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.
