Question

Last year a real estate agent earned an average of 40,250.00 month. suppose you recently selected a random sample of 25 real estate agents you determined how much each of them earns each month your Computations of their earning resulted to an average of 40,400.00 w/ a standard deviation of 225.00 using a 0.01 Level of significance can it be concluded that the average monthly earnings of estate agents has increased? Assume normality in the population.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\leq 40250$

$H_1:\mu>40250$

This corresponds to a right-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

Based on the information provided, the significance level is $\alpha = 0.01,$ $df=n-1=25-1=24$ degrees of freedom, and the critical value for a right-tailed test is $t_c = 2.492159.$

The rejection region for this right-tailed test is $R = \{t: t > 2.492159\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{40400-40250}{225/\sqrt{25}}\approx3.333333$$

Since it is observed that $t = 3.333333 >2.492159= t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for right-tailed $df=24$ degrees of freedom, $t=3.333333$ is $p= 0.001388,$ and since $p = 0.001388 < 0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

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

is greater than $40250,$ at the $\alpha = 0.01$ significance level.

Therefore, there is enough evidence to claim that the average monthly earnings of estate agents has increased, at the $\alpha = 0.01$ significance level.