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).

Sample proportion, = 0.7818

Test statistics, z = 5.9577

Step-by-step explanation:

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, : p = 45%   {means that the proportion of all Americans having brown eyes is 45%}

Alternate Hypothesis, : p 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.  =    ~ N(0,1)

where,   = sample proportion of Americans having brown eyes in a sample of 55 students = = 0.7818

             n = sample of students = 55

So, test statistics  =  

                               =  5.9577

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

0.85 or 85%

step-by-step explanation:

1) divide 23 by 27

2) round the decimal to the nearest 10

3) so your answer would be 0.85 or 85%

3 1/3 x 2 3/5 = 8 2/3

step-by-step explanation:

just multiply the first number by the second number to get your answer.