Answer to Question #278217 in Statistics and Probability for Satyo

Question

The head of BPS Jakarta stated that the average consumption/capita of his regional household was USD 10,500,000 with a standard deviation of USD 2500,000. An economic research institute conducted a survey by selecting a random sample of 1000 households. The survey results show that the average consumption/capita is USD 12,000,000. Test it with an alpha of 2%, can you conclude that in fact the average consumption/capita is higher than that stated by the Head of BPS?

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\leq10500000$

$H_1:\mu>10500000$

This corresponds to a right-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 $\alpha = 0.02,$ and the critical value for a right-tailed test is $z_c =2.0537.$

The rejection region for this left-tailed test is $R = \{z: z >2.0537\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{12000000−10500000}{2500000/\sqrt{1000}}\approx18.9737$$

Since it is observed that $z = 18.9737>2.0537=z_c ,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=P(Z>18.9737)=0,$ and since $p=0<0.01=\alpha,$  it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that in fact the average consumption/capita is higher than that stated by the Head of BPS, at the $\alpha = 0.02.$ significance level.