The mean monthly sale of a particular household item in department stores was observed as 133.3 units per store. After an advertising campaign, data on sales of this item from 25 stores were collected and it is found that the mean sales have increased to 141.5 units with a standard deviation of 15.2 units. Was the advertising campaign successful? Assume that level of significance is 5%.

Solution;

Using a one sided Hypothesis test;

Given;

$\bar{X_1}=133.3$

$\bar{X_2}=141.5$

n=25

$s=15.2$

$\alpha=0.05$

Take ;

$H_o:\bar{X_2}-\bar{X_1}<0$

$H_1:\bar{X_2}-\bar{X_1}>0$

Degrees of freedom;

$df=n-1=25-1=24$

From t tables;

$t_{cr}=t_{0.05/24}=1.711$

Computed t is;

$s_{\bar x}=\frac{s}{\sqrt n}=\frac{15.2}{\sqrt 25}=3.04$

$t_c=\frac{\bar X_2-\bar X_2}{s_{\bar x}}=\frac{141.5-133.3}{3.04}=2.697$

Since $t_{cr}<t_c$ ,we can't reject the null hypothesis ,there we say that there was improvement after advertisement.