Question

a) Use a t-distribution to find a confidence interval for the difference in means μd=μ1-μ2 using the relevant sample results from paired data. Assume the results come from random samples from populations that are approximately normally distributed, and that differences are computed using d=x1-x2.
A 99% confidence interval for μd using the paired data in the following table:

Case	1	2	3	4	5
Treatment 1	22	27	32	26	29
Treatment 2	18	29	25	20	20

Give the best estimate for μd, the margin of error, and the confidence interval.
Enter the exact answer for the best estimate, and round your answers for the margin of error and the confidence interval to two decimal places.
best estimate = ________________
margin of error = ________________
The 99% confidence interval is _______________to ___________________

b) Use a t-distribution and the given matched pair sample results to complete the test of the given hypotheses. Assume the results come from random samples, and if the sample sizes are small, assume the underlying distribution of the differences is relatively normal. Assume that differences are computed using d=x1-x2.
Test H0 : μd=0 vs Ha : μd<0 using the paired data in the following table:

Treatment 1	18	12	11	21	15	11	14	22
Treatment 2	18	19	25	21	19	14	15	20

Give the test statistic and the p-value.
Round your answer for the test statistic to two decimal places and your answer for the p-value to three decimal places.

test statistic = ______________________

p-value = ________________________

Give the conclusion using a 5% significance level.

Conclusion: Reject or Do not Reject?

Answer 1

(a) The formula to calculate the confidence interval for the difference in means is given as:

First we need to find the difference Treatment1 and Treatment2 :

Case	Treatment1	Treatment2
1	22	18	4	-0.8	0.64
2	27	29	-2	-6.8	46.24
3	32	25	7	2.2	4.84
4	26	20	6	1.2	1.44
5	29	20	9	4.2	17.64

standard deviation of difference

Best estimate:

Margin of error:

For 99% confidence interval,

99% Confidence interval for

               

               

So, the 99% confidence interval for is

______________________________________________

(b) The null and alternative hypothesis are:

So, at significance level we need to test the hypothesis for the true difference of means of two treatments.

Test-statistic:

Degrees of freedom:

Treatment1	Treatment2
18	18	0	3.375	11.390625
12	19	-7	-3.625	13.140625
11	25	-14	-10.625	112.890625
21	21	0	3.375	11.390625
15	19	-4	-0.625	0.390625
11	14	-3	0.375	0.140625
14	15	-1	2.375	5.640625
22	20	2	5.375	28.890625

Standard deviation for difference

Calculation for test-statistic:

So, the test-statistic is calculated as

P-value: Since it is a left-tailed hypothesis so,the p-value for the calculated test-statistic is given as:

So, the p-value is calculated as

Decision: significance level is given as and the p-value is calculated as .

Since,

So, at the sample data provides insufficient evidence to reject the null hypothesis, H₀ . Hence, we fail to reject the null hypothesis.

In other words, at given significance level of the sample data does not provide enough evidence to support the alternative hypothesis, so we fail to reject the null hypothesis.