SOLUTION: 16. A typical college student drinks an average of 96 ounces per day of various beverages that contain caffeine. A sample of 12 students at Wallace College revealed the following a
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Question 553577: 16. A typical college student drinks an average of 96 ounces per day of various beverages that contain caffeine. A sample of 12 students at Wallace College revealed the following amounts of beverages consumed containing caffeine:
108 96 84 84 120 96 108 132 72 120 72 96
Can we conclude that the average amount of beverages consumed containing caffeine at Ownes College is the same as the typical college student? Use the hypothesis testing procedure.
a. State the null and alternate hypotheses.
H0: __________________________________________________________________
H1: __________________________________________________________________
b. State the decision rule.
_______________________________________________________________________
Use the table to c. & d. Compute the mean and standard deviation.
x x-xbar and x-xbar^2
108
96
84
84
120
96
108
132
72
120
72
96
e. Compute the value of the test statistic.
f. Formulate the decision rule.
___________________________________________________________________________
___________________________________________________________________________
This feature of our site requires you to log in or register. Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! A typical college student drinks an average of 96 ounces per day of various beverages that contain caffeine. A sample of 12 students at Wallace College revealed the following amounts of beverages consumed containing caffeine:
108 96 84 84 120 96 108 132 72 120 72 96
Can we conclude that the average amount of beverages consumed containing caffeine at Ownes College is the same as the typical college student? Use the hypothesis testing procedure.
a. State the null and alternate hypotheses.
H0: u = 96
H1: u # 96
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b. State the decision rule.
Reject Ho is the test statistic is < -1.96 or > +1.96.
_______________________________________________________________________
Compute the mean and standard deviation.
x-bar = 99
std = 19.23
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e. Compute the value of the test statistic.
t(99) = (99-96)/[19.23/sqrt(12)] = 0.5404
f. Formulate the decision rule.
Since the test statistic is not in either reject interval
fail to reject Ho.
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Cheers,
Stan H.
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