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.

Answer 1

Let p1 represents proportion of females who are left-handed

Let p2 represents proportion of males who are left-handed

Given,

n1 = 815 and x1 = 66 so, p̂1 = x1/n1 = 66/815 = 0.08098

n2 = 575 and x2 = 60 so, p̂2 = x2/n2 = 60/575 = 0.10435

Pooled Proportion = (x1+x2)/(n1+n2) = (66+60)/(815+575) = 0.09065

Step 1: Null and alternative hypothesis are as follows:

Ho: p1 ≥ p2  

Ha: p1 < p2  

Step 2: Test statistic calculation

Test statistic z = (0.08098-0.10435)/(sqrt(0.09065*(1-0.09065)*(1/815+1/575))

Test statistic z = -1.4946

Step 3: P-value

P-value corresponding to z = -1.4946 is 0.0675 (Obtained using p-value calculator)

Step 4: Decision

Since p-value = 0.675 > α = 0.05, we fail to reject null hypothesis

Step 5: Conclusion

At 0.05 significance level, there is not enough evidence to conclude that the proportion of females who are left-handed is lower than the proportion of males who are left-handed