Question

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   Height (cm) of Main Opponent
181 165
171 179
169 175
190 183
202 192

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.

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?

Identify the test statistic. (Round to two decimal places as​ needed.)

What is the conclusion based on the hypothesis​ test?

Answer 1

Answer a:
Let x be the mean value of x , (where the variable x represents the height of the President)

Let y be the mean value of y, (where the variable y represents the height of the Main Opponent)

Let  z = x - y

The null hypothesis, H0 : z = 0

The alternative hypothesis, Ha : z > 0

Answer b:
The following table shows the calculations -

	Height of President(cm), (x)	Height of main Opponent(cm), (y)	di = x - y	di -	(di - )^2
	181	165	16	12.2	148.84
	171	179	-8	-11.8	139.24
	169	175	-6	-9.8	96.04
	190	183	7	3.2	10.24
	202	192	10	6.2	38.44
Total			19		432.8

The Mean of Differences,   = 19/5 = 3.8

Here, the number of observations, n = 5

The test statistic, t = (n ^ 0.5) ( -  z) / s'

where, s' is the standard deviation of the differences

The test statistic, t follows the t - distribution (with (n -1) degrees of freedom) under the Null Hypothesis

s' = {((di - )^2) / (n - 1)} ^ 0.5 = 10.4019

Substituting all the values,

The value of test statistic, t = 0.82 (up to 2 decimal places)

Answer c:

If the value of the test statistic is greater than the critical value, we reject the null hypothesis at 5% level of significance, else we fail to reject the null Hypothesis

Given, = 0.05

The critical value, t0.05, 4 = 2.1

Since, the critical value is less than the value of the test statistic we fail to reject the null hypothesis

(ASSUMPTION: We assume that the given data has properties similar to that of Normal Distribution)