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. 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 each gender was selected 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 \(\quad 43\) patients had high blood pressure Women 52 patients had high blood pressure

Short answer

No, there is not enough evidence to conclude that a higher percentage of women have high blood pressure than men.

Step-by-step solution

State the Hypotheses

Let's define our null and alternative hypotheses. The null hypothesis, denoted as \(H_0\), assumes that there is no difference in hypertension rates between men and women, i.e., the percentage of hypertensive men is the same as that of hypertensive women. The alternative hypothesis, denoted as \(H_1\), claims that the percentage of hypertensive women is higher than that of men. Thus, \(H_0: p_1 = p_2\) and \(H_1: p_1 > p_2\), where \(p_1\) is the proportion of hypertensive women, and \(p_2\) is the proportion of hypertensive men.

Find the Critical Value

Since we are conducting a one-tailed test (the claim is that the percentage of women is greater), we look up the critical value for \( \alpha = 0.05 \) in a standard normal distribution table. For a one-tailed test with \( \alpha = 0.05 \), the critical value is approximately 1.645.

Compute the Test Value

First, calculate the sample proportions: \( \hat{p_1} = \frac{52}{150} \approx 0.347 \) for women and \( \hat{p_2} = \frac{43}{150} \approx 0.287 \) for men. Then, calculate the pooled sample proportion: \( \hat{p} = \frac{52 + 43}{300} = 0.317 \). The test statistic is calculated using the formula: \( z = \frac{\hat{p_1} - \hat{p_2}}{\sqrt{\hat{p} (1 - \hat{p}) (\frac{1}{n_1} + \frac{1}{n_2})}} \), where \( n_1 = n_2 = 150 \). Substituting the values, we find \( z \approx \frac{0.347 - 0.287}{\sqrt{0.317 \times (1 - 0.317) \times (\frac{1}{150} + \frac{1}{150})}} \approx 1.174 \).

Make the Decision

Compare the computed test value to the critical value. The test value \( z = 1.174 \) is less than the critical value of 1.645. Since the test value does not exceed the critical value, we do not reject the null hypothesis.

Summarize the Results

Based on our calculations, we do not have enough evidence at the \( \alpha = 0.05 \) level to support the claim that a higher percentage of women have high blood pressure compared to men. Therefore, we fail to reject the null hypothesis.

Key concepts

Null Hypothesis

The null hypothesis is an essential starting point in hypothesis testing. It represents the default position that there is no effect or no difference. In our context of testing whether women have a higher rate of high blood pressure than men, the null hypothesis (\( H_0 \)) states that there is no difference in the percentage of hypertensive men and women. We express this mathematically as:\[ H_0: p_1 = p_2 \]where \( p_1 \) is the proportion of hypertensive women, and \( p_2 \) is the proportion of hypertensive men. This hypothesis assumes equality or no preferential condition, setting the stage for the test to either reinforce or challenge this assumption.
In hypothesis testing, failing to reject the null hypothesis does not prove it true, but suggests that there's not enough evidence against it given the data.

Alternative Hypothesis

Opposite to the null hypothesis, the alternative hypothesis proposes what we are looking to establish through evidence. In this example, the alternative hypothesis (\( H_1 \)) suggests that the proportion of hypertensive women is greater than that of men. Formally, it is written as:\[ H_1: p_1 > p_2 \]This is a one-tailed hypothesis, as it considers a single direction of comparison. Confidently stating an alternative hypothesis requires strong statistical evidence, which is investigated by analyzing sample data.
The essence of proposing an alternative hypothesis is to shift focus from what is generally presumed (as per the null hypothesis) to a different proposed reality, based on sample findings.

Critical Value

The critical value functions as a benchmark in hypothesis testing. It helps us decide whether to reject the null hypothesis. For a one-tailed test at a significance level (\( \alpha \)) of 0.05, the critical value is determined using a standard normal distribution table. This value, approximately 1.645 for our exercise, defines the threshold beyond which the null hypothesis will be rejected in favor of the alternative.
Critical values are derived from the significance level, which reflects how willing you are to mistakenly reject a true null hypothesis. Essentially, it helps frame how strong the evidence must be to support the alternative hypothesis. If our computed test value exceeds the critical value, it offers sufficient evidence to consider the alternative hypothesis legitimate.

Test Statistic

A test statistic is a calculated value from sample data used to determine whether to reject the null hypothesis. It standardizes the data difference in terms of standard error. In this exercise, we compute the test statistic using the formula:\[z = \frac{\hat{p_1} - \hat{p_2}}{\sqrt{\hat{p} (1 - \hat{p}) (\frac{1}{n_1} + \frac{1}{n_2})}} \]where \(\hat{p_1} \) and \( \hat{p_2} \) are the sample proportions, and \( \hat{p} \) is the pooled proportion from both samples.
The calculated test value (\( z \approx 1.174 \)) quantitatively represents how far the observed sample proportion is from the null hypothesis assumption under the normal distribution. Comparing this statistic to the critical value helps make the final decision about which hypothesis to accept.

Decision Rule

The decision rule in hypothesis testing is a clear procedure based on comparison between the test statistic and critical value. If the test statistic exceeds the critical value, we reject the null hypothesis in favor of the alternative hypothesis. Conversely, if it is less, as in our exercise where \( z = 1.174 \) is less than the critical value 1.645, we fail to reject the null hypothesis.
The decision rule provides a justifiable rationale to support statistical conclusions. It ties together all the calculated elements - the hypothesized, observed, and critical values - to confirm whether the sample data significantly contradicts the assumed default state, denoted by the null hypothesis.