Question

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. The overall U.S. public high school graduation rate is \(73.4 \% .\) For Pennsylvania it is \(83.5 \%\) and for Idaho \(80.5 \%-\) 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

Yes, 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

Define the null hypothesis, denoted as \(H_0\), and the alternative hypothesis, denoted as \(H_1\). In this context, we'll use: - \(H_0: p_1 = p_2\), meaning the proportion of students graduating in Pennsylvania is the same as in Idaho.- \(H_1: p_1 eq p_2\), meaning there is a difference in the proportion of students graduating between Pennsylvania and Idaho.

Identify the Claim

The claim made by the difference in graduation rates is that there is a difference between the proportions of students graduating in Pennsylvania and Idaho. This aligns with the alternative hypothesis \(H_1: p_1 eq p_2\).

Find the Critical Value(s)

As it is a two-tailed test with a significance level of 0.05, the critical z-values are found using standard normal distribution tables. These values are approximately \(z = \pm 1.96\).

Compute the Test Value

Calculate the test statistic using the formula for comparing two proportions:\[z = \frac{(\hat{p}_1 - \hat{p}_2)}{\sqrt{\hat{p}(1-\hat{p})(\frac{1}{n_1}+\frac{1}{n_2})}}\]where \(\hat{p}_1 = \frac{980}{1200}\), \(\hat{p}_2 = \frac{940}{1200}\), and \(\hat{p} = \frac{980+940}{1200+1200}\).Now compute:- \(\hat{p}_1 = 0.8167\)- \(\hat{p}_2 = 0.7833\)- \(\hat{p} = 0.80\)- Substitute these into the formula for \(z\) to get the test statistic.

Calculate the Test Statistic

Substitute and calculate:\[z = \frac{0.8167 - 0.7833}{\sqrt{0.8 \times 0.2 \times (\frac{1}{1200} + \frac{1}{1200})}} \approx \frac{0.0334}{0.0163} \approx 2.05\]

Make the Decision

Compare the computed test statistic (\(z = 2.05\)) with the critical value (\(\pm 1.96\)). Since \(2.05\) is greater than \(1.96\), the test statistic falls into the rejection region. Therefore, reject the null hypothesis \(H_0\).

Summarize the Results

At the 0.05 level of significance, we have enough evidence to reject the null hypothesis. Thus, we conclude that there is a significant difference in the proportion of students graduating between Pennsylvania and Idaho.

Key concepts

Critical Value

In hypothesis testing, the critical value acts as a crucial threshold. It determines whether the test statistic's result supports rejecting the null hypothesis or not. When you perform a test, you first decide on the level of significance, usually represented by \(\alpha\), to gauge the risk you're willing to take of rejecting a true null hypothesis. In this case, \(\alpha = 0.05\).
The critical value is based on the type of test. For a two-tailed test, like this one, the critical values cut off the extreme 5% of the distribution, both 2.5% in each tail. Using the standard normal distribution table, the critical values here are approximately \(+1.96\) and \(-1.96\). If your computed test statistic falls beyond these critical values, you reject the null hypothesis.

• Critical values indicate cut-off points for acceptance or rejection of hypotheses.
• Calculated using the significance level and the type of statistical test.
• In a two-tailed test with \(\alpha = 0.05\), it is typically \(\pm 1.96\).

Test Statistic

The test statistic is a value derived from your sample data during hypothesis testing. It directly influences the decision-making process. You calculate it based on the difference between sample statistics. In this scenario, it compares the proportion of high school graduates from two states.
For this exercise, we use the formula primarily for the difference in proportions:\[z = \frac{(\hat{p}_1 - \hat{p}_2)}{\sqrt{\hat{p}(1-\hat{p})(\frac{1}{n_1}+\frac{1}{n_2})}}\]Where:

• \(\hat{p}_1\) and \(\hat{p}_2\) are sample proportions for Pennsylvania and Idaho respectively,
• \(\hat{p}\) is the pooled proportion across both samples.

By substituting the values, we already computed, the test statistic \(z\) ended up as approximately \(2.05\). Since \(z\) lies outside the range of critical values, it suggests a significant difference.

Null Hypothesis

The null hypothesis is a foundational concept in hypothesis testing. It represents a statement or proposition that there is no effect or no difference. In its essence, it's the default or "no change" assumption.
In this context, the null hypothesis (\(H_0\)) asserts that the graduation rates between the two states, Pennsylvania and Idaho, are the same. Mathematically, this means: \[H_0: p_1 = p_2\]Here, \(p_1\) and \(p_2\) represent the proportions of graduates from Pennsylvania and Idaho, respectively.

• The null hypothesis is generally the claim you try to find evidence against.
• A failure to reject \(H_0\) implies no significant difference was found.
• Remember, rejecting it does not confirm the alternative hypothesis completely but shows a significant sample-based evidence against it.

Alternative Hypothesis

The alternative hypothesis challenges the null hypothesis, suggesting that there is indeed an effect or a difference. In many situations, this is the research hypothesis we're driven to support with statistical evidence.
For our problem, the alternative hypothesis (\(H_1\)) states there is a difference in the proportions of graduates between the states:\[H_1: p_1 eq p_2\]This hypothesis aligns with our initial claim and is interested in proving the disparity between graduation rates in Pennsylvania and Idaho.

• The alternative hypothesis is what you might suspect is true before collecting data.
• Success in rejecting \(H_0\) usually provides enough evidence to support \(H_1\).
• Here, it indicated that the graduation rates between the states are not equal.