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# Answer to Question #211459 in Engineering for joe

Question #211459

the mean weekly sales of soap bars in departmental store was 146.3 bars per store after an advertising campaign the mean weakly sale in 22 store for a typical week increased to153.7and showed a standard deviation of 17.2 was the advertising campaign successful?

1
2021-06-29T00:31:32-0400

Null hypothesis "H_0:\\mu=\\mu_0", i.e., the advertising campaign isn’t successful.Alternativehypothesis "H_1:\\mu>\\mu_0" (Right tail), the advertising campaign is successful.Under H0 the test statistic is

"T=\\frac{\\mu-\\mu_0}{s}\\sqrt{n}=\\frac{153.7-146.3}{17.2}\\sqrt{22}=2.02"

Since it is observed that

"t=2.02>1.72=t_c,t=2.02>1.72=t \n\nc"

, 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.

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.

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