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Oxana [17]
3 years ago
7

g A popular theory is that presidential candidates have an advantage if they are taller than their main opponents. Listed are he

ights​ (in centimeters) of randomly selected presidents along with the heights of their main opponents. Complete parts​ (a) and​ (b) below. Height (cm )of President 191 180 180 182 197 180 Height (cm )of Main Opponent 166 179 168 183 194 186 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, 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?
Mathematics
1 answer:
stira [4]3 years ago
5 0

Answer:

Step-by-step explanation:

Corresponding heights of presidents and height of their main opponents form matched pairs.

The data for the test are the differences between the heights.

μd = the​ president's height minus their main​ opponent's height.

President's height. main opp diff

191. 166. 25

180. 179. 1

180. 168. 12

182. 183. - 1

197. 194. 3

180. 186. - 6

Sample mean, xd

= (25 + 1 + 12 - 1 + 3 + 6)/6 = 5.67

xd = 5.67

Standard deviation = √(summation(x - mean)²/n

n = 6

Summation(x - mean)² = (25 - 5.67)^2 + (1 - 5.67)^2 + (12 - 5.67)^2+ (- 1 - 5.67)^2 + (3 - 5.67)^2 + (- 6 - 5.67)^2 = 623.3334

Standard deviation = √(623.3334/6 sd = 10.19

For the null hypothesis

H0: μd ≥ 0

For the alternative hypothesis

H1: μd < 0

The distribution is a students t. Therefore, degree of freedom, df = n - 1 = 6 - 1 = 5

The formula for determining the test statistic is

t = (xd - μd)/(sd/√n)

t = (5.67 - 0)/(10.19/√6)

t = 1.36

We would determine the probability value by using the t test calculator.

p = 0.12

Since alpha, 0.05 < than the p value, 0.12, then we would fail to reject the null hypothesis.

Therefore, at 5% significance level, we can conclude that for the population of heights for presidents and their main​ opponents, the differences have a mean greater than 0 cm.

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