Questions: The Pew Research Center has conducted extensive research on social media usage. One finding, reported in June 2018, was that 78% of adults aged 18 to 24 use Snapchat. Another finding was that 45% of those aged 18 to 24 use Twitter. Assume the sample size associated with both findings is 500. Round your answers to four decimal places. a. Develop a 95% confidence interval for the proportion of adults aged 18 to 24 who use Snapchat. to b. Develop a 99% confidence interval for the proportion of adults aged 18 to 24 who use Twitter. to c. In which case, part (a) or part (b), is the margin of error larger?

The Pew Research Center has conducted extensive research on social media usage. One finding, reported in June 2018, was that 78% of adults aged 18 to 24 use Snapchat. Another finding was that 45% of those aged 18 to 24 use Twitter. Assume the sample size associated with both findings is 500.
Round your answers to four decimal places.
a. Develop a 95% confidence interval for the proportion of adults aged 18 to 24 who use Snapchat.
to
b. Develop a 99% confidence interval for the proportion of adults aged 18 to 24 who use Twitter.
to
c. In which case, part (a) or part (b), is the margin of error larger?

Transcript text: The Pew Research Center has conducted extensive research on social media usage. One finding, reported in June 2018, was that $78 \%$ of adults aged 18 to 24 use Snapchat. Another finding was that $45 \%$ of those aged 18 to 24 use Twitter. Assume the sample size associated with both findings is 500 . Round your answers to four decimal places. a. Develop a $\mathbf{9 5 \%}$ confidence interval for the proportion of adults aged 18 to 24 who use Snapchat. $\qquad$ to $\square$ b. Develop a $99 \%$ confidence interval for the proportion of adults aged 18 to 24 who use Twitter. $\square$ to $\square$ c. In which case, part (a) or part (b), is the margin of error larger? $\square$

Solution

Solution Steps

To solve these problems, we will use the formula for a confidence interval for a proportion, which is given by:

\[ \text{CI} = \hat{p} \pm Z \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \]

where $\hat{p}$ is the sample proportion, $Z$ is the Z-score corresponding to the desired confidence level, and $n$ is the sample size.

a. For the 95% confidence interval for Snapchat users, use $\hat{p} = 0.78$, $n = 500$, and the Z-score for 95% confidence.

b. For the 99% confidence interval for Twitter users, use $\hat{p} = 0.45$, $n = 500$, and the Z-score for 99% confidence.

c. Compare the margin of error for both confidence intervals to determine which is larger.

Step 1: Calculate the 95% Confidence Interval for Snapchat Users

Given:

Sample proportion $ \hat{p}_{\text{Snapchat}} = 0.78 $
Sample size $ n_{\text{Snapchat}} = 500 $
Z-score for 95% confidence $ z_{0.95} \approx 1.960 $

The margin of error is calculated as: \[ \text{Margin of Error}_{\text{Snapchat}} = z_{0.95} \times \sqrt{\frac{\hat{p}_{\text{Snapchat}}(1 - \hat{p}_{\text{Snapchat}})}{n_{\text{Snapchat}}}} \approx 1.960 \times \sqrt{\frac{0.78 \times (1 - 0.78)}{500}} \approx 0.0363 \]

Thus, the 95% confidence interval for Snapchat users is: \[ \text{CI}_{\text{Snapchat}} = \left(0.78 - 0.0363, 0.78 + 0.0363\right) \approx (0.7437, 0.8163) \]

Step 2: Calculate the 99% Confidence Interval for Twitter Users

Given:

Sample proportion $ \hat{p}_{\text{Twitter}} = 0.45 $
Sample size $ n_{\text{Twitter}} = 500 $
Z-score for 99% confidence $ z_{0.99} \approx 2.576 $

The margin of error is calculated as: \[ \text{Margin of Error}_{\text{Twitter}} = z_{0.99} \times \sqrt{\frac{\hat{p}_{\text{Twitter}}(1 - \hat{p}_{\text{Twitter}})}{n_{\text{Twitter}}}} \approx 2.576 \times \sqrt{\frac{0.45 \times (1 - 0.45)}{500}} \approx 0.0573 \]

Thus, the 99% confidence interval for Twitter users is: \[ \text{CI}_{\text{Twitter}} = \left(0.45 - 0.0573, 0.45 + 0.0573\right) \approx (0.3927, 0.5073) \]

Step 3: Compare the Margins of Error

The margin of error for Snapchat users is approximately $ 0.0363 $, while for Twitter users it is approximately $ 0.0573 $. Therefore, the margin of error is larger for Twitter.