Question

6.4 Ounsted 1953] presents data about cases with convulsive disorders. Among the cases there were 82 females and 118 males. At the 5% significance level, test the hypothesis that a case is equally likely to be of either gender. The siblings of the cases were 121 females and 156 males. Test at the 10% significance level the hypothesis that the siblings represent 53% or more male births.

Answer 1

Solution:

Here, we have to use z test for population proportion.

Null hypothesis: H₀: A case is equally likely to be of either gender.

Alternative hypothesis: H_a: A case is not equally likely to be of either gender.

H₀: p = 0.5 versus H_a: p ≠ 0.5

This is a two tailed test.

Test statistic formula for one sample z test for population proportion is given as below:

Z = (P - p)/sqrt(p*(1 - p)/n)

Where, P is the sample proportion, p is the population proportion, Z is critical value, and n is sample size.

We are given

Level of significance = α = 0.05

n = 82+118 = 200

x = 82

P = x/n = 82/200 = 0.41

Z = (0.41 – 0.50)/sqrt(0.50*(1 – 0.50)/200)

Z = -0.09/ 0.0354

Z = -2.5456

P-value = 0.0109

(by using z-table)

P-value < α = 0.05

So, we reject the null hypothesis

There is insufficient evidence to conclude that a case is equally likely to be of either gender.

Now, we have to use another z test for population proportion for checking the following hypotheses:

Null hypothesis: H₀: The siblings represent 53% or more male births.

Alternative hypothesis: H_a: The siblings represent less than 53% of male births.

H₀: p ≥ 0.53 versus H_a: p < 0.53

This is a lower tailed test.

We are given

Level of significance = α = 0.10

n = 121 + 156 = 277

x = 156

P = x/n = 156/277 = 0.563176895

Z = (P - p)/sqrt(p*(1 - p)/n)

Z = (0.563176895 – 0.53)/sqrt(0.53*(1 – 0.53)/277)

Z = 0.033177 / 0.0300

Z = 1.1063

P-value = 0.8657

(by using z-table)

P-value > α = 0.10

So, we do not reject the null hypothesis

There is sufficient evidence to conclude that the siblings represent 53% or more male births.