Question

The mean weekly a sales of soap bars in different departmental stores was 146.3 bars per store. after an advertising campaign the mean weekly increased to 153.7 and showed a standard deviation of 72.2. Was the advertising campaign successful at 5% level of significance? You may like to use the values given at end.

Answer 1

Answer on Question #47511 – Math – Statistics and Probability

The mean weekly a sales of soap bars in different departmental stores was $\mu_0 = 146.3$ bars per store. after an advertising campaign the mean weekly increased to $\mu = 153.7$ and showed a standard deviation of $\sigma = 72.2$. Was the advertising campaign successful at $5\%$ level of significance? You may like to use the values given at end.

Solution

Null hypothesis $H_0: \mu = \mu_0$, i.e., the advertising campaign isn't successful.

Alternative hypothesis $H_1: \mu > \mu_0$ (Right tail), the advertising campaign is successful.

Under $H_0$ the test statistic is

$$z = \frac{\mu - \mu_0}{\sigma} = \frac{153.7 - 146.3}{72.2} = 0.10.$$

Since calculated value of $z = 0.10$ is lower than tabulated value of $z_{0.05} = 1.645$ at $5\%$ level of significance we don't reject $H_0$. The advertising campaign wasn't successful in promoting sales.

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