Questions: Suppose a poll is taken that shows 222 out of 400 randomly selected users of a certain social media website feel that the site should do more to decrease hateful and abusive content on the site. Test the hypothesis that the majority (more than 50% ) of the site's users feel the site should do more to decrease hateful and abusive content on the site. Use a significance level of 0.05. Determine the null and alternative hypotheses. H0: p = 0.5 Ha: p > 0.5 (Type integers or decimals. Do not round.)

Solution Steps

We are testing the following hypotheses:

• Null Hypothesis (\(H_0\)): \(p \leq 0.5\) (The proportion of users who feel the site should do more to decrease hateful and abusive content is less than or equal to 50%)
• Alternative Hypothesis (\(H_a\)): \(p > 0.5\) (The proportion of users who feel the site should do more to decrease hateful and abusive content is greater than 50%)

The test statistic is calculated using the formula: \[ Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \] where:

• \(\hat{p} = \frac{222}{400} = 0.555\)
• \(p_0 = 0.5\)
• \(n = 400\)

Substituting the values, we find: \[ Z = \frac{0.555 - 0.5}{\sqrt{\frac{0.5(1 - 0.5)}{400}}} = 2.2 \]

The P-value associated with the test statistic \(Z = 2.2\) is calculated to be: \[ \text{P-value} = 0.0139 \]

For a significance level of \(\alpha = 0.05\) in a one-tailed test, the critical value of \(Z\) is: \[ Z_{critical} = 1.6449 \] Since \(Z = 2.2 > Z_{critical} = 1.6449\), we reject the null hypothesis.

Given that the P-value \(0.0139 < 0.05\) and the test statistic \(Z = 2.2\) exceeds the critical value, we have sufficient evidence to reject the null hypothesis. Therefore, we conclude that the majority of users (more than 50%) feel that the site should do more to decrease hateful and abusive content.

Final Answer