Answer:
Since the calculated value of t= 2.249 does not falls in the rejection region we therefore accept the null hypothesis at 5 % significance level . On the basis of this we conclude that the bonus plan has not increased sales significantly.
Step-by-step explanation:
The null and alternative hypotheses as
H0: μd=0 Ha: μd≠0
Significance level is set at ∝= 0.05
n= 6
degrees of freedom = df = 6-1 = 5
The critical region is t ( base alpha by 2 with df=5) ≥ ± 2.571
The test statistic under H0 is
t = d/ sd/ √n
Which has t distribution with n-1 degrees of freedom
Sales Difference
Person After Before d = after - before d²
1 94 90 4 16
2 82 84 -2 4
3 90 84 6 36
4 76 70 6 36
5 79 80 -1 1
6 85 80 5 25
∑ 18 118
d`= ∑d/n= 18/6= 3
sd²= 1/6( 118- 18²/6) = 1/6 ( 118 - 54) = 10.67
sd= 3.266
t= 3/ 3.266/ √6
t= 2.249
Since the calculated value of t= 2.249 does not falls in the rejection region we therefore accept the null hypothesis at 5 % significance level . On the basis of this we conclude that the bonus plan has not increased sales significantly.
