SOLUTION: 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 18 randomly chosen cups of coffee f
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Question 191500: 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 18 randomly chosen cups of coffee from the machine were recorded. The sample mean and sample standard deviation were 6.16 ounces and 0.28 ounces, respectively. If we assume that the discharge amounts are normally distributed, is there enough evidence, at the 0.1 level of significance, to conclude that the true mean discharge,m , differs from 6 ounces?
Perform a two-tailed test. Then fill in the table below.
Carry your intermediate computations to at least three decimal places and round your answers as specified in the table. (If necessary, consult a list of formulas.)
The value of the test statistic (round to the nearest three decimals):
The p-value (Round to at least three decimals):
At the 0.1 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.
Ho: u = 6
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In a test of the machine, the discharge amounts in 18 randomly chosen cups of coffee from the machine were recorded. The sample mean and sample standard deviation were 6.16 ounces and 0.28 ounces, respectively. If we assume that the discharge amounts are normally distributed, is there enough evidence, at the 0.1 level of significance, to conclude that the true mean discharge,m , differs from 6 ounces?
Ha: u is not equal to 6
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Perform a two-tailed test. Then fill in the table below.
Carry your intermediate computations to at least three decimal places and round your answers as specified in the table. (If necessary, consult a list of formulas.)
The value of the test statistic (round to the nearest three decimals):
t(6.16) = (6.16-6)/[0.28/sqrt(18)] = 2.4244
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The p-value (Round to at least three decimals):
p-value = 2*P(t > 2.4244 with df = 17) = 2*0.134 = 0.027
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At the 0.1 level of significance can we conclude that the true mean discharge differs from 6 ounces?
Since the p-value is less than 10% reject Ho.
The test provides evidence the u is not 6.
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Cheers,
Stan H.
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