Question

A survey that was conducted showed the average American that visited a Target store spends $62 per visit. I believe the mean amount of money spent is higher than $62. I surveyed 12 of my friends and for the following amounts:

125, 75, 50, 100, 100, 130, 145, 300, 75, 60, 50, 50

Assume a random sample determine if the average American spends more than $62 per visit to Target. Use a significance level of .03.

PLEASE DO NOT HANDWRITE AS I NEED TO COPY THE ANSWER INTO A DISCUSSION BOARD. ALSO PLEASE TRY AND USE GRAPHS IN YOUR ANSWERS BY USING FORMULAS AND STAT CRUNCH.

Answer 1

Let mu be the average American that visited a target store spends per visit

we want to test,

H0: mu=62 Vs H1: mu>62

Test statistic under H0,

T=(xbar-mu)/(s/√n)

follows t distribution with n-1 degrees of freedom.

Where, s is sample mean square can be calculated as,

s^2=sum((xi-xbar)^2)/n-1

and xbar is mean.

Here n=12 , alpha=0.03

And we calculate

xbar=105

s^2=4872.727 so that s=69.805

Hence, T=(105-62)/(69.805/√12)

T=2.1339

Critical region for this test is

CR={T>T(n-1),alpha}

T(n-1),alpha is a table value

T(11),0.03=2.096

Here T>T(11),0.03

Therefore, we may reject H0 at 3 % l.o.s.

Therefore average American that visited a target store may spends more than $62 per visit.