Question 1073494: Trying to find the answer for this:
A coffee manufacturer is interested in whether the mean daily consumption of regular-coffee drinkers is less than that of decaffeinated-coffee drinkers. Assume the population standard deviation for those drinking regu- lar coffee is 1.20 cups per day and for those drinking decaffeinated coffee is 1.36 cups per day. A random sample of 50 regular-coffee drinkers showed a mean of 4.35 cups per day. A sample of 40 decaffeinated-coffee drinkers showed a mean of 5.84 cups per day. Use the 0.01 significance level. Compute and interpret the p-value.
Only if you can, some detail in the answer please so I can understand how to do it :)
Thank You!!!
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! This is a two independent sample test(regular-coffee drinkers vs decaffeinated-coffee drinkers), each sample size is > or = 40, so we can assume a normal distribution
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We are to test if the mean daily consumption of regular-coffee drinkers(u(1)) is less than that of decaffeinated-coffee drinkers(u(2))
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This is a one-tailed test, our null and alternate hypothesis tests are
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Ho: u(2) < or = u(1)
H1: u(2) > u(1)
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Since we know the standard deviation/s of the population, we can use a z-statistic
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u(1) of the sample is 4.35 and u(2) of the sample is 5.84
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n(1), this sample size is 50 and n(2), this sample size is 40
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s(1) standard deviation of population is 1.20 and s(2) standard deviation of the population is 1.36
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z = (5.84 - 4.35) / square root((1.36^2 / 40) + (1.20^2 / 50)) = 5.4393
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the significance level, alpha, is 0.01
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z results in a p-value of 0.00001 which is < 0.01, so we reject the null hypothesis(Ho), therefore
:
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the mean daily consumption of regular-coffee drinkers is greater than that of decaffeinated-coffee drinkers
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