Answer to Question #335285 in Statistics and Probability for sheshe

Question

A supermarket owner believes that the mean income of its customers is P50,000 per month.One-hundred customers are randomly selected and asked of their monthly income.The sample mean is P48,500 per month and the standard deviation is P3,200.Is there sufficient evidence to indicate that the mean income of the customers of the supermarket is P50,000 per month? Use a=0.05

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=50000$

$H_a:\mu\not=50000$

This corresponds to a two-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 $\alpha = 0.05,$ $df=n-1=99$ degrees of freedom, and the critical value for a two-tailed test is $t_c = 1.984217.$

The rejection region for this two-tailed test is $R = \{t: |t| > 1. 1.984217\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{48500-50000}{3200/\sqrt{100}}=-4.6875$$

Since it is observed that $|t| = 4.6875 >1.1984217= t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for two-tailed, $df=99$ degrees of freedom, $t=-4.6875$ is $p=0.000009,$ and since $p=0.000009<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

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

is different than 50000, at the $\alpha = 0.05$ significance level.