Answer to Question #210265 in Statistics and Probability for Michell

Question

A seller claimed that her lip tint has a mean organic content of 90%. Arrival seller ask 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 that the population is approximately normally distributed

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=90$

$H_1:\mu\not=90$

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 $\alpha=0.01,$ 

$df=n-1=60-1=59$ ​​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:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{85-90}{5/\sqrt{60}}\approx-7.7460$$

Since it is observed that $|t|=7.7460>2.661759=t_c,$ 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 90, at the $\alpha=0.01$ significance level.

Using the P-value approach: The p-value for two-tailed, $\alpha=0.01, df=59,$

$t=-0.10396$ 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.

Therefore, there is enough evidence to claim that the her lip tint has a mean organic content different than 90%, at the $\alpha=0.01$ significance level.