Information A machine fills bottles of chocolate milk that should weigh 16 ounce
ID: 3133774 • Letter: I
Question
Information A machine fills bottles of chocolate milk that should weigh 16 ounces. There is a small amount of (unavoidable) variability from bottle to bottle, so that the standard deviation for the actual bottle weights is = 0.22 ounces. If the bottles are underfilled, consumers will complain. If the bottles are overfilled, the bottling company will lose profit.
Here are the actual weights of a sample of bottles of chocolate milk:
15.9 15.9 16.5 15.9 16.2 16.2 16.2 16.3 15.6 15.9
15.7 16.4 16.4 16.3 16.2 16.4 16.6 15.9 15.3 16.1
“Do these data give good reason to think that the mean weight of actual chocolate milk in the bottles is different from the label value of 16 ounces?”
Question 1 We are going to answer this question by running a hypothesis test forµ = the mean .
Question 2 STATE the appropriate null and alternative hypotheses to be studied.
Question 3 Find the value of the sample mean x =
Question 4 Find the value of the test statistic z =
Question 5 Find your z-score in table A. The table A entry is
Question 6 1 – (the table A entry) is
Question 7 Consider the appropriate alternative hypothesis for this problem. The P-value for our hypothesis test is
Question 8 Is the sample statistically significant at the 10% level?
Question 9 Is the sample statistically significant at the 5% level?
Question 10 STATE a conclusion appropriate for your hypothesis test. This means one or two complete sentences on the topics of this question.
Question 11 What (other) assumption do we have to make about this sample data in order for the conclusion from the hypothesis test based on this data to be valid?
Question 12 What have you proven with this hypothesis test?
Explanation / Answer
2.
Formulating the null and alternative hypotheses,
Ho: u = 16
Ha: u =/ 16 [ANSWER]
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3.
X = sample mean = 16.095 [ANSWER]
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4.
Getting the test statistic, as
X = sample mean = 16.095 [ANSWER]
uo = hypothesized mean = 16
n = sample size = 20
s = standard deviation = 0.22
Thus, z = (X - uo) * sqrt(n) / s = 1.93 [ANSWER]
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5.
P(z<1.93) = 0.0268 [ANSWER]
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6.
P(z>1.93) = 1 - 0.0268 = 0.9732 [ANSWER]
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7.
As this is a two tailed test,
P value = 0.0268*2 = 0.0536 [ANSWER]
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8.
As P < 0.10, yes, it is significant.
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9.
As P > 0.05, no, it is not significant.
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10.
For 0.10 level:
There is significant evidence at 0.10 level that the true mean filling is not 16 oz.
For 0.05 level:
There is no significant evidence at 0.05 level that the true mean filling is not 16 oz.
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11.
The data should be normally distributed.
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12.
For 0.10 level:
There is significant evidence at 0.10 level that the true mean filling is not 16 oz.
For 0.05 level:
There is no significant evidence at 0.05 level that the true mean filling is not 16 oz.