Question

Suppose a government official wants to know what proportion of residents are infected with the an illness. There are a limited number of tests. So a random sample of 180 residents was taken, and these residents were tested for the illness. Of these 180 people, 36 were found to be infected with the illness. a) Find a 95% confidence interval for the true proportion of residents who are infected. b) a healthcare professional says they suspect that fewer than 30% of residents were infected at that time. Use a hypothesis test ( significance level=0.01) to determine if this statement is supported by the data and be sure to show the sample size is large enough to make estimates in the analysis.

Answer 1

Part a)

p̂ = X / n = 36/180 = 0.2
p̂ ± Z(α/2) √( (p * q) / n)
0.2 ± Z(0.05/2) √( (0.2 * 0.8) / 180)
Z(α/2) = Z(0.05/2) = 1.96
Lower Limit = 0.2 - Z(0.05) √( (0.2 * 0.8) / 180) = 0.1416
upper Limit = 0.2 + Z(0.05) √( (0.2 * 0.8) / 180) = 0.2584
95% Confidence interval is ( 0.1416 , 0.2584 )
( 0.1416 < P < 0.2584 )

Part b)

To Test :-

H0 :- P = 0.30
H1 :- P < 0.30

P = X / n = 36/180 = 0.2
Test Statistic :-
Z = ( P - P0) / √(P0 * q0 / n)
Z = ( 0.2 - 0.3 ) / √(( 0.3 * 0.7) /180)
Z = -2.928

Test Criteria :-
Reject null hypothesis if Z < -Z(α)
Z(α) = Z(0.01) = 2.326
Z < -Z(α) = -2.9277 < -2.326, hence we reject the null hypothesis
Conclusion :- We Reject H0

Decision based on P value
P value = P ( Z < -2.9277 )
P value = 0.0017
Reject null hypothesis if P value < α = 0.01
Since P value = 0.0017 < 0.01, hence we reject the null hypothesis
Conclusion :- We Reject H0

There is sufficient evidence to support the claim that fewer than 30% of residents were infected at that time.

n * P >= 10 = 180 * 0.2 = 36
n * (1 - P ) >= 10 = 180 * ( 1 - 0.2 ) = 144

Hence sample size is large enough.