Hypothesis Testing

Question

We are testing the claim that houses in a particular community average less than 90 days on the market. A random sample of 9 homes averaged 77.4 days on the market with a sample standard deviation of 29.6 days. Assume the population is normally distributed. Test the claim at the a = 0.05 significance level by comparing the calculator t-score to the critical t-store. 

Answer 1

This question is about rejecting a null hypothesis, the claim that the population mean is indistinguishable from 90.  

 

Okay, so let's use common sense here first.  If our sample says they average 77.4 days, and our estimate is 90 days, then we have a difference of 12.6 days.  With a sample SD for the 9 houses of 29.6 days, our sample is at the -.43 SD level.  

 

The central limit theorem says that the sample size will have shrunk our underlying population SD by the square root of the sample size.  

 

Therefore, because of our sample size of 9, we have to multiply by the square root of 9 to get .43*3 = 1.29 for our t value.  

 

This is in English why the formula looks like this - 

 

t = [ x - μ ] / [ s / sqrt( n ) ]

 

And we can plug the original numbers into that equation to get the same result - 

 

t = (77.4 - 90) / (29.6 / sqrt(9)) = 1.29

 

Now we have to check the charts for the critical value.  For 95% confidence -- which is the T.025 column on the table, if you're using tables -- with 8 degrees of freedom (n=9, then subtract 1) we get a critical value of 2.306.

 

Since 1.29 is not greater than 2.306, the claim passes the test.  We can't reject the claim that the average is 90.