A few days ago
sweet_candy

Hypothesis testing? pls help?

A manufacturer of salad dressing uses machines to dispense liquid ingredient into bottles that move along a filling line. The machine that dispenses dressings is working properly when the mean amount dispensed is 8 ounces. The population standard deviation of the amount dispensed is 0.15 ounce. A sample of 50 bottles is selected periodically, and the filling line is stopped if there is evidence that the mean amount dispensed is different from 8 ounces.

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?

Top 2 Answers
A few days ago
Mitch

Favorite Answer

Step 0 – State Given Information

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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4 years ago
borgmeyer
Use a t-attempt considering the fact which you have not got the inhabitants time-honored deviation and the pattern length is small (n<30) s=a million.3 n=26 H0: x=5.7 Ha: x<5.7 point magnitude=0.05 attempt statistic= (4.5-5.7)/(a million.3/sqrt(26))= -4.7 attempt calculating the rest i've got not got a t-table or image calculator with me
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