SOLUTION: The US Department of Agriculture claims that the mean consumption of coffee by a person in the US is 24.2 gallons. A random sample of 120 people shows that the mean consumption is

Question 459100: The US Department of Agriculture claims that the mean consumption of coffee by a person in the US is 24.2 gallons. A random sample of 120 people shows that the mean consumption is 23.5 gallons per year with a standard deviation of 3.2 gallons. At significance level 0.05 can you reject the claim?
I set Ho: u=24.2, Ha: u=23.5, found the z-score to be -2.4. I used that to find the p-value, which I got .0164, but I'm not sure if that's right.

Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Ha should be

Because we do NOT know the population standard deviation, we CANNOT use the z test statistic. We must use the t test statistic (note: the sample size of 120 is large enough to get away with using a normal distribution, since the T distribution approaches a normal distribution for large n, but it's better to stick with T if you don't know the population standard deviation)

That test statistic is

which rounds to . So you have the correct value, just the wrong distribution.

Now use a calculator, or a table, to find the area under the curve T distribution curve from negative infinity to (note: the degrees of freedom are equal to 120 - 1 = 119). This area is equal to 0.0089728

Double this area (this is a two tailed test) to get 2* 0.0089728 = 0.0179456

So the p-value is roughly 0.018 (rounded to three places)(so you're close)

Since the p-value is less than alpha at 0.05, this means that we reject the null hypothesis.

Remember the claim was "the mean consumption of coffee by a person in the US is 24.2 gallons.", so the claim was Ho: . Because we rejected Ho (the null hypothesis), this means that we reject the claim that "the mean consumption of coffee by a person in the US is 24.2 gallons."
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