Answer to Question #195051 in Statistics and Probability for Ecra

Question

Past experience indicates that the monthly long distance telephone bill per household in a particular community is normally distributed, with a mean of Sh. 1012 and a standard deviation of Sh. 327. After an advertising campaign that encouraged people to make long distance telephone calls more frequently, a random sample of 57 households revealed that the mean monthly long distance bill was Sh. 1098. Can we conclude at the 10% significance level that the advertising campaign was successful

Answer 1

$\mu = 1012 \\ \sigma= 327 \\ \bar{x}=1098 \\ n=57 \\ H_0: \mu=1012 \\ H_1: \mu > 1012 \\ z=\frac{\bar{x}-\mu}{\sigma/ \sqrt{n}} \\ = \frac{1098-1012}{327/ \sqrt{57}} = 1.985$

The calculated p-value can be calculated as follows.

$P= 1-P(Z<1.985) \\ = 1 -0.9764 = 0.0236$

The calculated p-value is 0.0236. From the p-value, the alternate hypothesis is accepted. We can conclude at the 10% significance level that the advertising campaign was successful.