Question

An article stated, "Surveys tell us that more than half of America's college graduates are avid readers of mystery novels." Let p denote the actual proportion of college graduates who are avid readers of mystery novels. Consider a sample proportion p̂ that is based on a random sample of 220 college graduates.

(a) If p = 0.5, what are the mean value and standard deviation of p̂? (Round your answers to four decimal places.) mean standard deviation If p = 0.6, what are the mean value and standard deviation of p̂? (Round your answers to four decimal places.) mean standard deviation Does p̂ have approximately a normal distribution in both cases? Explain. Yes, because in both cases np > 10 and n(1 − p) > 10. No, because in both cases np < 10 or n(1 − p) < 10. No, because when p = 0.5, np < 10. No, because when p = 0.6, np < 10. (b) Calculate P(p̂ ≥ 0.6) for p = 0.5. (Round your answer to four decimal places.) Calculate P(p̂ ≥ 0.6) for p = 0.6. (c) Without doing any calculations, how do you think the probabilities in Part (b) would change if n were 395 rather than 220? When p = 0.5, the P(p̂ ≥ 0.6) would decrease if the sample size was 395 rather than 220. When p = 0.6, the P(p̂ ≥ 0.6) would remain the same if the sample size was 395 rather than 220. When p = 0.5, the P(p̂ ≥ 0.6) would decrease if the sample size was 395 rather than 220. When p = 0.6, the P(p̂ ≥ 0.6) would decrease if the sample size was 395 rather than 220. When p = 0.5, the P(p̂ ≥ 0.6) would remain the same if the sample size was 395 rather than 220. When p = 0.6, the P(p̂ ≥ 0.6) would decrease if the sample size was 395 rather than 220. When p = 0.5, the P(p̂ ≥ 0.6) would remain the same if the sample size was 395 rather than 220. When p = 0.6, the P(p̂ ≥ 0.6) would remain the same if the sample size was 395 rather than 220.

Answer 1

a) The sample proportion is distributed with mean = p and variance = pq/n

When p =0.5 and n= 220 mean = 0,.5 and standard deviation = 0.03370 (square root of 0.5*0.5/220)

When p = 0.6 n= 220 mean = 0.6 and standard deviation =0.03302

b) yes because in both cases np >15 and n(1-p) >15 normal distribution can be assumed

P (p>=0.6) = p (z>=(0.6-0.5)/0.037) = P (z>0.37) = 0.35569

P (p>=0.6) = P (z>=0.6-0.6/0.03302) = P (z>=0) =0.5

c) If n = 395 then variance = 0.5*0.5/395 (when p=0.5) = 0.0006 standard deviation = 0.025

In this case P (p>=0.6) will decrease as the standard deviation is reducing and the value of z will shift to the right.

When p=0.6 no change is there when the sample size is changed and the probability remains 0.5