Question

Problem 3. (17 points A random sample of 80 in-coming college freshmen showed that were planning to bring a car with them to campus next fail. Suppose that we are interested informing a confidence interval at 90 percent for the proportion of all incoming college freshmen who are planning to bring a car to campus next fall Where appropriate, express your wer a proportion not a percentage. Round answers to three decimal places The estimates The standard errors The mutipli Note: You can comparacion this problem preview answers

Answer 1

Let x be the number of freshmen planning to bring a car with them to campus next fall = 36

Random sample n = 80

a) The estimate is  $1596223028072_blob.png$ = x /n = 36/ 80 = 0.45

b) The standard error =
p*(1-P) n
= $\sqrt{\frac{0.45*(1-0.45)}{80}}$

= 0.056

c) The multiplier :

We are given confidence level = 0.90

Therefore α = 1 - 0.90 = 0.1 , 1 - (α/2) = 0.95

So we have to find z score corresponding to area 0.9500 on z score table

So Multiplier  = 1.645