Question

A local winery bottles thousands of bottles of wine per season. The winery has a machine that automatically dispenses the amount of wine per bottle before aging. Each season the wine maker randomly samples 50 bottles of wine to ensure the amount of wine per bottle is 750 ml. If there is evidence that the amount is different than (less or more than) 750 ml then the winery will need to evaluate the machine and perhaps rebottle or consider selling the wine at a discount. The sample yields a mean and standard deviation of 752 ml and 11 ml. Use a significance level of 0.05.

State: Is there sufficient evidence that the average fill of the wine bottles is different than 750 milliliters?

Plan:

a.State the null and alternative hypotheses to answer the question of interest.

b. What type of test is appropriate to answer the question of interest and why?

c. State the level of significance.

Solve:

d. Calculate the test statistic. State the degrees of freedom and p-value.

e.Calculate a 95% confidence interval for µ.

Conclude:

f. Interpret the results from the hypothesis test and confidence interval in the context of the problem. Use the four step process described at the beginning of the activity to write a four part conclusion.   

Answer 1

a) Below are the null and alternate hypothesis
H0: mu = 750
H1: mu not equals to 750

b)
As we do not know the population std. dev. , we use t-test (two tailed)

c)
Significance level is 0.05

d)
Test statistics,
t = (752 - 750)/(11/sqrt(50)) = 1.2856

df = 50 - 1 = 49

p-value = 0.204608617 (=2*T.DIST.RT(1.2856,49))

e)
As p-value is greater than the significance level of 0.05, we failed to reject the null hypothesis.
This means there are not sufficient evidence to conclude that, average fill of wine is different than 750ml