For this problem, carry at least four digits after the decimal in your calculations. Answers may vary slightly due to rounding. In a random sample of 63 professional actors, it was found that 38 were extroverts. A button hyperlink to the SALT program that reads: Use SALT. (a) Let p represent the proportion of all actors who are extroverts. Find a point estimate for p. (Round your answer to four decimal places.) Incorrect: Your answer is incorrect. (b) Find a 95% confidence interval for p. (Round your answers to two decimal places.) lower limit Incorrect: Your answer is incorrect. upper limit Incorrect: Your answer is incorrect.

In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [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].

Question a:

In a random sample of 63 professional actors, it was found that 38 were extroverts.

We use this to find the sample proportion, which is the point estimate for p. So

[tex]\pi = \frac{38}{63} = 0.6032[/tex]

Question b:

Sample of 63 means that [tex]n = 63[/tex]

95% confidence level

So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].

The lower limit of this interval is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6032 - 1.96\sqrt{\frac{0.6032*0.3968}{63}} = 0.4824[/tex]

The upper limit of this interval is:

[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6032 + 1.96\sqrt{\frac{0.6032*0.3968}{63}} = 0.724[/tex]

Rounding to two decimal places:

Lower limit: 0.48

Upper limit: 0.72