A study reported that in a sample of 109 people who watch television news, 37 had elevated diastolic blood pressure levels (in millimeters of mercury, or mmHg). In a sample of 76 people who Do not watch television news, 12 had elevated diastolic blood pressure levels. Can you conclude that the proportion of people with elevated diastolic blood pressure levels differs between
News-watchers and those who do not watch news? Use the $\alpha = 0.05$ level of significance.

Question

Quizplus · Accepted Answer

To determine if there is a significant difference between proportions, we can use a two-sample proportion hypothesis test. We will use the following null and alternative hypotheses:

H0: p1 = p2 (the proportion of people with elevated diastolic blood pressure levels is the same for news-watchers and non-watchers)
Ha: p1 ≠ p2 (the proportion of people with elevated diastolic blood pressure levels differs between news-watchers and non-watchers)

We will use a significance level of α = 0.05. We can assume that the samples are independent and that the conditions for normality are met.

The test statistic is calculated as:

z = (p1 - p2) / sqrt(p * (1 - p) * (1/n1 + 1/n2))

where p = (x1 + x2) / (n1 + n2) is the pooled proportion, x1 and x2 are the number of people with elevated diastolic blood pressure levels in each sample, and n1 and n2 are the sample sizes.

Using the given values:

p1 = 37/109 = 0.339
p2 = 12/76 = 0.158
n1 = 109
n2 = 76

p = (37 + 12) / (109 + 76) = 0.248

z = (0.339 - 0.158) / sqrt(0.248 * 0.752 * (1/109 + 1/76)) = 2.55

The critical values for a two-tailed test at α = 0.05 with a z-test are ±1.96.

Since the calculated test statistic (z = 2.55) is greater than the critical value (±1.96), we reject the null hypothesis. Therefore, we can conclude that there is a significant difference between the proportion of people with elevated diastolic blood pressure levels for news-watchers and non-watchers.

A Study Reported That in a Sample of 109 People α=0.05\alpha = 0.05α=0.05

A Study Reported That in a Sample of 109 People $\alpha = 0.05$