Answer to Question #348825 in Statistics and Probability for Tino

Question

Determine the given and compute the test statistic of the problem below using Central Limit Theorem, and construct the rejection region for each.

A company claimed that their N95 face mask has a mean filtration efficiency rate of 95%. A group of student researcher wanted to verify this claim. They bought and tested 40 of their N95 face masks. They found out that the average filtration efficiency rate of these face mask was 90% with a standard deviation of 4%. Test the claim at 5% level of significance and assume that the population is approximately normally distributed.

Answer 1

If the population is normal, then the sampling distribution of the sample means will be approximately normal even for samples of size less than 30).

The following null and alternative hypotheses need to be tested:

$H_0:\mu=95$

$H_1:\mu\not=95$

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 $\alpha = 0.05,$ $df=n-1=39$ and the critical value for a two-tailed test is $t_c =2.022691.$

The rejection region for this two-tailed test is $R = \{t:|t|>2.022691\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{90-95}{4/\sqrt{40}}=-7.9057$$

Since it is observed that $|t|=7.9057>2.022691=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for two-tailed, $df=39$ degrees of freedom, $t=-7.9057$ is $p=0,$ and since $p= 0<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

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

is different than 95, at the $\alpha = 0.05$ significance level.