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 64 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.

Answer 1

A grocery store has an average sales of $8000 per day. The store introduced several advertising campaigns in order to increase sales. A sample of 64 days of sales was selected. It was found that the average was $8300 per day. It is known that the standard deviation of the population is $1200.

From given, we have,

The null hypothesis,

H0 : μ = 8000

The alternative hypothesis,

H1 : μ > 8000

Average = $\bar {x}$ = 8300

Population standard deviation = σ = 1200

Sample size = n = 64

z - test is given by,

$z = \dfrac{\bar {x} - \mu} {\frac{\sigma }{\sqrt n} }$

$z = \dfrac{8300 - 8000} {\frac{1200 }{\sqrt {64}} }$

$z = \dfrac{300} {\frac{1200 }{8} }\\\\z = \dfrac{300}{150}\\\\z = 2$

Therefore, the test statistics z = 2

The P value at α = 0.05 is given by,

P - value  = P (z > 2)

0.02275