Answer to Question #112547 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

The following null and alternative hypotheses need to be tested:

$H_0:\mu=60000$

$H_1:\mu\not=60000$

This corresponds to a two-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

Based on the information provided, the significance level is $\alpha=0.05,$ and the critical value for a two-tailed test is $z_c=1.96.$

The rejection region for this two-tailed test is $R=\{z:|z|>1.96\}$

The z-statistic is computed as follows:

$$z={\bar{X}-\mu\over \sigma/\sqrt{n}}={47000-60000\over 7000/\sqrt{130}}\approx-21.175$$

Since it is observed that $|z|=21.175>1.96,$ it is then concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is different than 60000, at the 0.05 significance level.

Using the P-value approach: The p-value is $p<0.00001,$ and since $p<0.00001<0.05,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is different than 60000, at the 0.05 significance level.