Question

A grocery store has an average sales of $8000 per day. The store introduced several advertising campaigns in order to increase sales. To determine whether or not the advertising campaigns have been effective in increasing sales, a sample of 100 days of sales was selected. It was found that the average was $8300 per day. From past information, it is known that the standard deviation of the population is $1200. The p-value is

Answer 1

0.0062

Step-by-step explanation:

p = 1 - P(Z < Zcal)

Zcal = (8300-8000)/ (1200/√100)

Zcal = 2.5

p = 1 - P(Z < Zcal)

= 1 - P(Z < 2.5)

= 1 - 0.9938

= 0.0062