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 left parenthesis cm right parenthesis of President 181 176 171 189 194 181 Height left parenthesis cm right parenthesis of Main Opponent 171 178 176 169 189 172 a. Use the sample data with a 0.01 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, mu Subscript d 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? Upper H 0​: mu Subscript d equals 0 0 cm Upper H 1​: mu Subscript d not equals 0 0 cm ​(Type integers or decimals. Do not​ round.) Identify the test statistic. tequals nothing ​(Round to two decimal places as​ needed.) Identify the​ P-value. ​P-valueequals nothing ​(Round to three decimal places as​ needed.) What is the conclusion based on the hypothesis​ test? Since the​ P-value is ▼ less than or equal to greater than the significance​ level, ▼ reject fail to reject the null hypothesis. There ▼ is is not sufficient evidence to support the claim that presidents tend to be taller than their opponents. b. Construct the confidence interval that could be used for the hypothesis test described in part​ (a). What feature of the confidence interval leads to the same conclusion reached in part​ (a)? The confidence interval is nothing cmless thanmu Subscript dless than nothing cm. ​(Round to one decimal place as​ needed.) What feature of the confidence interval leads to the same conclusion reached in part​ (a)? Since the confidence interval contains ▼ only negative numbers, zero, only positive numbers, ▼ reject fail to reject the null hypothesis.

Answer 1

The following table is obtained:

	Sample 1	Sample 2	Difference = Sample 1 - Sample 2
	181	171	10
	176	178	-2
	171	176	-5
	189	169	20
	194	189	5
	181	172	9
Average	182	175.833	6.167
St. Dev.	8.39	7.25	9.02
n	6	6	6

For the score differences we have

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μD​ = 0

Ha: μD​ > 0

This corresponds to a right-tailed test, for which a t-test for two paired samples be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.01, and the degrees of freedom are df=5.

Hence, it is found that the critical value for this right-tailed test is tc​=3.365, for α=0.01 and df=5.

The rejection region for this right-tailed test is R=t:t>3.365.

(3) Test Statistics

The t-statistic is computed as shown in the following formula:

The p-value is p=0.0774

(4) Decision about the null hypothesis

Since it is observed that t=1.675≤tc​=3.365, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.0774, and since p=0.0774≥0.01, it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean μ1​ is greater than μ2​, at the 0.01 significance level.

Confidence Interval

The 99% confidence interval is −8.682<μD​<21.015

Graphically

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