What is the test statistic and critical value?

Question

A movie theater company wants to see if there is a difference in the average price of movie tickets in St. Louis and Cleveland. They sample 20 ticket stubs from St. Louis and 25 from Cleveland. Test the claim using a 10% level of significance. Assume the population variances are unequal and that movie ticket prices are normally distributed. Give answer to at least 4 decimal places.

St Louis: 10, 9, 8, 8, 8, 10, 9, 8, 8, 8, 10, 11, 11, 9, 9, 10, 11, 10, 7, 12

Cleveland: 15, 11, 13, 10, 13, 9, 12, 9, 11, 11, 10, 12, 10, 14, 10, 10, 15, 13, 14, 13,10, 8, 11, 11, 10

Based on the hypotheses, find the following:

Test Statistic = 

Critical Values = ± (Just enter the positive CV.)

Answer 1

St Louis average ticket price = $9.50

average = sum divided by 20

Cleveland averrage ticekt price = $11.56

average = sum divided by 25

alpha = 10% = .10

2 tailed test

degrees of freedom is a complicated formula

df = (((s1^2/n1)^2 + (s2^2/n2)^2)/(s1^2/(n2-1) + (s2^2/(n2-1)

you need to calculate s1 and s2, the sample variances of the two samples

n1-1=19,n2-1=24

use t table to find critical t value

finding row with df

and column with alpha = .10

CV = t critical = that table entry

Ho: st louis mean = cleveland mean

Ha: means are not equal

if t observed > t critical, reject Ho

sample mean difference = $11.56-$9.50 = $2.06

assume population variances are not equal

then calculate

tdf = (sample mean difference -(pop variance difference)/sqr((s1^2/n1 + s2^2/n2)

= (2.06 - pop var diff)/sqr(s1^2/20 + s2^2/25)

use excel, as the calculations get very tedious