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
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:
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.
Comments
Leave a comment