Question

8.2 one - sample interval for the proportion. construct and interpret a confidence interval for the population proportion.
you recently sent out a survey to determine if the percentage of adults who use social media has changed from 66%, which was the percentage of adults who used social media five years ago. of the 2790 people who responded to survey, 1726 stated that they currently use social media.
a) use the data from this survey to construct a 98% confidence interval estimate of the proportion of adults who use social media. record the result below in the form of (#, #). the parenthesis and comma must be included. round your final answer to four decimal places.
b) can you conclude that the percentage of adults who use social media has changed? explain.
yes, because the proportion of adults who used social media five years ago is inside of the confidence interval.
yes, because the proportion of adults who used social media five years ago is not inside the confidence interval.
no, because the proportion of adults who used social media five years ago is inside the confidence interval.
no, because the proportion of adults who used social media five years ago is not inside the confidence interval.
c) if it has changed, has it increased or decreased?
the percentage increased.
the percentage decreased.
we cannot determine that the percentage has changed.
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Explanation:

Step1: Calculate sample proportion

The sample proportion $\hat{p}=\frac{x}{n}$, where $x = 1726$ (number of successes) and $n=2790$ (sample - size). So, $\hat{p}=\frac{1726}{2790}\approx0.6186$.

Step2: Find the z - value for 98% confidence interval

The confidence level is 98%, so the significance level $\alpha=1 - 0.98 = 0.02$. Then $\alpha/2=0.01$. The z - value $z_{\alpha/2}=z_{0.01}\approx2.3263$.

Step3: Calculate the margin of error

The formula for the margin of error $E$ for a proportion is $E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$. Substitute $\hat{p}=0.6186$, $n = 2790$, and $z_{\alpha/2}=2.3263$ into the formula.
First, calculate $1-\hat{p}=1 - 0.6186 = 0.3814$. Then $\frac{\hat{p}(1 - \hat{p})}{n}=\frac{0.6186\times0.3814}{2790}\approx\frac{0.2359}{2790}\approx0.00008455$.
$E = 2.3263\sqrt{0.00008455}\approx2.3263\times0.0092 = 0.0214$.

Step4: Calculate the confidence interval

The confidence interval is $\hat{p}-E$0.6186-0.0214The 98% confidence interval is $(0.5972,0.6400)$.

Step5: Answer part b

The proportion of adults who used social media five years ago is $p_0 = 0.66$. Since $0.66$ is not inside the confidence interval $(0.5972,0.6400)$, we can conclude that the percentage of adults who use social media has changed. The answer is: Yes, because the proportion of adults who used social media five years ago is not inside the confidence interval.

Step6: Answer part c

Since the upper - bound of the confidence interval $0.6400$ is less than $0.66$, the percentage has decreased.

Answer:

a) $(0.5972,0.6400)$
b) Yes, because the proportion of adults who used social media five years ago is not inside the confidence interval.
c) The percentage decreased.