Question

Who are taller, pro football players or pro basketball players? A random sample 45 pro football players resulted in a mean height of x = 6.179 feet. A random sample 40 pro basketball players resulted in a mean height of y = 6.453 feet. It is recognized that the true standard deviation of pro football players heights is σ_x = 0.47 feet while it is recognized that the true standard deviation of pro basketball players heights is σ_y = 0.55 feet. The true (unknown) mean of football players heights is μ_x feet, while the true (unknown) mean of basketball players heights is μ_y feet.

Type	Sample Size	Sample Mean	Standard Deviation
Football (X)	45	6.179	0.47
Basketball (Y)	40	6.453	0.55

f)Create a 94% confidence interval for μ_x - μ_y. ( , )

g) What is the length of your 94% confidence interval for μ_x - μ_y?  

h) If we used this data to test H₀:μ_x - μ_y=0 against the alternative H_a:μ_x - μ_y < 0 then what would the value of the calculated test statistic z have been?  

i) If we used this data to test H₀:μ_x - μ_y=0 against the alternative H_a:μ_x - μ_y<0 then what would the p value have been?  

j)If we used this data to test H₀:μ_x - μ_y=0 against the alternative H_a:μ_x - μ_y ≠0 then what would the p value have been?

Answer 1

(f)

n1 = 45

n2 = 40

x1-bar = 6.179

x2-bar = 6.453

s1 = 0.47

s2 = 0.55

% = 94

Degrees of freedom = n1 + n2 - 2 = 45 + 40 -2 = 83

Pooled s = √(((n1 - 1) * s1^2 + (n2 - 1) * s2^2)/DOF) = √(((45 - 1) * 0.47^2 + ( 40 - 1) * 0.55^2)/(45 + 40 -2)) = 0.509158294

SE = Pooled s * √((1/n1) + (1/n2)) = 0.50915829431985 * √((1/45) + (1/40)) = 0.110643533

t- score = 1.90684906

Width of the confidence interval = t * SE = 1.90684905968412 * 0.110643532565295 = 0.210980516

Lower Limit of the confidence interval = (x1-bar - x2-bar) - width = -0.274 - 0.210980516032262 = -0.484980516

Upper Limit of the confidence interval = (x1-bar - x2-bar) + width = -0.274 + 0.210980516032262 = -0.063019484

The confidence interval is [ -0.4850, -0.0630]

(g) It is 0.210980516

(h)

Data:

n1 = 45

n2 = 40

x1-bar = 6.179

x2-bar = 6.453

s1 = 0.47

s2 = 0.55

Hypotheses:

Ho: μ1 = μ2

Ha: μ1 < μ2

Test Statistic:

SE = √{(s1^2 /n1) + (s2^2 /n2)} = √((0.47)^2/45) + ((0.55)^2/40)) = 0.111675373

z = (x1-bar - x2-bar)/SE = (6.179 - 6.453)/0.11167537279494 = -2.453540052

(i) p- value = 0.007072889

(j)

Data:   

n1 = 45

n2 = 40

x1-bar = 6.179

x2-bar = 6.453

s1 = 0.47

s2 = 0.55

Hypotheses:   

Ho: μ1 = μ2   

Ha: μ1 ≠ μ2   

Test Statistic:   

SE = √{(s1^2 /n1) + (s2^2 /n2)} = √(((0.47)^2/45) + ((0.55)^2/40) = 0.111675373

z = (x1-bar -x2-bar)/SE = (6.179 - 6.453)/0.11167537279494 = -2.45354005

p- value = 0.014145779