Question

Hypothesis Testing- Problem (3): Problem (3): An article in the Washington Post on March 16, 1993 stated that nearly 45 percent of all Americans have brown eyes. A random sample of n - 82 UNG students found 30 with brown eyes. Suppose p is the true proportion of those who have brown eyes. Does this sample provide evidence at the 5 percent level that the proportion of brown eyes is different than the 45% proportion stated in 1993? Preparation Step: Statement concerning Step 2. Verify data conditions. Calculate the Test Statistics. population parameter- What do they(Researchers) want to prove Step 4. State the conclusion/Decision: Verification Step Step 5. Interpret the Step 1. State the Hypotheses:hepvalue. p-value means What assumptions are needed for the validity of the test calculations |Describe the sample statistic distribution using the graph. Indicate the that you performed. p-value on the graph.

Answer 1

From the article in the washington post we found that proportion of americans having brown eyes= P = 0.45

Proportion of americans not having brown eyes = Q= 0.55

No. of students selected at random =n=82

No. of students in the sample having brown eyes =x=30

sample proportion of students having brown eyes =p= x/n =30/82 = 0.3689

sample proportion of students not having brown eyes = q= 1- p = 0.6341

STEP 1

To test whether the sample is representing the popualtion with proportion P = 0.45

For testing this we can apply normal test for single proportion

set up null hypothesis H0: sample is representing the population with proportion of brown eyes P=.45

Against

set up alternative hypothesis H1: sample is not representing the population with proportion P= 0.45

i.e H1 : P $\tiny \not\equiv$ 0.45

STEP 2

To test the validity of null hypothesis construct the normal test statistic Z= $\tiny \frac{p-P}{\frac{\sqrt{\sqrt{PQ}}}{n}}$ and it follows normal distribution with mean 0 and standard deviation 1

STEP 3

from the given data under the null hypothesis H0 the value of test stistic Z= $\tiny \frac{0.3649-0.45}{\frac{\sqrt{\sqrt{0.45x0.55}}}{82}}$ = - 1.5310

modulus of Z = 1.5310 since it is two tailed test

STEP4

critical value of Z at 5% level of significance is 1.96

we observe that calculated value of Z =1.5310 < the critical value of Z = 1.96

hence we conclude that the null hypothesis is true i.e the sample is representing the population with proportion of brown eyed people in america is P= 0.45

or The P value is 0.1257 at 5% level of significance hence we accept H0

STEP 5 conclusion

from the above normal test we confirm the statement given in washington post

i.e 45% of the americans having brown eye