Question

Researchers wanted to compare the proportion of females who are left-handed to the proportion of males who are left-handed. The researchers randomly selected 815 females and 575 males. They observed that 66 of the females and 60 of the males were left-handed.

Test the claim that the proportion of females who are left-handed is lower than the proportion of males who are left-handed. Test the claim at the 5% significance level.

Credit will be given for a complete analysis that includes

1. The null and alternative hypotheses for this test.

2. Check the normality criteria.

3. Compute the sample difference in proportions (female-male)

4. Calculate the pooled proportion and use it to calculate standard error, and the test statistic Z.

5. Determine the P-value and interpret it in the context of this problem.

6. Write a conclusion explaining whether the sample data supports the claim. Use complete sentences.

Answer 1

p1cap = X1/N1 = 66/815 = 0.081
p1cap = X2/N2 = 60/575 = 0.1043

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

(p1cap - p2cap) = (0.081-0.1043) = -0.0233

pcap = (X1 + X2)/(N1 + N2) = (66+60)/(815+575) = 0.0906

Test statistic
z = (p1cap - p2cap)/sqrt(pcap * (1-pcap) * (1/N1 + 1/N2))
z = (0.081-0.1043)/sqrt(0.0906*(1-0.0906)*(1/815 + 1/575))
z = -1.49

P-value Approach
P-value = 0.0681
As P-value >= 0.05, fail to reject null hypothesis.

There is not sufficient evidence to conclude that the the proportion of females who are left-handed is lower than the proportion of males who are left-handed.