Answer :
We start by letting the sample size be
[tex]$$ n = 1000 $$[/tex]
and the sample proportion of U.S. adults in favor be
[tex]$$ \hat{p} = 0.460. $$[/tex]
Before constructing the confidence interval, it is important to verify the conditions for inference. In particular, we need the numbers of successes and failures to be sufficiently large (each at least 15):
- The number of successes is
[tex]$$ n\hat{p} = 1000 \times 0.460 = 460, $$[/tex]
- and the number of failures is
[tex]$$ n(1-\hat{p}) = 1000 \times (1 - 0.460) = 540. $$[/tex]
Since both [tex]$460$[/tex] and [tex]$540$[/tex] are greater than [tex]$15$[/tex], the conditions for inference are met.
Next, we compute the standard error (SE) for the sample proportion using the formula:
[tex]$$ SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}. $$[/tex]
Substituting the values, we get:
[tex]$$ SE = \sqrt{\frac{0.460 \times 0.540}{1000}} \approx 0.015761. $$[/tex]
For a [tex]$95\%$[/tex] confidence interval, the critical value is approximately [tex]$z^{\ast} = 1.96$[/tex]. The margin of error (ME) is then calculated as:
[tex]$$ \text{ME} = z^{\ast} \times SE \approx 1.96 \times 0.015761 \approx 0.030891. $$[/tex]
Now, we can determine the lower and upper limits of the confidence interval:
[tex]$$ \text{Lower bound} = \hat{p} - \text{ME} \approx 0.460 - 0.030891 \approx 0.429109, $$[/tex]
[tex]$$ \text{Upper bound} = \hat{p} + \text{ME} \approx 0.460 + 0.030891 \approx 0.490891. $$[/tex]
Rounded to three decimal places, the confidence interval is:
[tex]$$ (0.429,\, 0.491). $$[/tex]
Interpretation: We are [tex]$95\%$[/tex] confident that the true proportion of all U.S. adults who favor making Washington D.C. a separate state is between [tex]$0.429$[/tex] and [tex]$0.491$[/tex].
[tex]$$ n = 1000 $$[/tex]
and the sample proportion of U.S. adults in favor be
[tex]$$ \hat{p} = 0.460. $$[/tex]
Before constructing the confidence interval, it is important to verify the conditions for inference. In particular, we need the numbers of successes and failures to be sufficiently large (each at least 15):
- The number of successes is
[tex]$$ n\hat{p} = 1000 \times 0.460 = 460, $$[/tex]
- and the number of failures is
[tex]$$ n(1-\hat{p}) = 1000 \times (1 - 0.460) = 540. $$[/tex]
Since both [tex]$460$[/tex] and [tex]$540$[/tex] are greater than [tex]$15$[/tex], the conditions for inference are met.
Next, we compute the standard error (SE) for the sample proportion using the formula:
[tex]$$ SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}. $$[/tex]
Substituting the values, we get:
[tex]$$ SE = \sqrt{\frac{0.460 \times 0.540}{1000}} \approx 0.015761. $$[/tex]
For a [tex]$95\%$[/tex] confidence interval, the critical value is approximately [tex]$z^{\ast} = 1.96$[/tex]. The margin of error (ME) is then calculated as:
[tex]$$ \text{ME} = z^{\ast} \times SE \approx 1.96 \times 0.015761 \approx 0.030891. $$[/tex]
Now, we can determine the lower and upper limits of the confidence interval:
[tex]$$ \text{Lower bound} = \hat{p} - \text{ME} \approx 0.460 - 0.030891 \approx 0.429109, $$[/tex]
[tex]$$ \text{Upper bound} = \hat{p} + \text{ME} \approx 0.460 + 0.030891 \approx 0.490891. $$[/tex]
Rounded to three decimal places, the confidence interval is:
[tex]$$ (0.429,\, 0.491). $$[/tex]
Interpretation: We are [tex]$95\%$[/tex] confident that the true proportion of all U.S. adults who favor making Washington D.C. a separate state is between [tex]$0.429$[/tex] and [tex]$0.491$[/tex].
