Question

(1) 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 70 professional actors, it was found that 45 were extroverts.

(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.)


(b) Find a 95% confidence interval for p. (Round your answers to two decimal places.)

lower limit
upper limit

(2) For this problem, carry at least four digits after the decimal in your calculations. Answers may vary slightly due to rounding.

A random sample of 5,100 permanent dwellings on an entire reservation showed that 1,641 were traditional hogans.

(a) Let p be the proportion of all permanent dwellings on the entire reservation that are traditional hogans. Find a point estimate for p. (Round your answer to four decimal places.)

(b) Find a 99% confidence interval for p. (Round your answer to three decimal places.)

lower limit ____

upper limit ____

(3) 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 520 judges, it was found that 282 were introverts.

(a) Let p represent the proportion of all judges who are introverts. Find a point estimate for p. (Round your answer to four decimal places.) ______


(b) Find a 99% confidence interval for p. (Round your answers to two decimal places.)

lower limit
upper limit

Answer 1

Question 1

Part a)
p̂ = X / n = 45/70 = 0.6429

Part b)
p̂ ± Z(α/2) √( (p * q) / n)
0.6429 ± Z(0.05/2) √( (0.6429 * 0.3571) / 70)
Z(α/2) = Z(0.05/2) = 1.96
Lower Limit = 0.6429 - Z(0.05) √( (0.6429 * 0.3571) / 70) = 0.5306
upper Limit = 0.6429 + Z(0.05) √( (0.6429 * 0.3571) / 70) = 0.7551
95% Confidence interval is ( 0.53 , 0.76 )
( 0.53 < P < 0.76 )

Question 2

Part a)
p̂ = X / n = 1641/5100 = 0.3218

Part b)
p̂ ± Z(α/2) √( (p * q) / n)
0.3218 ± Z(0.01/2) √( (0.3218 * 0.6782) / 5100)
Z(α/2) = Z(0.01/2) = 2.576
Lower Limit = 0.3218 - Z(0.01) √( (0.3218 * 0.6782) / 5100) = 0.3049
upper Limit = 0.3218 + Z(0.01) √( (0.3218 * 0.6782) / 5100) = 0.3386
99% Confidence interval is ( 0.30 , 0.34 )
( 0.30 < P < 0.34 )

Question 3

Part a)
p̂ = X / n = 282/520 = 0.5423

Part b)
p̂ ± Z(α/2) √( (p * q) / n)
0.5423 ± Z(0.01/2) √( (0.5423 * 0.4577) / 520)
Z(α/2) = Z(0.01/2) = 2.576
Lower Limit = 0.5423 - Z(0.01) √( (0.5423 * 0.4577) / 520) = 0.486
upper Limit = 0.5423 + Z(0.01) √( (0.5423 * 0.4577) / 520) = 0.5986
99% Confidence interval is ( 0.49 , 0.60 )
( 0.49 < P < 0.60 )