Question

Conduct the following test at the a=0.10 level of significance by determining

​(a) the null and alternative​ hypotheses,

​(b) the test​ statistic, and​

(c) the​ P-value. Assume that the samples were obtained independently using simple random sampling.

Test whether p1≠p2. Sample data are x1=30​, n1=254​, x2=36​, and n2=301.

Answer 1

Solution:

a)

The  null and alternative​ hypotheses are

H₀ :   = vs H_a :     

b)

₁ = x1 / n1 = 30/254 = 0.1181

₂ = x2 / n2 = 36/301 = 0.1196

Let be the pooled proportion.

= (x1 +x2)/(n1 + n2) = (30+36)/(254+301) = 0.1189

1 - = 1 - 0.1189 = 0.8811

The test statistic z is

z =  

   = (0.1181 - 0.1196)/[0.1189*0.8811*((1/254)+(1/301))]

= -0.345

Test statistic z = -0.054

c)

Here , TWO  tailed test (because sign in Ha)

For two tailed test ,

p value = P(Z < -0.054) + P(Z > +0.054)

Due to symmetry , we can write ,

p value = 2 * P[Z < -0.345)

using z table ,

= 2 * 0.4785

= 0.9570

p value = 0.9570

d)

Decision:

Fail to reject H₀

(Because p value is greater than given alpha level 0.10)

e)

Conclusion:

There is not sufficient evidence to support the claim that