Question

_____ A certain brand of apple juice is supposed to have 64 ounces of juice. Because the penalty for underfilling 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 underfilling. She randomly samples 22 bottles of juice, measures the contents, and obtains the following data.

63.03	63.94	63.95	63.95	63.95	63.97	63.97	63.98	63.98	63.99	64.00
64.01	64.01	64.01	64.01	64.02	64.05	64.05	64.05	64.06	64.08	64.10

1. Produce a boxplot for this data, and comment on whether you think the data is normally distributed (this is one of the requirements of the test).
2. Find the mean and standard deviation of the sample.
3. What are the observational units?
4. What is the variable collected? What type of variable is it?
5. Write out the null and alternative hypotheses for testing whether this brand of apple juice meets the target mean of 64.05 ounces.
6. Find the test statistic for the test.
7. Find the p-value for the test.
8. Using significance level 0.01, what is your conclusion? Do you reject the null or fail to reject the null?
1. State your conclusion in words.

Answer 1

(a)

The data appears to be normally distributed (Has a slight right skew)

(b) Mean = 63.96, Standard deviation = 0.213

(c) Observational units are the bottles of juice

(d) Variable collected are the amounts of juice filled by the machine. It is a numerical variable on interval/ratio level.

(e) Ho: μ = 64.05 and Ha: μ ≠ 64.05

(f to i)

Data:    

n = 22   

μ = 64.05   

s = 0.213   

x-bar = 63.96   

Hypotheses:   

Ho: μ = 64.05   

Ha: μ ≠ 64.05   

Decision Rule:   

α = 0.01   

Degrees of freedom = 22 - 1 = 21

Lower Critical t- score = -2.831359554

Upper Critical t- score = 2.831359554

Reject Ho if |t| > 2.831359554

Test Statistic:   

SE = s/√n = 0.213/√22 = 0.045411753

t = (x-bar - μ)/SE = (63.96 - 64.05)/0.045411752583745 = -1.98187

p- value = 0.060734414   

Decision (in terms of the hypotheses):

Since 1.981865814 < 2.83136 we fail to reject Ho

Conclusion (in terms of the problem):

There is no sufficient evidence that the mean fill amount is different from 64.05 ounces.