Homework Sourse4 One method for estimating the availability of office space
4. One method for estimating the availability of office space in large cities is to conduct arandom sample of offices, and calculate the proportion of offices currently being used. Suppose that realestate agents believe that p = 0.70 of all offices are currently occupied, and decide to take a sample toassess their belief. They are considering a sample size of n = 40.a)
The shape of the sampling distribution of p
The shape will become more normal as the proportion increases
a. Show that this sample size is large enough to justify using the normal approximation to thesampling distribution of p.
Whenever a sample size is above 30, by the central limit theorem you can assume that thedata is enough to justify a normal distribution. Correct.
b) What is the mean of the sampling distribution of p if the real estate agents are correct?
0.70 Correct.
c) What is the standard deviation of the sampling distribution of p if the real estate agents arecorrect?
0.0724 Correct
d)
If the real estate agents are correct, what is the probability that a sample proportion, p,would differ from p = 0.70 by as much as 0.05?
The answer here for 4d is0.4902, but I don\'t understand how that is the right answer. Please provide a detailed answer and all steps. I have the answers for a, b, and c. Thanks
Solution
a)
We want
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 0.7 - 0.05 = 0.65
x2 = upper bound = 0.7 + 0.05 = 0.75
u = mean = 0.7
s = standard deviation = 0.0724
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = -0.69
z2 = upper z score = (x2 - u) / s = 0.69
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.2451
P(z < z2) = 0.7549
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.5098
Thus, those outside this interval is the complement = 1- 0.5098 = 0.4902 [ANSWER]
