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 45 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 748 ml and 13 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) As we are trying to test whether the mean is different from 750, therefore this is a two tailed test and the null and the alternate hypothesis here are given as:

b) This is a two tailed test and as we are not given the population standard deviation, therefore this would be a t test which would be done here.

c) The level of significance is 0.05 ( it is given in the problem )

d) The test statistic here is computed as:

Therefore -1.0320 is the test statistic value here.

The degrees of freedom is df = n - 1 = 44 degrees of freedom.

As this is a two tailed test, the p-value here is computed from the t distribution tables as:

p = 2P( t₄₄ < -1.0320 ) = 2*0.1538 = 0.3076

Therefore 0.3076 is the p-value here.

e) From t distribution tables, we get:

P( -2.015 < t₄₄ < 2.015 ) = 0.95

Therefore the confidence interval here is obtained as:

This is the required 95% confidence interval.

f) As the p-value here is greater than the level of significance, therefore the test is not significant and we cannot reject the null hypothesis here. Therefore we dont have sufficient evidence here to support the statement that the mean has changed from 750.

Also as the confidence interval contains 750, again it proves the same thing.