Question

Given two independent random samples with the following results:

n1=462 p^1=0.24   n2=599=0.41 p^2=0.41

Can it be concluded that the proportion found in Population 2 exceeds the proportion found in Population 1?  Use a significance level of α=0.01 for the test.

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Step 2 of 5 :  

Compute the weighted estimate of p, p‾p‾. Round your answer to three decimal places.

Answer 1

Solution:

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

H₀: p₁ = p₂ versus H_a: p₁ < p₂

This is a lower tailed test. (Left tailed)

We are given

Level of significance = α = 0.01

n1 = 462, p̂1 = 0.24, n2 = 599, p̂2 = 0.41

p̄ = (n1p̂1 + n2p̂2) / (n1 + n2)

p̄ = (462*0.24 + 599*0.41) / (462 + 599)

p̄ = 0.335975

p̄ = 0.3360

Test statistic formula is given as below:

Z = (p̂1 - p̂2) / sqrt(p̄*(1 - p̄)*((1/n1)+(1/n2)))

Z = (0.24 – 0.41) / sqrt(0.3360*(1 - 0.3360)*((1/462)+(1/599)))

Z = -5.8127

P-value = 0.00

(by using z-table)

P-value < α = 0.01

So, we reject the null hypothesis

There is sufficient evidence to conclude that the proportion found in Population 2 exceeds the proportion found in Population 1.