Thread: T-Test in Hypothesis Testing help!

A brewery claims that the consumers are getting a mean volume equal to 32 ounces of beer in their quart bottles. The Bureau of Weights and Measures randomly selects 36 bottles. Test the claim of the brewery about the mean volume equal to 32 oz. at the .05 significance level.

--What I found: : 31.9458333, Σx: 1150.05, Σx^2: 36749.2643, Sx: .5331416883, n= 36

--Now, after this I know i have to write down its original claim and the null hypothesis (μ = 32) with its alternate hypothesis. My problem is I don't know to how many digits I'm supposed to round these numbers to. I've already gotten 3 different answers. Can anyone enlighten me? The calc gives me t= -.6093 with the p-value of .5463 (rounded). Should I just put this down?

Don't put aything down if you don't know what it refers to. Your null hypothesis is correct, although you could also say the mean weight is equal to 32 or MORE. The alternate hypothesis would then be that the mean weight is actually LESS than 32.

When you plugged your values into your calculator, and got your test-statistic of -.6093; then somehow you got a P-Value of 0.5. Do you know this this means? What this is essentially saying is that 50% of the time you would expect to get a mean weight of less than 32. That can not be right. How did you go about letting the calculator computer your P-value for you?

Hi. Well the p-value of .05 was automatically given to me by my professor. The calculator(TI-83) cannot and will not give me a p-value. What I was concerned about was the x bar being 31.9458….should I round this off to 32? IT will total my test statistic to 0…is that right? And as of right now, I have no idea what the complete picture or graph is supposed to look like, it's two-tailed correct?

oh wait, my mistake! I got the p-value confused with the critical value.....Yes, my calculator actually did give me a p-value of .546....the .05 is not a p-value, its supposed to be the significance level which is the area of the region of rejection.

You absolutely can not round 31.9458 to 32. By rounding it to 32 you are saying that the mean weight isn't 31.9458 but 32 - which is not true. Now you are going to have to round sure, but you can not round to such a degree. The standard deviaiton of the sample means may be such, that 31.9458 lies far outside of the 95% confidence interval. By rounding to 32, you essentially put your sample mean within the confidence interval of your test, and thus fail to reject your null hypothesis - letting the brewery off the hook (this may or may not be a good thing).

As to your P-value, do you understand what the P-value refers to? It is a number between 0 and 1 (well it'll never be zero or one. . .i suppose it can be but that would be a boring test), and it tells your the probability that you will get a sample mean (or proportion or variance or whatever you are testing for) that is as extreme as the observed value. So say your mean is 0, and you get a test statistic of -1.69; looking at a z-chart (since we're testing the mean and our sample size is appropriately large), we can see that the chances of us getting a value of -1.69 is 0.0455. THIS is your P-Value. If you ran the tests, you would expect to observe a sample mean (from a size N) of -1.69 0.0455 times. Depending on the experiment you have set up, this should be enough evidence to reject whatever null hypothesis you have set up - since its not very likely that the sample you have selected occured by chance. Similarly, if you have a P-value of 0.5, this is saying that HALF of the time, you CAN expect to get a value of the sample mean you observed from your test. In your case, you have a 50% chance of observing a mean weight of 31.9458. Since your significance level is 0.05 (or 5% level of significance), you FAIL to reject. A P-Value of 0.5 tells us nothing (useful) about a test.

Also remember for a two-tailed test, that if you are testing at a significance level of 0.05, then the area in ONE-tail is 0.025.

I see what you're saying...the professor actually gave us a choice to run this through the Z-Test method or the T-Test method. I chose the T-Test method on my TI-83 calc. It did give me a slightly different answer than using the Z-Test. As of right now, because I had rounded the mean weight to 32, I have µ ≠ 32, t=0, p=1 and xbar= 32....
Using the mean weight of 31.9458333 gave me a t= -.609 with a p-value of .546....although I'm still not sure if this is correct since earlier I had tested it using the Z-Test method and it gave me a z=-.617 with a p-value of .536....
Any suggestions as to what method is more precise perhaps?

What are the individual values you received for this hypothesis test?

so i need to find:
a. Null Hypothesis: µ = 32
b. Alternate Hypothesis: µ ≠ 32
c. Test statistic: t= x̅- µ/s/√n = -.609 (two-tailed)
P-value: .5463
Statistical Conclusion: Fail to reject Ho: there is not sufficient evidence to warrant rejection of the original claim that u= 32.

You said you calculated the sample mean yourself. What I wanted was the individual values of the beer bottles that you used to calculate the sample mean and standard deviation.