Testing Null Hypothesis

[thordogg24]

Hi everyone,

Thanks for any help you can offer ahead of time. I am struggling with this stuff and could really use any help available.

Here's the question

A beer distributor claims that a new display, featuring a life-size picture of a well-known athlete, will increae product sales in supermarkets by an average of 40 cases a week. For a random sample of 25 supermarkets, the average sales increase was 31.3 cases and the sample standard deviation was 12.2 cases. Test, at the 5 percent level, the null hypothesis that the population mean sales increase is at least 40 cases, stating any assumptions you make.

I believe that you use the chi-square goodness of fit test. x2 = (O-E)/E but not sure what to do with this. Anyone have any ideas on how to test this problem?

Thanks again,
Thor

 

[galactus]

This is not a \(\displaystyle {\chi}^{2}\) test. Just a regular ol' hypothesis test for the mean.

What alpha level are you to use?. You didn't say. I will assume .05 then. You can easily change it.

\(\displaystyle H_{0}:{\mu}\geq 40, \;\ claim\)

\(\displaystyle H_{a}:{\mu}<40\)

Run the test using \(\displaystyle z=\frac{(x-{\mu})\sqrt{n}}{\sigma}\)

I get a test stat of -3.57, critical value of -1.71, p-value of .0008

Since the test stat is in the rejection region you choose to reject the null.

Also, the p value is less than the alpha, so again we choose to reject the null.

Since the p-value is so low, it would appear we will reject no matter what we use. Normally, we never go below an alpha of .001.

.001 would be the most significant alpha level in this case.

How would you summarize this?.

 

[thordogg24]

Thanks for your help, the reason I mentioned the chi-square goodness of fit test is the question mentioned it as a hint so I thought that it was what should be used. What numbers do you plug in to get the test stat, critical value and the p-value. The numbers make sense but I guess not sure how you got them.

Thanks,
Thor

 

[galactus]

You're given the numbers. Plug them into the formula to get the test stat.

\(\displaystyle x=31.3, \;\ {\mu}=40, \;\ n=25, \;\ {\sigma}=12.2\)

To find the critical value, this is a t distribution because the sample size is less than 30.

You must look it up in the t table. Do you have a stats book with it in?.

Degrees of freedom is 24 and it is a right tailed test with alpha = .05

I ran it all through my handy dandy Excel sheet I have for this purpose.