Question

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.

Answer 1

Answer:

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:

$E = z_\frac{\alpha}{2} *\frac{\sigma}{\sqrt{n} } \\\\E=1.96*\frac{200}{\sqrt{100} } =39.2$

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

$E = z_\frac{\alpha}{2} *\frac{\sigma}{\sqrt{n} } \\\\E=1.96*200=392$

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

The confidence interval is between $1108 and $1892.