Question

1. a total of 10,000 professionals attend a medical conference, and a random sample of 120 of these conference attendees are surveyed. when asked if they were satisfied with the chosen conference venue, 75% of those who were surveyed said “yes.” which one of the following statements will be correct if a 90% confidence interval is constructed based on this data?

a. because the margin of error is 0.065, this means that if we add and subtract 6.5 from 0.75, we’ll obtain an interval that has a good chance of including the true population proportion.
b. we can be 90% confident the interval from 0.685 to 0.815 includes the true sample proportion.
c. increasing the size of the sample will reduce the width of the interval and probably reduce the center of the interval as well.
d. because ( n = 10000 ), the margin of error will be approximately 0.010.
e. none of the above answer options are correct.

Explanation:

Step1: Analyze Option A

Margin of error is added and subtracted from the sample proportion (not the percentage value). Here, sample proportion \(p = 0.75\). Margin of error \(E\) is added and subtracted from \(0.75\), not \(75\). So, Option A is incorrect.

Step2: Analyze Option B

Confidence intervals are constructed for the population proportion, not the sample proportion. The sample proportion is the center of the confidence interval. So, Option B is incorrect.

Step3: Analyze Option C

Increasing the sample size \(n\) reduces the margin of error \(E=\ z_{\alpha/2}\sqrt{\frac{p(1 - p)}{n}}\) (since \(E\) is inversely proportional to \(\sqrt{n}\)), which reduces the width of the interval. But the center of the interval is the sample proportion \(\hat{p}\), and increasing \(n\) does not change the value of \(\hat{p}\) (if the sampling is random and representative). So, Option C is incorrect.

Step4: Analyze Option D

The formula for margin of error for proportion is \(E = z_{\alpha/2}\sqrt{\frac{p(1 - p)}{n}}\). For a \(90\%\) confidence interval, \(z_{\alpha/2}=1.645\), \(p = 0.75\), \(n = 120\) (sample size, not the population size \(N = 10000\)). \(E=1.645\sqrt{\frac{0.75\times(1 - 0.75)}{120}}\approx1.645\times\sqrt{\frac{0.75\times0.25}{120}}\approx1.645\times0.0395\approx0.065\). The population size \(N\) is not used in the formula for margin of error when \(n/N\leq0.1\) (here \(n = 120\), \(N=10000\), \(n/N=0.012\leq0.1\)). So, Option D is incorrect.

Answer:

E. None of the above answer options are correct.