Answer to Question #111824 in Statistics and Probability for Mikhail Reony

Question

A market research company informs a prospective mini marker entrepreneur that the average income per household in the region is RM60 000 per annum. The household income is assumed to be normally distributed with standard deviation of RM7000, based on an early study. For a random sample of 130 households, the mean household income is found to be RM47 000. Test the null hypothesis that the population mean household income is RM60 000 at 5 percent significance level.

Answer 1

Null hypothesis $H_0:\mu=60000$

Alternative hypothesis $H_a:\mu\ne60000$

Test statistic: $z=\frac{\bar x-\mu}{\frac{\sigma}{\sqrt{n}}}=\frac{47000-60000}{\frac{7000}{\sqrt{130}}}=-21.17$

P-value: $p=P(Z<-21.17)+P(Z>21.17)<0.0001$

Since the P-value is less than 0.05, reject the null hypothesis

There is a sufficient evidence that the mean income is different from RM60000