I. About 75% of young adult Internet users (ages 18 to 29) watch online v survey
ID: 2935499 • Letter: I
Question
I. About 75% of young adult Internet users (ages 18 to 29) watch online v survey contacts an SRS of 1000 young adult Internet users and calcuia sample who watch online video. a. What is the mean of the sampling distribution of p? b. Find the standard deviation of the sampling distribution of p c. Is the sampling distribution Check that the 10% condition is met. approximately Normal? Check that the Normal conditions are met. d. If the sample size were 9000 rather than 1000, how would this change the sampling distr of p? e. What is the probabili ity, in an SRS of 1000 young adults, that between 71% and 77% of them watch videos online? 2. A USA Today Poll asked a random sample of 1012 U.S. adults what they do with the milk in the bowl after they have eaten the cereal. Of the respondents, 70% said that they drink it. Let p be the proportion of people in the sample who drink the cereal milk. a. What is the mean of the sampling distribution of p? b. Find the standard deviation of the sampling distribution of p. Check that the 10% condition is met. c. Is the sampling distribution approximately Normal? Check that the Normal conditions are met d. Find the probability ofobtaining a sample of 1012 adults in which 67% or fewer say they drink the cereal milk. 3 Your mail-order company advertises that it ships 90% of its orders within three working days. You select an SRS of 100 of the 5000 orders received in the past week for an audit. a. What is the mean of the sampling distribution of p? b. Find the standard deviation of the sampling distribution of p. Check that the 10% condition is met. is the sampling distribution approximately Normal? Check that the Normal conditions are met. c. eals that 86 oft ses orders were shipped on time (that is 86%). What is the probability that the proportion of on time orders is 86% or less?Explanation / Answer
a) Mean = 0.75
b) Standard error, SE = sqrt(p*(1-p)/n) = sqrt(0.75*0.25/1000) = 0.013693
c) Finally, the shape of the distribution of p-hat will be approximately normal as long as the sample size n is large enough. The convention is to require both np and n(1 – p) to be at least 10.
np = 1000*0.75 = 750, n(1-p) = 1000*0.25 = 250
Yes it is normal
d) Mean will remain same at 0.75
Standard deviation will change SE = sqrt(0.75*0.25/9000) = 0.0045644
e) P(0.71 <p < 0.77) = P(p < 0.77) - P(p < 0.71) = P(Z < (0.77-0.75)/0.013693) - P(Z < (0.71-0.75)/0.013693)
= P(Z < 1.46) - P(Z < -2.92) = 0.9279 - 0.0018 = 0.9261