Question

(3) The Eastern Publishing Company hires college students to sell books during the summer The company wants to evaluate the effectiveness of measure its efficiency, a company representative randomly records for each person the number of sales completed during the past summer, and indicates whether the salesperson participated in the new training program. Assume equal variances. The results are the new training program. To selects a group of salespeople, Sample 1 Mean 482 books Std Dev 43 books n 22 (training) Sample 2 Mean 465 books Std Dev 39 books n 27 (no training) a. State the null and alternative hypotheses. b. State the Decision Rule for a .05 level of significance. Do these data present sufficient evidence to indicate a difference in the average number of books sold at the .05 significance level? c.

Answer 1

a.

$\small \mu _{1}$ : Population average number of books sold by sales person who has participated in the new training

$\small \mu _{2}$ : Population average number of books sold by sales person who has not participated in the new training

a. Null hypothesis : H_o: There is no difference in the average number of books sold ;$\small \mu _{1}-\mu _{2}=0$

Alternate Hypothesis : H_a : There is difference in the average number of books sold ;$\small \mu _{1}-\mu _{2}\neq 0$ (Two Tailed test)

Equal Variance assumed.

Given,

Given
n1 : Sample Size of Sample 1(Training)	22
n1 : Sample Size of Sample 2 (No Training)	27
$\small \overline{x}_{1}$: Sample Mean of Sample 1	482
$\small \overline{x}_{2}$ : Sample Mean of Sample 2	465
s1 : Sample Standard Deviation of Sample 1	43
s2 : Sample Standard Deviation of Sample 2	39
Level of Significance :	0.05
Degrees; of; freedom : ( n1+n2-2=22+27-2=47)	47

b. Decision rule

For two tailed test :

For Two tailed test; Fail to reject null hypothesis if Calculated Value of 't' is within critical values i.e  $\small t_{\alpha /2}$  and $\small t_{1-\alpha /2}$

$\small t_{\alpha /2,df} = t_{0.025,47} = -2.0117$

$\small t_{1-\alpha /2} = t_{0.975,47} = 2.0117$

c.

$\small Pooled\; \; Sample\; \; Standard\; deviation\; \; :s_{p}=\sqrt{\frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2}}$

$\small s_{p}= \sqrt{\frac{(22-1)43^{2}+(27-1)39^{2}}{22+27-2}}=\sqrt{\frac{78375}{47}}=\sqrt{1667.5532}=40.8357$

$\small Test\; \; Statistic : t=\frac{\overline{x_{1}}-\overline{x_{2}}}{s_{p}\sqrt{\frac{1}{n_{1}}+ \frac{1}{n_{2}}}}$

$\small =\frac{(482-465)}{40.8357\times \sqrt{\frac{1}{22}+\frac{1}{27}}}=\frac{17}{11.7286}=1.4495$

As Calculated Value is with in the Critical Values i.e.( -2.0117 < 1.4495 < 2.0117 ) Fail To Reject Null Hypothesis

There is not sufficient evidence to indicate a difference in the average number of books sold at the 0.05 significance level.