A certain brand of apple juice is supposed to have 64 ounces

A certain brand of apple juice is supposed to have 64 ounces per bottle. Because the penalty for under-filling bottles is severe, the target mean amount of juice is 64.05 ounces. However, the filling machine is not precise, and the exact amount of juice varies from bottle to bottle. The quality-control manager wishes to verify that the mean amount of juice in each bottle is 64.05 ounces so that she can be sure that the machine is not over- or under-filling the bottles. She randomly selects a sample of 22 bottles of juice, measures the content. And obtains the following data: a) Because the sample size is small, she must verify that the amount of juice is normally distributed and that the sample contains no outliers. Construct a normal probability plot and a modified box and whisker plot (attach if you wish). Can normality be assumed without reservation? b) Test the claim that the true average is equal to 64.05 ounces at the 0.01 level of significance. In your hypothesis test, be sure to include the following: Null and alternative hypothesis The type of test you used, The calculated test statistic The P-Value The decision c) Construct an appropriate confidence interval for this situation and comment how it supports your decision in (b). d) Based on (b) and (c). should the assembly line be shut down so that the machine can be re-calibrated? Explain your answer. e) Why was a 0.01 level of significance selected as opposed to 0.05 or 0.10? Explain your answer.

Solution

a) Normal probability plot is attached and we see that the points are forming a staright line (approx), so we can say that the data is normally distributed

b) Null hypothesis H0: true average=64.05

Alternative hypothesis H1: true average not equal to 64.05

We will use t test for single mean (two tailed) as the sample is small.

We put the data in SPSS, and find the following result table

           One-Sample Test
           Test Value = 64.05   
99% Confidence Interval of the Difference
t   df   Sig. (2-tailed)   Mean Difference   Lower   Upper
data   -4.490   21 .000   -.04273 -.0697   -.0158

The calcluated t statistic is -4.490

P value is 0.000 which is less than 0.01, so we reject the null hypothesis and conclude that true average is not equal to 64.05.

c) The 99% confidence interval is (63.98,64.03)

d) Based on the results in b) and c), we can say that assembly line should be shut down so that the machine can be re-caliberated. In testing also, we accepted the fact that true average is not 64.05 and confidence intervals also show that out 100, 99 times the true average will lie between (63.98,64.03) which does not even consist of 64.05.

e) We have used 0.01 level of significance beacuse in such precise situation, one can\'t afford to take the risk of 5% or 10%.