Answer to Question #180646 in Statistics and Probability for Joshua Cornfield

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

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