Question

question 2 of 10
which confidence level will produce the widest confidence interval, given a sample proportion of 0.4?
a. 90%
b. 95%
c. 99.7%
d. 68%

Explanation:

Step1: Recall the relationship between confidence level and confidence interval width

The formula for a confidence interval for a proportion is \(\hat{p}\pm z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\), where \(\hat{p}\) is the sample proportion, \(n\) is the sample size, and \(z_{\alpha/2}\) is the critical value. As the confidence level increases, the value of \(z_{\alpha/2}\) increases.

Step2: Compare the critical values for each confidence level

• For a \(68\%\) confidence level, \(z_{\alpha/2}\approx1\) (since for a normal distribution, about \(68\%\) of the data lies within \(1\) standard deviation of the mean).
• For a \(90\%\) confidence level, \(z_{\alpha/2}= 1.645\).
• For a \(95\%\) confidence level, \(z_{\alpha/2}=1.96\).
• For a \(99.7\%\) confidence level, \(z_{\alpha/2}=3\) (since for a normal distribution, about \(99.7\%\) of the data lies within \(3\) standard deviations of the mean).

Since the margin of error \(E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\) and \(\hat{p}=0.4\) (fixed) and assuming \(n\) is fixed (not given as changing), the larger the \(z_{\alpha/2}\), the wider the confidence interval.

Answer:

C. 99.7%