Question 329773: A coin-operated drink machine was designed to discharge a mean of 6 ounces of coffee per cup. In a test of the machine, the discharge amounts in 20 randomly chosen cups of coffee from the machine were recorded. The sample mean and sample standard deviation were 6.13 ounces and 0.23 ounces, respectively. If we assume that the discharge amounts are normally distributed, is there enough evidence, at the 0.05 level of significance, to conclude that the true mean discharge, µ, differs from 6 ounces?
Perform a one-tailed test
Null Hypothesis: Ho
Alternative Hypothesis: H1
Type of Test Statistic:
The Value of the Test Statistic:
The two critical values at the 0.05 level of significance.
At the 0.05 level of significance, can we conclude that the true mean discharge differs from 6 ounces?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A coin-operated drink machine was designed to discharge a mean of 6 ounces of coffee per cup. In a test of the machine, the discharge amounts in 20 randomly chosen cups of coffee from the machine were recorded. The sample mean and sample standard deviation were 6.13 ounces and 0.23 ounces, respectively. If we assume that the discharge amounts are normally distributed, is there enough evidence, at the 0.05 level of significance, to conclude that the true mean discharge, µ, differs from 6 ounces?
Perform a one-tailed test
Null Hypothesis:
Ho: u = 6 oz
Ha: u is not equal to 6 oz
Alternative Hypothesis: H1
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Type of Test Statistic: t-value
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The Value of the Test Statistic:
t(6.13) = (6.13-6)/[0.23/sqrt(20)] = 2.5277
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The two critical values at the 0.05 level of significance.
invT(0.025 when df= 19) = -2.0930
invT(0.975 when df= 19) = 2.0930
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At the 0.05 level of significance, can we conclude that the true mean discharge differs from 6 ounces?
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Ans: Yes
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Cheers,
Stan H.
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