Question

1. To determine the mean cost of groceries in a certain city with known standard deviation of $2, an identical grocery basket of food is purchased at each store in a random sample of nine stores. If the average cost is $47.52, the 95% confidence interval estimate for the cost of these groceries in the city is (assume cost of groceries is normally distributed).

A) $47.52 ± $1.22

B) $47.52 ± $1.31

C) $47.52 ± $1.42

D) $47.52 ± $4.49

2. Continuing Problem 1, the underlying sampling distribution is

A) Z B) Student t C) Binomial D) Chi-square 16.

3. Continuing Problem 1, if the maximal allowable estimation error is half of the population standard deviation, then the sample size shall be

A) 4 B) 9 C) 16 D) 25

Answer 1

Solution :

Given that,

1) Point estimate = sample mean = = 47.52

Population standard deviation =    = 2

Sample size = n = 9

At 95% confidence level

= 1 - 95%  

= 1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z_{0.025 = 1.96}

Margin of error = E = Z/2 * ( /n)

= 1.96 * ( 2 /  9 )

= 1.31

At 90% confidence interval estimate of the population mean is,

  ± E

47.52 ± 1.31   

( 46.21, 48.83 )  

correct option is = B

2) Z distribution , because Population standard deviation is known

correct option is = A

3) margin of error = E = 1

sample size = n = [Z_/2* / E] ²

n = [ 1.96 * 2 / 1 ]²

n = 15.36

Sample size = n = 16

correct option is = C