Question

11. Assume that you plan to use a significance level of α = 0.05 to test the claim that Pl = P2· Use the given sample sizes and numbers of successes to find the P-value for the hypothesis test.
n1=60, n2=75, x1=20, x2=35

Answer 1

Given Data

For sample 1

sample size is N1 ​= 60

Number of favorable cases is X1​=20

sample proportion is

$p_{1}'=\frac{X1}{N1}=\frac{20}{60}=0.333$

For sample 1

sample size is N2 ​= 75

Number of favorable cases is X2​=35

sample proportion is

$p_{2}'=\frac{X2}{N2}=\frac{35}{75}=0.467$

pooled proportion is computed as

$p'=\frac{X1+X2}{N1+N2}=\frac{20+35}{60+75}=0.407$

Since we plan to use a significance level of α = 0.05 to test the claim that Pl = P2.Null and alternative hypotheses need to be tested is

Ho : p1 ​= p2​ (claim)

Ha : p1 ​≠ p2

This corresponds to a two-tailed test

The significance level is α=0.05, and the critical value for a two-tailed test is

zc​ (lower)= -1.96 using excel formula =NORM.S.INV(0.025)

zc​ (upper)= 1.96 using excel formula =NORM.S.INV(0.975)

The rejection region for this two-tailed test is

R = z < -1.96 & z > 1.96

Test statistics

$z = \frac{p_{1}'-p_{2}'}{\sqrt{p'(1-p')\left ( \frac{1}{n1}+ \frac{1}{n2} \right )}}$

0.333 -0.467 0.407(1 -0.407)( + )

Since z = - 1.567 and fall in non rejection region, null hypothesis is not rejected.

The p-value is p = 0.1172 using excel formula =2*NORM.S.DIST(-1.567,TRUE)

and since p = 0.1172 > 0.05, It is concluded that the null hypothesis is not rejected.

Answer is p value = 0.1172