Answer to Question #334844 in Statistics and Probability for van

Question

A sociologist believes that it costs more than Php90,000 to raise a child from birth to age one. A random sample of 49 families, each with a child is selected to see if this figure is correct. The average expenses for these families reveal a mean of Php92,000 with a standard deviation of Php4,500. 

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\le90000$

$H_a:\mu>90000$

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 $\alpha = 0.05,$ $df=n-1=49-1=48$ degrees of freedom, and the critical value for a right-tailed test is $t_c = 2.010635.$

The rejection region for this right-tailed test is$R = \{t: t > 2.010635\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{92000-90000}{4500/\sqrt{49}}\approx3.111111$$

Since it is observed that $t =3.111111> 2.010635= t_c ,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for right-tailed, $df=48$ degrees of freedom, $t=3.111111$ is $p=0.001567,$ and since $p=0.001567<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is greater than 90000, at the $\alpha = 0.05$ significance level.