Question 1182224: 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 is 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 by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! Despite the proportion, this is not a 1-sample proportion test but a 1-sample t-test. The mean is presumably the mean of the percentages, rather than the percent 90% or more.
Ho: mu=90%
Ha: mu NE 90%
alpha=0.01 p{reject Ho|Ho true}
test is a t(0.995, df=59)
critical value is |t|>2.664
calculation t=(85-90)/5/sqrt(60)
=-5*sqrt(60)/5
=-7.75
reject Ho, the mean is not 90%, p-value about 1 x 10^-10
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