Question

A marketing consultant was hired to visit a random sample of five sporting goods stores across the state of California. Each store was part of a large franchise of sporting goods stores. The consultant taught the managers of each store better ways to advertise and display their goods. The net sales for 1 month before and 1 month after the consultant's visit were recorded as follows for each store (in thousands of dollars):_________.
Before visit: 57.1 94.6 49.2 77.4 43.2After visit: 63.5 101.8 57.8 81.2 41.9Do the data indicate that the average net sales improved? (Use a= 0.05)

Answer 1

Answer:

Step-by-step explanation:

Corresponding net sales before 1 month and after 1 month form matched pairs.

The data for the test are the differences between the net sales before and after 1 month.

μd = the​ net sales before 1 month minus the​ net sales after 1 month.

Before after diff

57.1 63.5 - 6.4

94.6 101.8 - 7.2

49.2 57.8 - 8.6

77.4 81.2 - 3.8

43.2 41.9 1.3

Sample mean, xd

= (- 6.4 - 7.2 - 8.6 - 3.8 + 1.3)/5 = - 4.94

xd = - 4.94

Standard deviation = √(summation(x - mean)²/n

n = 5

Summation(x - mean)² = (- 6.4 + 4.94)^2 + (- 7.2 + 4.94)^2 + (- 8.6 + 4.94)^2+ (- 3.8 + 4.94)^2 + (1.3 + 4.94)^2 = 60.872

Standard deviation = √(60.872/5

sd = 3.49

For the null hypothesis

H0: μd ≥ 0

For the alternative hypothesis

H1: μd < 0

The distribution is a students t. Therefore, degree of freedom, df = n - 1 = 5 - 1 = 4

2) The formula for determining the test statistic is

t = (xd - μd)/(sd/√n)

t = (- 4.94 - 0)/(3.49/√5)

t = - 3.17

We would determine the probability value by using the t test calculator.

p = 0.017

Since alpha, 0.05 > than the p value, 0.017, then we would reject the null hypothesis. Therefore, at 5% significance level, the data indicate that the average net sales improved.