A researcher constructs a confidence interval for a population proportion using a sample of size 50. The value of pË† is .3, and the resulting confidence interval is determined to be (.1323, .4677). What's the level of confidence for this interval?a. 80%b. 90%c. 95%d. 99%e. Not enough information to answer

In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence interval [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

Z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].

For this problem, we have that:

[tex]p = 0.30, n = 50[/tex]

Let's start from the higher confidence levels, since the higher the confidence level, the higher the width of the interval.

d. 99%

So [tex]\alpha = 0.01[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.905[/tex], so [tex]Z = 2.575[/tex].

The lower limit is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.30 - 2.575\sqrt{\frac{0.3*0.7}{50}} = 0.133[/tex]

The upper limit is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.30 + 2.575\sqrt{\frac{0.3*0.7}{50}} = 0.467[/tex]

This is very close to the interval found, so d. is the correct answer