## Question 751:

1

No answer provided yet.We perform a 2-sample t-test on the raw values. We get means of 89.8 and 71.2 for West and East respectively. The standard deviations are 14.865 and 10.4 and the sample sizes are equal at 6.

The ratio of the variances is a bit more than 2 to 1 108 vs 220 (recall that the variances are just the standard deviation squared). This suggest that there are different variances and so we need to adjust the formula a bit (by making the degrees of freedom more conservative).  If you are using software it will make the adjustment for you.

The difference between the means is 18.667 rating points. To determine if this is significant, we divide it by the standard error of the difference (SE). The SE is found as a weighted values of the standard deviations divided by the square root of the sample sizes. We get 7.41. Dividing the difference by this we get 18.667/7.41 = 2.51, which is the test statistic. That means there is about a 2.5 standard error difference. We evaluate this for significance by using a t-table on  8 degrees of freedom (this adjustment is called the Welch-Swathmore procedure). We get the p-value of .0358 which is less than .05.

So we reject the Null Hypothesis, which is that there is no difference in ratings. We conclude that there is a difference between ratings between the regions, with the West having a higher rating.