Answer to Question #342415 in Statistics and Probability for "Ning"

Question

a seller claim that her lip tint has a mean organic content of 90%. A rival seller asked 60 users of that lip tint and found that it has a mean organic content of 85% with a standard deviation of 5%. test the claim at 1% level of significance and assume tha the population is approximately normally distributed

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=90$

$H_a:\mu\not=90$

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 $\alpha = 0.01,$ $df=n-1=59$ degrees of freedom, and the critical value for a two-tailed test is $t_c =2.661759.$The rejection region for this two-tailed test is $R = \{t:|t|> 2.661759\}.$

The t-statistic is computed as follows:

$$t=\dfrac{85-90}{5/\sqrt{60}}=-7.746$$

Since it is observed that $|t| = 7.746>2.661759=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for two-tailed $df=59$ degrees of freedom, $t=-7.746$ is $p=0,$ and since $p=0<0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is different than 90, at the $\alpha = 0.01$ significance level.