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 186 171 167 176 185 172 Height (cm) of Main Opponent 174 189 168 177 188 186 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. 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? cm Hoito H₂: Ho cm t= (Type integers or decimals. Do not round.) Identify the test statistic (Round to two decimal places as needed) Identify the P-value. P-value=(Round to three decimal places as needed.) What is the conclusion based on the hypothesis test? ud Since the P-value is the significance level, the null hypothesis. There 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 cm < a < cm. (Round to one decimal place as needed.) What feature of the confidence interval leads to the same conclusion reached in part (a)? in Since the confidence interval contains the null hypothesis Enter your answer in each of the answer boxes ? un sy

Answer 1

a)

null Hypothesis:μd		=	0
alternate Hypothesis: μd		>	0

S. No	President	Main Opponent	diff:(d)=x1-x2	d2
1	186	174	12	144.00
2	171	189	-18	324.00
3	167	168	-1	1.00
4	176	177	-1	1.00
5	185	188	-3	9.00
6	172	186	-14	196.00
	total	=	Σd=-25	Σd2=675
	mean dbar=	d̅     =	-4.1667
	degree of freedom =n-1                            =		5
	Std deviaiton SD=√(Σd2-(Σd)2/n)/(n-1) =		10.684880
	std error=Se=SD/√n=		4.3621
	test statistic            =	(d̅-μd)/Se         =	-0.96
	p value	=	0.808	from excel: tdist(0.955,5,1)

since the p value is greater than significance level, fail to reject the null hypothesis. There is not sufficient evidence to ……………

b)

for 98% CI; and 5 degree of freedom, value of t=			3.365
therefore confidence interval=sample mean -/+ t*std error
margin of errror          =t*std error=		14.6784
lower confidence limit                     =		-18.845
upper confidence limit                    =		10.512
from above 98% confidence interval for population mean =(-18.8 <µd <10.5)

since the confidence interval contains zero, fail to reject the null hypothesis