Question

Suppose a company sells liquid soap in bottles that should contain 7 ounces of soap. A quality control manager takes a random sample of 40 bottles to check whether the machine which fills the bottles is putting in the proper amount of soap. They will do a hypothesis test. The null hypothesis says that the mean amount of soap going into the bottles equals 7 ounces. The alternative hypothesis says that the mean is different than 7 ounces. a. If you had the sample information, which procedure would you use on your calculator to obtain the P-value for the hypothesis test? (Just tell me what you would pick from the stat, TESTS menu.)

b. Suppose you did that procedure and got a P-value of 0.1678. Would you reject the null hypothesis using a 5% significance level? (In other words, do you have enough evidence that the mean amount of soap going into the bottles is different than 7 ounces?) YES or NO

Answer 1

Solution-A:

Ho:mu=7

Ha:mu not =7

t=xbar-mu/s/sqrt(n)

t=xabr-7/s/sqrt(40)

since population standard deviation is not known use t distribution'

STAT>TESTS>T TEST

Solution-b:

level of significance =alpha=0.05

p=0.1678

p>alpha

Do not reject null hypothesis

we do do have enough evidence that the mean amount of soap going into the bottles is different than 7 ounces

NO