Question

For Exercises 7 through \(27,\) perform these steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. High School Graduation Rates The overall U.S. public high school graduation rate is \(73.4 \% .\) For Pennsylvania it is \(83.5 \%\) and for Idaho \(80.5 \%-\mathrm{a}\) difference of \(3 \%\). Random samples of 1200 students from each state indicated that 980 graduated in Pennsylvania and 940 graduated in Idaho. At the 0.05 level of significance, can it be concluded that there is a difference in the proportions of graduating students between the states?

Short answer

There is a significant difference in graduation rates between Pennsylvania and Idaho at the 0.05 significance level.

Step-by-step solution

State the Hypotheses

To solve this problem, we begin by stating the null and alternative hypotheses. \( H_0: p_1 = p_2 \) (There is no difference in graduation rates between Pennsylvania and Idaho). \( H_1: p_1 eq p_2 \) (There is a difference in graduation rates between Pennsylvania and Idaho). Here, \( p_1 \) represents the proportion of graduates in Pennsylvania and \( p_2 \) represents the proportion in Idaho. The claim we are testing is the alternative hypothesis \( H_1 \).

Find the Critical Values

Use the chosen significance level of \( \alpha = 0.05 \). Since this is a two-tailed test, we split \( \alpha \) into two tails, giving \( \alpha/2 = 0.025 \) for each tail. The critical z-values for a significance level of 0.05 (in a two-tailed test) are approximately -1.96 and 1.96.

Compute the Test Statistic

First, calculate the proportions: For Pennsylvania: \( \hat{p_1} = \frac{980}{1200} \approx 0.8167 \)For Idaho: \( \hat{p_2} = \frac{940}{1200} \approx 0.7833 \)Next, compute the pooled proportion: \[ \hat{p} = \frac{980 + 940}{1200 + 1200} = \frac{1920}{2400} = 0.8 \]The standard error (SE) is given by:\[ SE = \sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)} = \sqrt{0.8 \times 0.2 \times \left(\frac{1}{1200} + \frac{1}{1200}\right)} \approx 0.0163 \]Now, compute the test statistic:\[ z = \frac{\hat{p_1} - \hat{p_2}}{SE} = \frac{0.8167 - 0.7833}{0.0163} \approx 2.052 \]

Make the Decision

Compare the calculated test statistic \( z = 2.052 \) with the critical values -1.96 and 1.96. Since 2.052 falls outside the range of -1.96 to 1.96, we reject the null hypothesis \( H_0 \).

Summarize the Results

At the 0.05 level of significance, we have sufficient evidence to conclude that there is a statistically significant difference in the high school graduation rates between Pennsylvania and Idaho. Therefore, the claim that the proportions of graduation rates are different is supported by the test.

Key concepts

Null Hypothesis

In hypothesis testing, the null hypothesis, denoted as \( H_0 \), is a statement that there is no effect or difference in the context of the experiment. It serves as the default claim that we test against. For our exercise, the null hypothesis is formulated as \( H_0: p_1 = p_2 \), where \( p_1 \) is the proportion of high school graduates in Pennsylvania and \( p_2 \) is the proportion in Idaho.

The null hypothesis essentially states that there is no difference in the graduation rates between these two states. It's crucial because it provides a baseline or starting point for the statistical test. The assumption here is that any observed difference in the sample is due to random chance rather than a real difference in the populations.

Rejecting the null hypothesis means we have found enough evidence to support that there is a statistically significant difference. On the other hand, failing to reject it implies that the data did not show a significant effect, and any observed difference could merely be due to random sampling variability.

Alternative Hypothesis

The alternative hypothesis, denoted as \( H_1 \), is the statement we want to test for. It is the opposite of the null hypothesis and suggests that there is a real effect or difference. In the context of our current problem, the alternative hypothesis is \( H_1: p_1 eq p_2 \).

This claim indicates that the graduation rates for Pennsylvania and Idaho are not equal, suggesting a difference exists between the two. The alternative hypothesis is crucial because it represents the scenario that the research is attempting to provide evidence for.

The hypothesis is tested by performing a statistical test that calculates a test statistic to compare against a critical value. If our test statistic falls in the critical region, we have enough evidence to reject the null hypothesis in favor of the alternative. Remember, in hypothesis testing, the alternative hypothesis is what the researcher would likely want to demonstrate, or show that a change or difference is present.

Critical Value

Critical values are the threshold boundaries that determine when we should reject the null hypothesis in favor of the alternative hypothesis. These values are derived from the statistical distribution of the test being used and the chosen significance level, often denoted as \( \alpha \).

In our two-tailed test example, using a significance level \( \alpha = 0.05 \), the critical values for a standard normal distribution (z-test) are approximately -1.96 and 1.96. These boundaries divide the distribution into regions such that if the test statistic lies beyond these critical values, we reject the null hypothesis.

Let's break this down further:

• - **Left critical value**: This refers to the lower end cut-off of the tails of the distribution (-1.96 in this case).
• - **Right critical value**: This refers to the upper end cut-off of the tails (1.96 here).

If the test statistic falls outside this range, it suggests that the observed data are statistically significant, thus leading us to reject the null hypothesis.

Test Statistic

The test statistic is a standardized value that results from the process of converting your sample data into a form that can be compared against a known distribution. It's calculated using your sample data and helps you determine if your sample results are typical or unusual under the null hypothesis.

For our exercise, the test statistic is computed as a z-score using the formula:\[ z = \frac{\hat{p_1} - \hat{p_2}}{SE} \]where \( \hat{p_1} \) and \( \hat{p_2} \) are the sample proportions, and SE is the standard error, calculated using the pooled proportions.

In this case, the calculated z-value was 2.052. To determine the significance, we compare this z-score to the critical values of -1.96 and 1.96 for a two-tailed test at the 0.05 level of significance.

Since 2.052 exceeds these critical boundaries, we can say that the test statistic leads us to reject the null hypothesis. The test statistic is key in hypothesis testing as it helps us measure how far away our sample statistic is from what we would expect under the null hypothesis, providing insight into the likelihood of the observed sample if the null hypothesis were true.