An article in the Washington Post on March 16, 1993 stated that nearly 45% of all Americans have brown eyes. A random sample of 55 students found 43 with brown eyes. Give the numerical value of the sample proportion, p^p^, and find the value of the test statistic. Sample Proportion p^=p^= (round to four decimal places). Test Statistic z=z= (round to four decimal places).

We are given that an article in the Washington Post on March 16, 1993 stated that nearly 45% of all Americans have brown eyes. A random sample of 55 students found 43 with brown eyes.

Let p = population proportion Americans having brown eyes

SO, Null Hypothesis, [tex]H_0[/tex] : p = 45% {means that the proportion of all Americans having brown eyes is 45%}

Alternate Hypothesis, [tex]H_a[/tex] : p [tex]\neq[/tex] 45% {means that the proportion of all Americans having brown eyes is different from 45%}

The test statistics that will be used here is One-sample z proportion statistics;

T.S. = [tex]\frac{\hat p -p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ~ N(0,1)

where, [tex]\hat p[/tex] = sample proportion of Americans having brown eyes in a sample of 55 students = [tex]\frac{43}{55}[/tex] = 0.7818

n = sample of students = 55

So, test statistics = [tex]\frac{0.7818-0.45}{\sqrt{\frac{0.7818(1-0.7818)}{55} } }[/tex]

= 5.9577

Therefore, the value of test statistics, z = 5.9577.