Question

In a random sample of 480 males, it was found that 14.4% write with their left hand. In a random sample of 1040 females, it was found that 9.6% write with their left hand. We want to use a 95% significance level to test the claim that the rate of left-handedness among females is less than that among males. Provide a hypothesis test with all six steps AND use a p-value to make your decision.

Answer 1

Sample 1 be male

Sample 2 be female

For sample 1, we have that the sample size is N_1= 480, the = 14.4% = 0.144

For sample 2, we have that the sample size is N_2 = 1040, the = 96% = 0.096

The value of the pooled proportion is computed as

Also, the given significance level is α=0.05.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho:p1​=p2​

Ha:p1​<p2​

This corresponds to a left-tailed test, for which a z-test for two population proportions needs to be conducted.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the critical value for a left-tailed test is z_c = -1.64

(3) Test Statistics

The z-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that z = 2.744 > z_c = -1.64 , it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p = 0.997 , and since p = 0.997 > 0.05, it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population proportion p1​ is less than p2​, at the 0.05 significance level.