Question

An article published in the Washington Post claims that 45 percent of all Americans have brown eyes. A random sample of n=82 college students students found 30 who had brown eyes.
Consider testing

?0:?=.45H0:p=.45
??:?≠.45Ha:p≠.45

(a) The test statistic is z =

(b) P-value =

Answer 1

(a)

The test statistic is z = -1.53

(b)

P-value = 0.1260

Solution:

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

The null and alternative hypotheses for this test are given as below:

H₀: p = 0.45 versus H_a: p ≠ 0.45

This is a two tailed test.

Test statistic formula for this test is given as below:

Z = (p̂ - p)/sqrt(pq/n)

Where, p̂ = Sample proportion, p is population proportion, q = 1 - p, and n is sample size

x = number of items of interest = 30

n = sample size = 82

p̂ = x/n = 30/82 = 0.365853659

p = 0.45

q = 1 - p = 0.55

Z = (p̂ - p)/sqrt(pq/n)

Z = (0.365853659 - 0.45)/sqrt(0.45*0.55/82)

Z = -1.5316

The test statistic is z = -1.53

P-value = 0.1260

(by using z-table)

P-value > α = 0.05

So, we do not reject the null hypothesis at 5% level of significance.

There is sufficient evidence to conclude that 45 percent of all Americans have brown eyes.