Question

Listed below are systolic blood pressure measurements (mm Hg) taken fromthe right and left arms of the same woman. Assume that the paired sample data is asimple random sample and that the differences have a distribution that is approximatelynormal. Use a 0.01 significance level to test for a difference between the measurementsfrom the two arms. What can be concluded? Round to 3 decimal places.Right arm148 151 139 135 138Left arm175 168 190 145 143In this exampleμdis the mean value of the differences d for the population of all pairsof data, where each individual difference d is defined as the measurement from the rightarm minus the measurement from the left arm.(a) Test the claim using a hypothesis test.Identify the null and alternative hypothesis,test statistic, p-value and final conclusion that addresses the original claim.(b) Test the claim by constructing an appropriate confidence interval.Write a conclu-sion for the confidence interval.

Answer 1

We are to test the difference in the measurements of the right and left arms of the same individuals. So the two data sets are dependent. Therefore this is a dependent samples t-test.

	Right	Left	d	d2
1	148	175	-27	729
2	151	168	-17	289
3	139	190	-51	2601
4	135	145	-10	100
5	138	143	-5	25
Total			-110	3744
Mean			-22
SD			18.1934

Mean =

SD =

.(a) Test the claim using a hypothesis test.Identify the null and alternative hypothesis,test statistic, p-value and final conclusion that addresses the original claim.

Claim is the null hypothesis

Test stat:

Where the null difference = 0

Test Stat = -2.7039

p - value = 2 P( > |Test Stat |) ................since it is two tailed

=2P(t₄ > 2.70)

=2 * 0.02694 .............using t-dist tables

p-value= 0.05388

Since p-value > 0.01

We do not reject the null hypothesis. There is insufficient evidence to conclude that there is difference between measurement of arms.

(b) Test the claim by constructing an appropriate confidence interval.Write a conclu-sion for the confidence interval.

(1- )% is the confidence interval for population mean difference

Where = 0.01 therefore confidence level =1 - 0.01 = 99%

Critical value =

= t_4,0.005

= 4.6041 .............using t-dist tables

Margin of error =

Subsituting the values

(-59.4605, 15.4605)

This means we are 99% confident that the true mean difference between measurements lie within this interval.

Since this interval includes '0', we can do not reject the null hypothesis.