting the null hypothesis. Since 0.204 is quite large, it suggests that there is a high probability of observing our sample results under the assumption that the null hypothesis is true.

This step is critical when deciding on the validity of our hypothesis test.
Significance Level
The significance level, denoted by \( \alpha \), is the threshold at which we decide whether to accept or reject the null hypothesis. It's predefined before conducting the test and represents our tolerance for making a Type I error (rejecting a true null hypothesis).

Generally, a common choice for \( \alpha \) is 0.05, as you've seen in this exercise. This means we're okay with a 5% chance of incorrectly rejecting the null hypothesis.

To determine the outcome of the hypothesis test, we compare the p-value to this significance level. If the p-value is less than or equal to \( \alpha \), we reject the null hypothesis; otherwise, we fail to reject it. In the given exercise, our p-value is 0.204 which is greater than 0.05, leading us to "fail to reject" the null hypothesis.

Understanding the significance level is essential as it directly influences the reliability of our test conclusions.

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Most popular questions from this chapter

Education influences attitude and lifestyle. Differences in education are a big factor in the "generation gap." Is the younger generation really better educated? Large surveys of people age 65 and older were taken in \(n_{1}=32\) U.S. cities. The sample mean for these cities showed that \(\bar{x}_{1}=15.2 \%\) of the older adults had attended college. Large surveys of young adults (age \(25-34\) ) were taken in \(n_{2}=35\) U.S. cities. The sample mean for these cities showed that \(\bar{x}_{2}=19.7 \%\) of the young adults had attended college. From previous studies, it is known that \(\sigma_{1}=7.2 \%\) and \(\sigma_{2}=5.2 \%\) (Reference: American Generations, S. Mitchell). Does this information indicate that the population mean percentage of young adults who attended college is higher? Use \(\alpha=0.05\).

REM (rapid eye movement) sleep is sleep during which most dreams occur. Each night a person has both REM and non-REM sleep. However, it is thought that children have more REM sleep than adults (Reference: Secrets of Sleep by Dr. A. Borbely). Assume that REM sleep time is normally distributed for both children and adults. A random sample of \(n_{1}=10\) children ( 9 years old) showed that they had an average REM sleep time of \(\bar{x}_{1}=2.8\) hours per night. From previous studies, it is known that \(\sigma_{1}=0.5\) hour. Another random sample of \(n_{2}=10\) adults showed that they had an average REM sleep time of \(\bar{x}_{2}=2.1\) hours per night. Previous studies show that \(\sigma_{2}=\) \(0.7\) hour. Do these data indicate that, on average, children tend to have more REM sleep than adults? Use a \(1 \%\) level of significance.

The U.S. Department of Transportation, National Highway Traffic Safety Administration, reported that \(77 \%\) of all fatally injured automobile drivers were intoxicated. A random sample of 27 records of automobile driver fatalities in Kit Carson County, Colorado, showed that 15 involved an intoxicated driver. Do these data indicate that the population proportion of driver fatalities related to alcohol is less than \(77 \%\) in Kit Carson County? Use \(\alpha=0.01\).

The following is based on information taken from Winter Wind Studies in Rocky Mountain National Park, by D. E. Glidden (Rocky Mountain Nature Association). At five weather stations on Trail Ridge Road in Rocky Mountain National Park, the peak wind gusts (in miles per hour) for January and April are recorded below. \(\begin{array}{l|ccccc} \hline \text { Weather Station } & 1 & 2 & 3 & 4 & 5 \\ \hline \text { January } & 139 & 122 & 126 & 64 & 78 \\ \hline \text { April } & 104 & 113 & 100 & 88 & 61 \\ \hline \end{array}\) Does this information indicate that the peak wind gusts are higher in January than in April? Use \(\alpha=0.01\).

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