Question

An article published in the Washington Post claims that 45 percent of all Americans have brown eyes. A random sample of n=80n=80 college students students found 30 who had brown eyes.
Consider testing

H0:p=.45H0:p=.45
Ha:p≠.45Ha:p≠.45

(a) The test statistic is zz =

(b) P-value =

Answer 1

Solution;
Given: n=80,X=30
p^ =X/n

=30/80

=0.375
Now , H0: p=0.45 against Ha: p≠ 0.45
Since, the alternative hypothesis is two-tailed test.
α = 0.05
α/2=0.025
Zα/2= 1.96 .... from standard normal table.
Critical value is 1.96
Under the null hypothesis ,the test statistic is ,
Z=(p^ - p)/✓[p(1-p)/n]
Z=(0.375 - 0.45)/✓[0.45(1-0.45)/80]
Z=(−0.075)/0.0556
= −1.3489
Z= -1.35 ....upto 2 decimal place.
Since ,H1 is the two-tailed ,we shall apply
two-tailed test for the test. The significant value of Z at 5% level of significance for two- tailed test is 1.96 since the calculated value of Z= -1.35 is less than the table value of Z= 1.96 so it is significant we do not reject null hypothesis at 5% level of significance.
b) P-value= P(Z< -1.35)+P(Z> 1.35)
=0.0885 +[1- P(Z<= 1.35)]
=0.0885 +[1- 0.9115]
=0.0885 + 0.0885
=0.117
Since, P-value is greater then the level of significance so we failed to reject the null hypothesis i.e. we accept
the null hypothesis.
Note:
A large p-value (> 0.05) indicates weak evidence against the null hypothesis, so you fail to reject the null hypothesis