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. Hypertension It has been found that \(26 \%\) of men 20 years and older suffer from hypertension (high blood pressure) and \(31.5 \%\) of women are hypertensive. A random sample of 150 of women are hypertensive. from recent hospital records, and the following results were obtained. Can you conclude that a higher percentage of women have high blood pressure? Use \(\alpha=0.05 .\) Men 43 patients had high blood pressure Women 52 patients had high blood pressure

Short answer

No, the evidence is insufficient to conclude that a higher percentage of women have high blood pressure.

Step-by-step solution

State the Hypotheses

The null hypothesis (\(H_0\)) is that the proportion of women with hypertension \(p_1\) is less than or equal to the proportion of men \(p_2\), i.e., \(p_1 \leq p_2\). The alternative hypothesis (\(H_a\)) is that the proportion of women \(p_1\) is greater than the proportion of men \(p_2\), i.e., \(p_1 > p_2\). This is an upper-tailed test.

Identify the Claim

The claim is that a higher percentage of women have high blood pressure than men.

Find the Critical Value

Since \(\alpha = 0.05\) for a one-tailed test, we look up the critical z-value for 0.05 significance level in a standard normal distribution table. The critical value \(z_c\) is approximately 1.645.

Compute the Test Value

Calculate the test statistic using the formula for two proportions: \[ z = \frac{(\hat{p}_1 - \hat{p}_2) - 0}{\sqrt{\hat{p}(1 - \hat{p})(\frac{1}{n_1} + \frac{1}{n_2})}} \]where \(\hat{p}_1\) is the proportion of women with high blood pressure (52/150), \(\hat{p}_2\) is the proportion of men with high blood pressure (43/150), and \(\hat{p}\) is the pooled sample proportion based on the combined samples:\[ \hat{p} = \frac{52 + 43}{150 + 150} \= \frac{95}{300} = 0.3167 \]The test statistic calculation becomes:\[ z = \frac{(\frac{52}{150} - \frac{43}{150})}{\sqrt{0.3167 \times 0.6833 \times (\frac{1}{150} + \frac{1}{150})}}\]Carrying out the calculation gives \(z\approx 1.06\).

Make the Decision

Compare the test statistic \(z = 1.06\) to the critical value \(z_c = 1.645\). Since \(1.06 < 1.645\), we do not reject the null hypothesis.

Summarize the Results

There is not enough statistical evidence to support the claim that a higher percentage of women suffer from hypertension than men at the \(\alpha = 0.05\) significance level.

Key concepts

Critical Value

In hypothesis testing, the critical value is a threshold or boundary that divides the acceptance region from the rejection region for the null hypothesis. This value helps determine whether the test statistic falls within the rejection region, which would indicate that the null hypothesis should be rejected.

When performing a hypothesis test, you set a significance level, typically denoted as \( \alpha \). This determines how far from the mean of the distribution your test statistic must fall to reject the null hypothesis.

• For a one-tailed test, the critical value represents the point beyond which only \( \alpha \) percent of the data would lie.
• In this exercise, the critical value is determined by a \( z \)-distribution because we're comparing proportions.

For a one-tailed test with \( \alpha = 0.05 \), the critical \( z \)-value is found from standard normal distribution tables and is approximately \( 1.645 \). If your test statistic exceeds this value, you would reject the null hypothesis.

Test Statistic

The test statistic is a standardized value that measures the degree of divergence between sample data and a null hypothesis. It's calculated using statistical formulas suited to the specific conditions of your hypothesis test. In the case of proportions, like this, it's determined using the formula for the difference between two sample proportions.

For the exercise provided, the formula used was:\[ z = \frac{(\hat{p}_1 - \hat{p}_2)}{\sqrt{\hat{p}(1 - \hat{p})(\frac{1}{n_1} + \frac{1}{n_2})}} \]Where:

• \( \hat{p}_1 \) and \( \hat{p}_2 \) are the sample proportions for women and men, respectively.
• \( \hat{p} \) is the pooled sample proportion calculated from both groups.

Once you perform the calculations, you get a test statistic that is then compared to the critical value. In this instance, the test statistic \( z \approx 1.06 \) did not exceed the critical value, suggesting the null hypothesis was not rejected.

Null and Alternative Hypotheses

Hypotheses are foundational to conducting hypothesis testing. They provide statements about population parameters that we aim to test statistically against our sample data.

• The **Null Hypothesis** \((H_0)\) represents a statement of no effect or no difference. In this problem, it states that the proportion of hypertensive women is less than or equal to that of men \( ( p_1 \leq p_2 ) \).
• The **Alternative Hypothesis** \((H_a)\) suggests otherwise; it is often what researchers hope to prove. Here, it claims that the proportion of women with hypertension is greater than that of men \( ( p_1 > p_2 ) \).

Identifying these hypotheses correctly is crucial as they guide the test's direction (one-tailed or two-tailed) and influence the determination of critical values and the interpretation of results.

Significance Level

The significance level, denoted as \( \alpha \), is the probability threshold used to determine when to reject the null hypothesis. It signifies the level of risk you're willing to take that you'll reject a true null hypothesis, often referred to as a Type I error.

• Commonly, a significance level of \(0.05\) is used, meaning there's a 5% chance of incorrectly rejecting the null hypothesis.
• In hypothesis testing, a smaller \( \alpha \) level reduces the risk of a Type I error but requires a stronger statistical signal to reject the null hypothesis.

In this test involving hypertension proportions, \( \alpha = 0.05 \) sets a conservative threshold, necessitating that results must be significant enough beyond this level to assert that women have a higher proportion with confidence. Here, the test did not exceed this threshold, causing retention of the null hypothesis and indicating insufficient evidence to support the claim.