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

**H _{0} : = 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 H_{0}

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

e)

Conclusion:

There is not sufficient evidence to support the claim that

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