1. A poll conducted in 2018 found that 54% of US adult Twitter users get at least some news on Twitter. The standard error for this estimate was 2.4%, and a normal distribution can be used to model the sample proportion. Construct a 99% confidence interval for the fraction of US adult Twitter users who get some news on Twitter, and interpret the confidence interval in context.

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Answer:

The 99% confidence interval for the fraction of US adult Twitter users who get some news on Twitter is (0.4872, 0.6018). It means that we are 99% sure that the true proportion of US adult Twitter users who get some news on Twitter is between 0.4872 and 0.6018.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm zs[/tex]

In which

z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex], and s is the standard error.

54% of US adult Twitter users get at least some news on Twitter.

This means that [tex]\pi = 0.54[/tex]

The standard error for this estimate was 2.4%

This means that [tex]s = 0.024[/tex]

99% confidence level

So [tex]\alpha = 0.01[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.01}{2} = 0.995[/tex], so [tex]Z = 2.575[/tex]

The lower bound is:

[tex]\pi - zs = 0.54 - 2.575*0.024 = 0.4782[/tex]

The upper bound is:

[tex]\pi + zs = 0.54 + 2.575*0.024 = 0.6018[/tex]

The 99% confidence interval for the fraction of US adult Twitter users who get some news on Twitter is (0.4872, 0.6018). It means that we are 99% sure that the true proportion of US adult Twitter users who get some news on Twitter is between 0.4872 and 0.6018.