Question

A swimming school wants to determine whether a recently hired instructor is working out. 18 out of 30 of Instructor A's students passed the lifeguard certification test on the first try. In comparison, 50 out of 72 of more experienced Instructor B's students passed the test on the first try. Is Instructor A's success rate worse than Instructor B's? Use α = 0.05

1. A hypothesis in symbols and words

2. condition check

3. test statistics (z) and p value

4. interpretation

Answer 1

1)
Hypothesis:
H0 ; p1 = p2
Ha : p1 < p2

2)

Reject H0 if z < -1.64

3)
p1= 18/30 = 0.60 , n1 = 30
p2 = 50/72 = 0.694 , n2 =72

p = (p1 * n1 + p2 * n2) / (n1 + n2)
= (0.60 * 30 + 0.694 * 72) /( 30 + 72)
= 0.6664

SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }

= sqrt( 0.6664 * (1-0.6664) *((1/30) + (1/72)))
= 0.1025

z = ( p1 -p2)/ Se
= ( 0.60 - 0.694)/0.1025
= -0.9171

p value = 0.1796

4)

Do not reject H0

There is not sufficient evidence to support the claim