Question

Shiraz is trying to flip her sneakers. She checks prices of 40 pairs of new in box limited edition Jordan's in the pink-teal colourway on several reseller sites and finds that they have a mean asking price of $213.00 with a standard deviation of $51.00. Develop a 99% confidence statement for the resale market of these shoes.

The margin of error for the 99% confidence interval is 

Answer for part 1

.

We can be 99% confident that the average asking price of pink-teal Jordan's is between $

Answer for part 2 and coordinate 1

 and $

Answer for part 2 and coordinate 2

.

Answer 1

$The \ sample \ size \ of \ sneakers \ n = 40\\ Mean \ of \ asking \ price \ \bar{x} = \$ 213 \\ Standard \ deviation \ of \ asking \ price \ s \ = \$ 51\\ Standard \ Error \ of \ sample \ mean \ is \ (S.E)=\frac{s}{\sqrt{n}} \\ \Rightarrow S.E= \frac{51}{\sqrt{40}} =8.06\\ The \ critical \ value \ of \ 99\% \ confidence \ interval \ for \ population \ mean (\mu) \\ is \ z_\frac{\alpha}{2}=2.58 \\ \because n=40>30, we \ use \ Z-distribution \ to \ build \ the \ confidence\ interval\\ The \ 99\% \ confidence \ interval \ for\ population \ mean (\mu) \ of \ asking \ price \ is \\ \bar{x}-(\ z_\frac{\alpha}{2})S.E\le\mu\le \bar{x}+(\ z_\frac{\alpha}{2})S.E\\ Substituting \ the \ values \ \ z_\frac{\alpha}{2}=2.58\ \& \ S.E =8.06 \ we \ get \\ 213-(2.58)(8.06)\le\mu\le \ 213+(2.58)(8.06) \Rightarrow 213-20.7948\le\mu\le213+20.7948\\ \Rightarrow 192.2052\le\mu\le233.7948\\ \approx \ 192.21\le\mu\le233.79\\ \therefore The \ average \ market \ asking \ price \ of \ the \ shoes \ will \ be \\ \ 192.21\le\mu\le233.79 \ with \ 99\%\ confidence.\\ Coordinate 1 = \$ 192.21\\ Corrdinate 2 = \$ 233.79$