Question

in a survey of 2133 adults in a recent year, 1381 say they have made a new years resolution
construct 90% and 95% confidence intervals for the population proportion. interpret the results and compare the widths of the confidence intervals
the 90% confidence interval for the population proportion p is (□□)
(round to three decimal places as needed.)
the 95% confidence interval for the population proportion p is (□□)
(round to three decimal places as needed.)
with the given confidence, it can be said that the population proportion of adults who say they have made a new years resolution is between the endpoints of the given confidence interval
compare the widths of the confidence intervals. choose the correct answer below
a. the 90% confidence interval is wider
b. the 95% confidence interval is wider
c. the confidence intervals are the same width
d. the confidence intervals cannot be compared

Explanation:

Step1: Calculate sample proportion $\hat{p}$

Sample proportion $\hat{p}=\frac{x}{n}$, where $x = 1381$ and $n=2133$. So, $\hat{p}=\frac{1381}{2133}\approx0.647$.

Step2: Find $z$-values for 90% and 95% confidence intervals

For 90% confidence interval, $z_{\alpha/2}=z_{0.05}=1.645$ (from standard normal distribution table).
For 95% confidence interval, $z_{\alpha/2}=z_{0.025}=1.96$.

Step3: Calculate margin of error $E$ for 90% confidence interval

Margin of error formula is $E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$.
Substitute $\hat{p}=0.647$, $n = 2133$, $z_{\alpha/2}=1.645$ into the formula:

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90% confidence interval: $\hat{p}-E

Step4: Calculate margin of error $E$ for 95% confidence interval

Substitute $\hat{p}=0.647$, $n = 2133$, $z_{\alpha/2}=1.96$ into the margin of error formula:

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95% confidence interval: $\hat{p}-E

Answer:

The 90% confidence interval for the population proportion $p$ is $(0.630,0.664)$.
The 95% confidence interval for the population proportion $p$ is $(0.627,0.667)$.
With the given confidence, it can be said that the population proportion of adults who say they have made a New Year's resolution is between the endpoints of the given confidence interval.
For comparing the widths:
Width of 90% confidence interval $=0.664 - 0.630=0.034$.
Width of 95% confidence interval $=0.667 - 0.627 = 0.040$.
So, the 95% confidence interval is wider. The answer is B.