91Ó°ÊÓ

Please provide the following information for Problems. (a) What is the level of significance? State the null and alternate hypotheses. (b) Check Requirements What sampling distribution will you use? What assumptions are you making? Compute the sample test statistic and corresponding \(z\) or \(t\) value as appropriate. (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) Interpret your conclusion in the context of the application. Note: For degrees of freedom \(d . f .\) not in the Student's \(t\) table, use the closest \(d . f .\) that is smaller. In some situations, this choice of \(d . f .\) may increase the \(P\) -value a small amount and therefore produce a slightly more "conservative" answer. Art Funding: Politics 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 \(\alpha=0.05\)

Step by step solution

01

State the Hypotheses

The level of significance is set to \(\alpha = 0.05\). Let's define the null and alternative hypotheses. - Null Hypothesis \( (H_0) \): The population proportion of conservative voters inclined to spend more on the arts is equal to the proportion of moderate voters. \[ H_0: p_1 = p_2 \]- Alternative Hypothesis \( (H_a) \): The population proportion of conservative voters inclined to spend more on the arts is less than that of moderate voters. \[ H_a: p_1 < p_2 \]

02

Check Requirements and Assumptions

For comparing two proportions, we will use the normal distribution since both sample sizes seem big enough (\(n_1 = 93\), \(r_1 = 21\), \(n_2 = 83\), \(r_2 = 22\)). Assume the samples are independent and randomly sampled.

03

Calculate Test Statistic

First, calculate the sample proportions: \\[ \hat{p}_1 = \frac{r_1}{n_1} = \frac{21}{93} \approx 0.2258 \] \\[ \hat{p}_2 = \frac{r_2}{n_2} = \frac{22}{83} \approx 0.2651 \] \The pooled sample proportion is: \\[ \hat{p} = \frac{r_1 + r_2}{n_1 + n_2} = \frac{21 + 22}{93 + 83} \approx 0.244 \] \Calculate the standard error: \\[ SE = \sqrt{\hat{p}(1 - \hat{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)} \approx \sqrt{0.244(1 - 0.244)\left(\frac{1}{93} + \frac{1}{83}\right)} \approx 0.0653 \] \Calculate the z-score: \\[ z = \frac{\hat{p}_1 - \hat{p}_2}{SE} = \frac{0.2258 - 0.2651}{0.0653} \approx -0.6014 \]

04

Find the P-value

Using the z-score of -0.6014, find the P-value from the standard normal distribution table. This P-value is approximately 0.2743 for a one-tailed test. The area to the left of the z-value on the standard normal curve represents the P-value.

05

Make a Decision

Compare the P-value (0.2743) to the level of significance (0.05). Since 0.2743 > 0.05, we fail to reject the null hypothesis.

06

Conclusion

The data do not provide sufficient evidence to conclude that the population proportion of conservative voters inclined to spend more on the arts is less than the proportion of moderate voters. At the \(\alpha = 0.05\) significance level, there is not enough statistical significance.

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Key Concepts

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

Null and Alternative Hypotheses

In statistical hypothesis testing, the null and alternative hypotheses are crucial components. The **null hypothesis** (H_0) is a statement of no effect or no difference. It's what you assume is true until you have enough evidence against it. In our exercise, the null hypothesis states that the proportion of conservative voters who favor spending more on the arts is equal to the proportion of moderate voters, mathematically represented as \( H_0: p_1 = p_2 \).

The **alternative hypothesis** (H_a) is what you want to test. It indicates the presence of an effect or difference. Here, it suggests that the proportion of conservative voters in favor is less than that of moderate voters, shown as \( H_a: p_1 < p_2 \). This is a one-tailed test, as we are only interested in whether the proportion of conservative voters is less, not just different.

Proportion Comparison

In tests that involve proportions, making a comparison between two or more proportions is essential. To compare these proportions reliably, we start by calculating sample proportions for each group.

For the conservative voters, the sample proportion \( \hat{p}_1 \) is calculated as\[ \hat{p}_1 = \frac{r_1}{n_1} = \frac{21}{93} \approx 0.2258 \]For the moderate voters, the sample proportion \( \hat{p}_2 \) is given by\[ \hat{p}_2 = \frac{r_2}{n_2} = \frac{22}{83} \approx 0.2651 \]

A pooled sample proportion is often used when sample sizes differ slightly to offer a combined view of the general proportion between both groups. This is calculated as\[ \hat{p} = \frac{r_1 + r_2}{n_1 + n_2} = \frac{21 + 22}{93 + 83} \approx 0.244 \]This pooled proportion aids in assessing how much variance exists between the sample proportions.

P-value Calculation

The P-value plays a pivotal role in hypothesis testing, serving as a measure of the strength of evidence against the null hypothesis. It is the probability of observing a test statistic as extreme as the one obtained, assuming the null hypothesis is true. A small P-value indicates that the observed data is unlikely under the null hypothesis, leaning towards supporting the alternative hypothesis.

To find the P-value in a proportion comparison test, you calculate the z-score, which is a standard normal variable. For the given problem:\[ z = \frac{\hat{p}_1 - \hat{p}_2}{SE} = \frac{0.2258 - 0.2651}{0.0653} \approx -0.6014 \]The standard normal distribution table helps find the P-value corresponding to the z-score. Here, a z-score of -0.6014 yields a P-value of approximately 0.2743. This value indicates the probability that we would observe such a sample difference if the true population proportions were the same.

Significance Level

The significance level \( \alpha \) is a threshold set by the researcher to decide when to reject the null hypothesis. It represents the probability of committing a Type I error, which is rejecting a true null hypothesis. In many social science contexts, a common choice for \( \alpha \) is 0.05.

In this exercise, the significance level is set at \( \alpha = 0.05 \). When the obtained P-value is less than the significance level, there is enough evidence to reject the null hypothesis in favor of the alternative.
If the P-value is greater than \( \alpha \), as it is in our exercise (0.2743), we don't have sufficient evidence to reject the null hypothesis, implying that the data support the absence of a significant difference between the proportions.

Z-test

A Z-test is performed when sample sizes are large enough to assume that the sampling distribution of the difference in proportions is approximately normal. It's particularly useful in comparing means and proportions when the variances are known.

For our proportion comparison, because both sample sizes (93 for conservatives and 83 for moderates) are reasonably large, we utilize the Z-test. The test statistic helps determine the likelihood that the difference in sample proportions could occur under the assumption that the population proportions are equal. Calculating the z-score provides a better understanding of where the observed statistic stands relative to the hypothesized situation, thus allowing for the calculation of the associated P-value.

This Z-test confirms that with a z-score of -0.6014, the observed difference is not statistically significant at the 0.05 level, and we fail to reject the null hypothesis.