Answer to Question #165162 in Statistics and Probability for Mohanakrishnan

Question

The mean weekly sales of soap bars in departmental stores was 146.3 bars per store.       

   After 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

We have that

$\mu=146.3$

$n=22$

$\bar x=153.7$

$s=17.2$

Let the significance level be 5% in this test.

$\alpha=0.05$

$H_0:\mu = 146.3$

$H_a:\mu >146.3$

The hypothesis test is right-tailed.

Since the population standard deviation is unknown we use the t-test.

The critical value for 5% significance level and 21 df is 1.721

(degrees of freedom df = n – 1 = 22 – 1 = 21)

The critical region is t > 1.721

Test statistic:

$$t=\frac{\bar x - \mu}{\frac{s}{\sqrt n}}=\frac{153.7- 146.3}{\frac{17.2}{\sqrt {22}}}=2.018$$

Since 2.018 > 1.721 thus t falls in the rejection region we reject the null hypothesis.

At the 5% significance level the data do provide sufficient evidence to support the claim. We are 95% confident to conclude that that the advertising campaign successful.