A researcher is conducting a matched-pairs study. She gathers data on each pair in the study resulting in the table below. Assume that the data are normally distributed in the population.
Pair Group1 Group2
1 10 12
2 8 9
3 11 11
4 8 10
5 9 12
a) At 10% level of significance and using a 2-tailed test, what is your conclusion?
Answer: Is this correct. thnx^_^
D (Group1 - Group2) D^2
2 4
1 1
0 0
2 4
3 9
D= 8 D^2=18
X= 8/5= 1.6
S^2 = 18- ((8^2)/5)/(5-1) = 1.3--> S= 1.14
H0: M=0
H1: M>0
Reject H0 if t>2.1319
t= (1.6-0) / (1.14/5(sqroot)) = 3.14
There is enough evidence.
To determine the conclusion of this matched-pairs study, the researcher compared the data between Group 1 and Group 2 by calculating the difference (D) for each pair in the study. The squared differences were also calculated (D^2).
After finding the mean difference (X) and the variance (S^2) of the differences, the researcher set up the null and alternative hypotheses. The null hypothesis (H0) states that there is no difference between the mean values of the populations represented by Group 1 and Group 2. The alternative hypothesis (H1) suggests that there is a positive difference between the means.
The significance level chosen was 10%, which corresponds to a critical value of 2.1319 for a two-tailed test. If the absolute value of the calculated t-statistic exceeds this critical value, the null hypothesis is rejected.
Using the formula for calculating the t-statistic, which is (X - μ) / (S / √n), where μ is the hypothesized population mean difference (0 in this case), S is the estimated standard deviation of the differences (1.14), and n is the number of pairs (5), the value of the t-statistic was found to be 3.14.
Since the calculated t-statistic (3.14) is greater than the critical value (2.1319), there is enough evidence to reject the null hypothesis. It can be concluded, at a 10% level of significance, that there is a statistically significant positive difference between the means of Group 1 and Group 2 in the population.