Whole Foods is considering opening two new stores in a city

Whole Foods is considering opening two new stores in a city. Since the prices for their products tends to be higher, they want to know the income levels of the different locations they are considering. The mean annual family income for a sample of 15 people in the first location is $155583, with a standard deviation of $42164. A corresponding sample of 27 people at a secondlocation had a mean of $171497, with a standard deviation of $29840.

Given the result from part (a) on variances, test whether there is a difference in the average income between the two locations. Use a 0.01 significance level. (If you need it, the ugly degrees of freedom are 49.) Show work as stated in the quiz directions.

(If you have difficulty with large numbers, you can divide the incomes by 1000. For example, if income is 55,000, then use 55.000 for your calculations.) You just have to do it to mean and stdev - not n.)

Solution

Null Hypothesis : Mean1 = Mean2

Alternate Hypothesis : Mean1 != Mean2

n1 = 15 , n2 =27

SD1 = 42164 , SD2 = 29840

Mean1 = 155583 , Mean2 = 171497

z = (Mean1 - Mean2)/ sqrt(SD^2/n1 + SD^2/n2)

= ( 155583 - 171497 ) / ((42164)^2/ 15 + (29840^2)/27)

= -0.0001

Alpha = 0.02

From z table since it is two tailed test :

P = 2*(1 - 0.5) = 1

Since P > 0.02

Failed to reject the null hypothesis.