Question

Homework: 9-3: Two Dependent Samples (Matched Pairs) Save Score: 0 of 20 pts 3 of 3 (2 complete) HW Score: 51.28%, 32.31 of 63 pts 9.3.6-T D Assigned Media Question Help A popular theory is that presidential candidates have an advantage if they are taller than their main opponents. Listed are heights (in centimeters) of randomly selected presidents along with the heights of their main opponents. Complete parts (a) and (b) below. Height (cm) of President 178 184 178 176 185 170 Height (cm) of Main Opponent 162 181 171 172 183 186 a. Use the sample data with a 0.05 significance level to test the claim that for the population of heights for presidents and their main opponents, the differences have a mean greater than 0 cm. In this example, He is the mean value of the differences d for the population of all pairs of data, where each individual difference d is defined as the president's height minus their main opponent's height. What are the null and alternative hypotheses for the hypothesis test? Enter your answer in the edit fields and then click Check Answer. ?
Identify the test statistic.

Identify the​ P-value.

What is the conclusion based on the hypothesis​test?

Answer 1

The null and alternate hypothesis are:

H0: $\mu_{d}=0$

Ha: $\mu_{d}>0$

							TOTAL
President	176	184	178	176	185	170
Opponent	162	181	171	172	183	186
d	14	3	7	4	2	-16	14
d2	196	9	49	16	4	256	530

$\overline{x}_{d}=\frac{\sum x_{d}}{n}=\frac{14}{6}=2.3333$

$S_{d}=\sqrt{\frac{\sum x_{d}^2-n\overline{x}_{d}^2}{n-1}} =\sqrt{\frac{530-6(2.3333)^2}{6-1}}=9.9733$

The test statistic is given by:

$t=\frac{\overline{x}_{d}}{\frac{S_{d}}{\sqrt{n}}}=\frac{2.3333}{\frac{9.9733}{\sqrt{6}}}=0.57$

Since this is a right-tailed test, so the p-value is given by:

$P(t_{n-1}>0.57)=P(t_{5}>0.57)=0.296667$

Since p-value is greater than 0.05, so we do not have sufficient evidence to reject the null hypothesis
H0.

Thus we cannot say that $\mu_{d}>0$ i.e. difference in greater than 0 cm.