Question

2.Have you ever heard the saying…College students are short? Well, I think so at least. In other words, more than half of college students are less than 5’8” (68 inches). Use our class data as a sample to perform a test to determine if I am correct. Test at a .20 level of significance

Heights(inches) total of 192 height 67,66,63,70,69,64,63,64,68,72,63,65,63,67,70,71,62,64,66,74,62,70,64,65,64,69,60,70,68,65,74,64,64,64,63,68,66,61,67,69,72,62,60,76,65,67,64,74,56,62,69,72,72,69,62,74,76,62,59,79,68,65,72,66,65,68,68,73,68,73,71,67,68,76,65.5,61,62,60,70,65,68,66,65,67,66,66,70,70,72,62,62,63,70,70,72,62,65,61,72,64,71,66,60,61,72,70,69,69,73,70,72,71,74,71,68,66,68,74,64,69,65,68,64,64,62,70,68,68,67,71,73,65,63,66,67,61,71,61,70,71,62,72,69,70,62,64,64,62,71,70,69,69,71,70,72,76,60,66,56,83,68,70,68,68,60,66,63,67,67,65,73,67,63,67,68,59,67,63,70,72,63,75,62,68,73,64,59,74,72,64

66,66

Answer 1

H0: µ = 68

H0: µ < 68

µ = 68

n = 192

x: Sample mean = sum of all values/n

= 12892.5/192

= 67.15

s: Standard deviation of sample = (Σ(xi-x)2)/(n-1)

Σ(xi-x)2 = 3790.02

n-1=191

s =3790.02/191 = 19.84

Test statistic: t = (x-µ)/(s/n^0.5)

= (67.15-68)/(19.84/192^0.5)

= -0.59

Degrees of freedom = n-1 =191

Critical value (Using Excel function T.INV(probability,degrees of freedom) = T.INV(0.2,191) = -2.07

Since test statistic is more than critical value, we reject the null hypothesis and conclude that more than half of college students are less than 68 inches tall.