Question

Two different simple random samples are drawn from two different populations. The first sample consists of 20 people with 9 having a common attribute. The second sample consists of 2000 people with 1440 of them having the same common attribute. Compare the results from a hypothesis test of p 1equalsp 2 ​(with a 0.01 significance​ level) and a 99​% confidence interval estimate of p 1minusp 2.

Identify the test statistic.

Identify the critical​ value(s).

Answer 1

Solution:

Here, we have to use z test for two population proportions.

H₀: p₁ = p₂ versus H_a: p₁ ≠ p₂

Test statistic formula is given as below:

Z = (P1 – P2) / sqrt(P*(1 – P)*((1/N1) + (1/N2)))

Where,

X1 = 9

X2 = 1440

N1 = 20

N2 = 2000

P = (X1+X2)/(N1+N2) = (9 + 1440) / (20 + 2000) = 0.7173

P1 = X1/N1 = 9/20 = 0.45

P2 = X2/N2 = 1440/2000 = 0.72

Critical Values = -2.5758 and 2.5758

(by using z-table)

Z = (P1 – P2) / sqrt(P*(1 – P)*((1/N1) + (1/N2)))

Z = (.45 – .72) / sqrt(0.7173*(1 – 0.7173)*((1/20) + (1/2000)))

Test statistic = Z = -2.6682

P-value = 0.0076

(by using z-table)

P-value < α = 0.01

So, we reject the null hypothesis

Confidence interval for difference between two population proportions:

Confidence interval = (P1 – P2) ± Z*sqrt[(P1*(1 – P1)/N1) + (P2*(1 – P2)/N2)]

Where, P1 and P2 are sample proportions for first and second groups respectively.

Confidence level = 99%

Z = 2.5758

Confidence interval = (.45 – .72) ± 2.5758* sqrt[(0.45*(1 – 0.45)/20) + (0.72*(1 – 0.72)/2000)]

Confidence interval = -0.27 ± 0.2877

Lower limit = -0.27 - 0.2877 = -0.5577

Upper limit = -0.27 + 0.2877 = 0.0177

Confidence interval = (-0.5577, 0.0177)