Answer to Question #345525 in Statistics and Probability for Tine

Question

A taxi driver claims that his average monthly income is Php 2, 800.00 with a standard deviation of Php 250.00 . A sample of 25 drivers were surveyed and found that their average monthly income is Php 3,500.00with a standard deviation of Php 350.00.Test the hypothesis at 1% level of significance.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=2800$

$H_1:\mu\not=2800$

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 $\alpha = 0.01,$ and the critical value for a two-tailed test is $z_c = 2.5758.$

The rejection region for this two-tailed test is $R = \{z:|z|>2.5758\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{3500-2800}{250/\sqrt{25}}=14$$

Since it is observed that $|z|=14>2.5758=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=2P(z>14)= 0,$ and since $p= 0<0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$

is different than 2800, at the $\alpha = 0.01$ significance level.