A marketing researcher wants to estimate the mean amount spent (S) on Amazon. com by Amazon Prime member shoppers. Suppose a random sample of 100 Amazon Prime member shoppers who recently made a purchase on Amazon. com yielded a mean of $1,500. a. Suppose the standard deviation of the amount spent ($) on Amazon. com is $200. Construct a 95% confidence interval estimate for the mean spending for all Amazon Prime member shoppers.
b. Suppose the standard deviation of the sample of 100 Amazon Prime member shoppers is $200. Construct a 95% confidence interval estimate for the mean spending for all Amazon Prime member shoppers.

The answer is below

Explanation:

a)

Given that mean (μ) = $1500, standard deviation (σ) = $200, sample size (n) = 100

confidence (C) = 95% = 0.95

α = 1 -  C = 1 - 0.95 = 0.05

α/2 = 0.05 / 2 = 0.025

The z score that corresponds with 0.475 (0.5 - 0.025) is 1.96. Therefore the margin of error (E) is:

The confidence interval = (μ ± E) = (1500 ± 39.2) = (1500 - 39.2, 1500 + 39.2) = (1460.8, 1539.2)

The confidence interval is between $1460.8 and $1539.2.

b) Given that mean (μ) = $1500, standard deviation for 100 samples =  σ /√n = $200,

confidence (C) = 95% = 0.95

The confidence interval = (μ ± E) = (1500 ± 392) = (1500 - 392, 1500 + 392) = (1108, 1892)

The confidence interval is between $1108 and $1892.

there's 11 people for the 1st position. then, no matter who gets it, there's only 10 left for the second, and then 9 left for the third. 11 x 10 x 9 = 990.   the answer is there are 990 different ways the positions can be filled by these applicants