Question

a random sample of 150 worker with children in day care show a man day a care cost of r. s 3700 a standard deviation is r. s 600 verified the department claim that the mean is equal to r. s3500 at the 0.05 level of significance

Answer 1

The provided sample mean is $\bar{x}=3700$ and the sample standard deviation is $s=600.$

The sample size is $n=150.$ The number of degrees of freedom are $df=n-1=149,$ and the significance level is $\alpha=0.05.$

The following null and alternative hypotheses need to be tested:

$H_0: \mu=3500$

$H_1: \mu\not=3500$

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

Based on the information provided, the critical value for a two-tailed test is $t_c=1.976013.$

The rejection region for this two-tailed test is $R=\{t:|t|>1.976013\}$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{3700-3500}{600/\sqrt{150}}=4.082483$$

Since it is observed that $t=4.082483>1.976013=t_c,$ it is then concluded that the null hypothesis is rejected. Therefore, there is enough evidence to claim that the population mean $\mu$ is different than 3500, at the 0.05 significance level.

Using the P-value approach: The p-value $df=149, t=4.082483,$ $two-tailed,$ is $p=0.000072,$ and since $p=0.000072<0.05,$ it is concluded that the null hypothesis is rejected. Therefore, there is enough evidence to claim that the population mean $\mu$ is different than 3500, at the 0.05 significance level.