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Ne4ueva [31]
4 years ago
15

The following shows the monthly sales in units of six salespersons before and after a bonus plan was introduced. At 95% confiden

ce, determine whether the bonus plan has increased sales significantly. (For the following matched samples, let the difference "d" be: d = after - before.)
Salesperson After Before
1 94 90
2 82 84
3 90 84
4 76 70
5 79 80
6 85 80
Mathematics
1 answer:
Anarel [89]4 years ago
7 0

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

<u>6                  85          80                 5                       25   </u>

∑                                                       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.

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