Answer to Question #154307 in Statistics and Probability for Ali Ahmed

Question

Eleven sales executive trainees are assigned selling jobs right after their recruitment.
After a fortnight they are withdrawn from their field duties and given a month’s
training for executive sales. Sales executed by them in thousands of rupees before
and after the training, in the same period are listed below:
Sales Before
Training
Sales After
Training
23 24
20 19
19 21
21 18
18 20
20 22
18 20
17 20
23 23
16 20
a. Do these data indicate that the training has contributed to their performance?
b. Find the 95 per cent confidence Interval of the training

Answer 1

beforeafter d d-mean(d) (d-d(mean)^2

23 24 -1 0.2 0.04

20 19 1 2.2 4.84

19 21 -2 -0.8 0.64

21 18 3 4.2 17.64

18 20 -2 -0.8 0.64

20 22 -2 -0.8 0.64

18 20 -2 -0.8 0.64

17 20 -3 -1.8 3.24

23 23 0 1.2 1.44

16 20 -4 -2.8 7.84

total -12 0 37.6

The mean of the differences is ;

$\bar d=\frac{\sum d}{n}$

$=\frac{-12}{10}$

=-1.2

the sample standard deviation for the mean differences is;

$s=\sqrt\frac{\sum(d_i-\bar d)^2}{n-1}$

=$\sqrt\frac{37.6}{10-1}$

=2.044

a) to test whether training has contributed to their performance, we test the following hypotheses;

H0: the mean difference is equal to zero

H1: the mean difference is not equal to zero

t=$\frac{\bar d-0}{\frac{s}{\sqrt n}}$

=$\frac{-1.2}{\frac{2.044}{\sqrt 10}}$

=-1.856

we assume a 5% level of significance;

$t_{\alpha/2},_9=2.262$ from the t distribution tables.

if the t-critical is greater than the test statistic the null hypothesis is rejected.

|-1.856|<2.262 .

we fail to reject the null hypothesis at 5% level of significance since the test statistic is less than the critical value.

b) 95% confidence interval

$CI= \bar d ±t_{\alpha/2}*\frac{s}{\sqrt n}$

=$-1.2±2.262*\frac{2.044}{\sqrt{10}}$

=(-2.662, 0.262)

we are 95% confident that the mean differences from the training is between -2.662 and 0.262.