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.