Question

A company is trying out a new marketing plan. Prior to the new campaign, store sales per week were $5000. The new method is trialled in 15 randomly selected stores. The sample provided sales figures with a sample mean of $5500 and a standard deviation of $800. Can the company conclude that the new marketing plan works? Use ? = 0.01 and the following template to answer the question.

HO:

H1:
Level of Significance:

Test Statistic and Associated Sampling Distribution:

Calculation and Decision:

Conclusion:

What assumptions need to be made in order to undertake this test?

Indicate the test statistic and p-value on a bell curve.

 

Answer 1

H₀: = 5000

H₁: > 5000

Level of significance = 0.01

The test statistic t = ()/(s/sqrt(n))

                             = (5500 - 5000)/(800/sqrt(15))

                             = 2.42

DF = 15 - 1 = 14

P-value = P(T > 2.42)

             = 1 - P(T < 2.42)

             = 1 - 0.9851 = 0.0149

As the P-value is greater than the alpha value (0.0149 > 0.01), so the null hypothesis is not rejected.

So the company cannot conclude that the new marketing plan works.

We should assume that the population from which the samples are drawn must be normally distributed.