Hypothesis testing? pls help?
Suppose that the mean amount dispensed in a particular sample of 50 bottles is 7.983 ounces.
a.) Is there evidence that the population mean amount is different from 8 ounces? (use 0.05 level of significance)
b.) compute the p-value and interpret its meaning.
c.)What is your answer in (a) if the sample mean is 7.952 ounces and the standard deviation is .15 ounce?
Favorite Answer
µ = Population Mean = 8.0 oz.
σ = Population Standard Deviation = 0.15 oz.
xbar = 7.983 oz.
n = 50 observations
Since n > 30, we assume a normal distribution.
(P.S – Because n>30, we can use a Zcalc instead of a t-test)
……………………
Step 1 – Setup Hypothesis
a) Is there evidence that the population mean amount is different from 8 ounces? (Use a 0.05 level of significance)
Ho: µ = 8.0 oz. <--- Null hypothesis Ha: µ ≠ 8.0 oz. <--- Alternate Hypothesis (Claim) ........................ Step 2 - Select Test Statistic Since the population standard deviation is given and n >30, we assume a normal distribution.
Therefore, the test statistic is:
Zcalc = (xbar – μ)/(σ/√n)
……………………
Step 3 – Find Zcrit and State Decision Rule
Zcrit = 0.5 – .025 = .4750
Do a table lookup on 0.4750, which gives a z-value of 1.960
The table should be in the appendix of your stats book.
If not, you can view it here:
http://rvgs.k12.va.us/statman/Table-A3.jpg
This is a two-tailed test.
Since α/2 is in both tails, the decision rule is to
reject Ho if Zcalc > Zcrit of 1.960, or
reject Ho if -Zcalc < -Zcrit of -1.960, otherwise accept Ho ........................ Step 4 - Compute Zcalc Zcalc = (xbar - μ)/(σ/√n) Zcalc = (7.983 - 8.0)/(0.15/√50) Zcalc = (-0.017)/(0.15/7.071) Zcalc = -.801388 ........................ Step 5 - State Decision Since -Zcalc of -0.801388 > -Zcrit of -1.960, we accept Ho and reject the claim that the mean amount dispensed is significantly different from 8.0 ounces.
……………………
Ok – Now for the p-value:
Go back to the table and lookup a z-value of 0.801388
The p-value equals 0.422907 (I used Excel to find this value).
Since p-value of 0.422907 > α of .05, we accept Ho
……………………
What is your answer in (a) if the sample mean is 7.952 ounces and the standard deviation is .15 ounce?
Zcalc = (xbar – μ)/(σ/√n)
Zcalc = (7.952 – 8.0)/(0.15/√50)
Zcalc = -2.26274
Since -Zcalc of -2.2627 < -Zcrit of -1.960, we reject Ho and accept the claim that the mean amount dispensed is significantly different from 8.0 ounces. Good luck in your studies, ~ Mitch ~ P.S. - If you have a TI-83 or TI-84, you can compute this directly on your calculator: http://www.stat.wmich.edu/s216/book/node49.html I believe it will also return a p-value also. This is a really neat way to save time on tests! (Personally, I use a TI-89 Titanium, so I've never done it on a TI-83/84) Edit: I made a small error, which I believe I've corrected. Please double-check my calculations carefully to make sure they match yours!
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