Answer to Question #103121 in Statistics and Probability for Biraj Chhetri

Question

The mean weekly sales of soap bars in departmental stores was 146.3 bars per store. After an advertising campaign the mean weekly sales in 22 stores for a typical week increased to 153.7 and showed a standard deviation of 17.2. Was the advertising campaign successful?

Answer 1

The provided sample mean is $\bar{x}=153.7$ and the sample standard deviation is $s=17.2,$ and the sample size is $n=22.$

The following null and alternative hypotheses need to be tested:

$H_0: \mu=146.3$

$H_1: \mu>146.3$

This corresponds to a right-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used.

Based on the information provided, the significance level is $\alpha=0.05, df=n-1=22-1=21$ and the critical value for a right-tailed test is $t_c=1.721$

The rejection region for this right-tailed test is $R=\{t:t>1.721\}.$

The t-statistic is computed as follows:

$$t={\bar{x}-\mu_0 \over s/\sqrt{n}}={153.7-146.3 \over 17.2/\sqrt{22}}\approx2.018$$

Since it is observed that $t=2.018>1.7221=t_c,$ then it is concluded that the null hypothesis is rejected. Therefore, there is enough evidence to claim that the population mean is greater than 146.3, at the 0.05 significance level.

Using the P-value approach: 

$t=2.018, df=22-1, \alpha=0.05, one-tailed$

The p-value is $p=0.028274,$ and since $p=0.028274<0.05=\alpha,$ it is concluded that the null hypothesis is rejected. Therefore, there is enough evidence to claim that the population mean is greater than 146.3, at the 0.05 significance level.

Hence the advertising campaign was successful in promoting sales.