Question

In a study of 200 randomly selected adult female and 250 randomly selected adult male Internet users, 30% of the females and 38% of the males said that they plan to shop online at least once during the next months. At alpha = 0.10, test the claim that there is a difference between the proportion of female and the proportion of male Internet users who plan to shop online by specifically following and addressing the questions below:

1. Identify the claim and state the null and alternative hypotheses, Ho and Ha
2. Specify the level of significance, alpha
3. Find the critical values and identify the rejection region
4. Find the weighted estimate for p and q
5. Verify that np>=5 and nq>=5 for each sample
6. Use the two-sample z-test to find the standardized test statistic z
7. Determine whether to reject or fail to reject the null hypothesis

Interpret your decision in the context of the original claim

Answer 1

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: p1 = p2
Alternate Hypothesis, Ha: p1 < p2

0.10 is alpha

Rejection Region
This is left tailed test, for α = 0.1
Critical value of z is -1.28.
Hence reject H0 if z < -1.28

p1cap = 0.3
p1cap = 0.38
pcap = (X1 + X2)/(N1 + N2) = (60+95)/(200+250) = 0.3444

200 * 0.30 = 60 >= 5 and 200 * 0.7 = 140 > =5

250 * 0.38 = 95 >= 5 and 250 * 0.62 = 155 >= 5

Test statistic
z = (p1cap - p2cap)/sqrt(pcap * (1-pcap) * (1/N1 + 1/N2))
z = (0.3-0.38)/sqrt(0.3444*(1-0.3444)*(1/200 + 1/250))
z = -1.77

Reject H0

There is sufficient evidence to conclude that there is a difference between the proportion of female and the proportion of male Internet users who plan to shop online