Question

Conduct a test at the a=0.01 level of significance by determining (a) the null and alternative hypotheses, (b) the test statistic, and (c) the P-value. Assume the samples were obtained independently from a large population using simple random sampling. Test whether P1 >P2. The sample data are x1 = 116, n1 = 243, X2 = 139, and n2 = 302.

Answer 1

We are given the following sample data and are asked to test the hypothesis that p1>p2.

11 116,ni = 243, C2 139, n2 = 302, a = 0.01

a) Null hypothesis

$H_0:p_1=p_2$

i.e., there is no significant difference between the two sample proportions

Alternate hypothesis

$H_1:p_1>p_2$(one-tailed test)

i.e., there is a significant difference between the sample proportions.

b) Test statistic under H0

The test statistic for p1 and p2 is given by the formula,

$z=\frac{p_1-p_2}{\sqrt{\widehat{P}\widehat{Q}(\frac{1}{n_1}+\frac{1}{n_2})}}$

Since P is not known, we use the following formula to estimate P.

21+22 ni + n2

$=\frac{116+139}{243+302}$

=0.47(0.4678899083 actually).

$\widehat{Q}=1-\widehat{P}$

  $=1-0.47$

=0.53.

We also have to find p1 and p2

$p_1=\frac{x_1}{n_1}=\frac{116}{243}$

=0.48(0.4773662551 actually)

$p_2=\frac{x_2}{n_2}=\frac{139}{302}$

=0.46

Now we have all the required values to find the test statistic

$z=\frac{0.48-0.46}{\sqrt{0.47*0.53(\frac{1}{243}+\frac{1}{302})}}$

=0.47(0.46499984327 actually)

c) Now we find the p-value

The p-value for the test statistic is 0.3192(from areas of standard normal distribution after subtracting the value obtained from the table at 0.47 by 1 i.e., 1-0.6808 since this is a one-tail test and normal distribution is symmetric).

d) Inference

Since p-value of 0.3192 is gretaer than significance level of 0.01, we accept or in other words fail to reject the null hypothesis.

Hence $p_1=p_2$ i.e., there is no significant difference between the sample proportions.