Question

Efforts to find a genetic explanation for why certain people are right-handed and others left-handed have been largely unsuccessful. Reliable data are difficult to find because of environmental factors that also influence a child's "handedness." To avoid that complication, researchers often study the analogous problem of "pawedness" in animals, where both genotypes and the environment can be partially controlled. In one such experiment (27), mice were put into a cage having a feeding tube that was equally accessible from the right or the left. Each mouse was then carefully watched over a number of feedings. If it used its right paw more than half the time to activate the tube, it was defined to be "right-pawed." Observations of this sort showed that 67% of mice belonging to strain A/J are right-pawed. A similar protocol was followed on a sample of thirty-five mice belonging to strain A/HeJ. Of those thirty-five, a total of eighteen were eventually classified as right-pawed. Test whether the proportion of right-pawed mice found in the A/HeJ sample was significantly different from what was known about the A/J strain.

Required:
Use a two-sided alternative and let 0.05 be the probability associated with the critical region.

Answer 1

The null hypothesis is rejected.

There is enough evidence to support the claim that the proportion of right-pawed mice found in the A/HeJ sample differs significantly from what was known about the A/J strain.

Explanation:

This is a hypothesis test for the difference between proportions.

The claim is that the proportion of right-pawed mice found in the A/HeJ sample differs significantly from what was known about the A/J strain.

Then, the null and alternative hypothesis are:

`H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2\\eq 0`

where π1: proportion of right-pawed in strain A/J, and π2: proportion of right-pawed strain A/HeJ.

The significance level is 0.05. We will use the P-value approach to reject the null hypothesis or not.

The sample 1, of size n1=27 has a proportion of p1=0.67.

The sample 2, of size n2=35 has a proportion of p2=0.51.

`p_2=X_2/n_2=18/35=0.51`

The difference between proportions is (p1-p2)=-0.514285714285714.

p_d=p_1-p_2=0.67-0.51=-0.51429

The pooled proportion, needed to calculate the standard error, is:

`p=(X_1+X_2)/(n_1+n_2)=(0+18)/(27+35)=(18)/(62)=0.29032`

The estimated standard error of the difference between means is computed using the formula:

`s_(p1-p2)=\sqrt{(p(1-p))/(n_1)+(p(1-p))/(n_2)}=\sqrt{(0.29032*0.70968)/(27)+(0.29032*0.70968)/(35)}\\\\\\s_(M_d)=√(0.00763+0.00589)=√(0.01352)=0.1163`

Then, we can calculate the z-statistic as:

`z=(p_d-(\pi_1-\pi_2))/(s_(p1-p2))=(-0.51429-0)/(0.1163)=(-0.51429)/(0.1163)=-4.423`

This test is a two-tailed test, so the P-value for this test is calculated as (using a z-table):

`P-value=2\cdot P(t<-4.423)=0.000042`

As the P-value (0.000042) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the proportion of right-pawed mice found in the A/HeJ sample differs significantly from what was known about the A/J strain.