Question

A certain fishing lodge advertises that 75% of its guests catch a pike over 20 pounds. Last summer 60 out of a random sample of 83 guests did catch a pike over 20 pounds. Does this indicate that the population proportion is different from 75% ? use alpha = .05

Answer 1

null Hypothesis:              Ho:   p=		0.750
alternate Hypothesis:    Ha: p ≠		0.750

for 0.05 level and two tailed test , critical z=			1.96	(from excel:normsinv(0.025)
Decision rule :   reject Ho if absolute value of test statistic |z|>1.96

sample success x   =		60
sample size          n    =		83
std error σp =√(p*(1-p)/n) =		0.0475
sample prop p̂ = x/n=60/83=		0.7229
z =(p̂-p)/σp=(0.723-0.75)/0.723=		-0.57

since test statistic does not falls in rejection region we fail to reject null hypothesis

we do not have have sufficient evidence to conclude that the population proportion is different from 75%