Question

a seller claimed 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 assumes that the population is approximately normally distributed

Answer 1

Hypothesized Population Mean $\mu=90$

Sample Standard Deviation $s=5$

Sample Size $n=60$

Sample Mean $\bar{x}=85$

Significance Level $\alpha=0.01$

Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

$H_0: \mu\geq90$

$H_1: \mu<90$

This corresponds to left-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 fredom, and the critical value for two-tailed test i$t_c=2.661759.$

The rejection region for this left-tailed test is $R=\{t:t<-2.391229\}.$

The $t$ - statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{85-90}{5/\sqrt{60}}\approx-7.745967$$

Since it is observed that $t=-7.745967<-2.391229=t_c,$ it is then concluded that the null hypothesis is rejected.

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

Using the P-value approach: The p-value for left-tailed, the significance level $\alpha=0.01, df=59, t=-4.47214,$ is $p<0.00001,$ and since $p<0.00001<0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

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