Question

# An article published in the Washington Post claims that 45 percent of all Americans have brown...

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 =

(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 Ha: 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.

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