Question

A marketing researcher wants to estimate the mean amount spent​ ($) on a certain retail website by members of the​ website's premium program. A random sample of 99 members of the​ website's premium program who recently made a purchase on the website yielded a mean of ​$1700 and a standard deviation of ​$150.

Complete parts​ (a) and​ (b) below.

a. Construct a 90​%confidence interval estimate for the mean spending for all shoppers who are members of the​ website's premium program.

b. Interpret the interval of (a)

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 1700

Population standard deviation =    = 150

Sample size = n =99

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

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

= 1.645 * ( 150 /  99 )

= 24.80
At 90% confidence interval
is,

- E < < + E

1700- 24.80 <   < 1700+24.80

1675.2 <   < 1724.8

(b)

sample mean lies between (1675.2 , 1724.8)

lower bound 1675.2

upper bound 1724.8