Question

Some experts believe that 21​% of all freshwater fish in a country have such high levels of mercury that they are dangerous to eat. Suppose a fish market has 250 fish​ tested, and 42 of them have dangerous levels of mercury. Test the hypothesis that this sample is not from a population with 21​% dangerous​ fish, assuming that this is a random sample. Use a significance level of 0.05. Comment on your conclusion.

Answer 1

null Hypothesis:               Ho:   p		=	0.21
alternate Hypothesis:    Ha: p		≠	0.21
for 0.05 level with two tailed test , critical z=			1.960
Decision rule :   reject Ho if absolute value of test statistic |z|>1.96
sample success x   =		42
sample size          n    =		250
std error   se =√(p*(1-p)/n) =		0.0258
sample proportion p̂ = x/n=		0.1680
test stat z =(p̂-p)/√(p(1-p)/n)=		-1.63

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 this sample is not from a population with 21​% dangerous​ fish