The owner of Bun N Run Hamburgers wishes to compare the sale

The owner of Bun \'N\' Run Hamburgers wishes to compare the sales per day at two regions of the country. The sample mean number sold for 85 randomly selected days at the East Coast sites were 85.55 with a sample standard deviation was 15.85. For a random sample of 95 days at the West Coast locations, the sample mean sold was 79.85 and the sample standard deviation was 12.85. At the .05 significance level, is there a difference in the mean number of hamburgers sold at the two locations?

show all 5 steps to test hypotheses test

Solution

Given n1=85, xbar1=85.55, s1=15.85
n2=95, xbar2=79.85, s2=12.85

The test hypothesis is
Ho:1-2=0
Ha:1 - 2 not equal to 0

The test statistic is

Z=(xbar1- xbar2)/[s1^2/n1+s2^2/n2]

=(85.55-79.85)/sqrt(15.85^2/85 +12.85^2/95 )

=2.63

Given a=0.05, the critical value is |Z(0.025)|=1.96 (check standard normal table)

Since Z=2.63 is larger than 1.96, we reject Ho.

So we can conclude that there is a difference in the mean number of hamburgers sold at the two locations