Question

According to an exit poll in the 2015 Kentucky gubernatorial election, 456 of the 837 total poll respondents reported voting for the Republican candidate. You have been asked to determine if there enough evidence to predict if the Republican candidate won. Test that the proportion of exit poll respondents reported voting for the Republican candidate is greater than 50% in the population of all voters at the 5% significance level.

A)  What are the null and alternative hypotheses?

B) Based on the significance level at which you are testing, what is (are) the critical value(s) for the test?

- A one-sided z-test with z_crit = 1.645

- A two-sided z-test with z_crit = 1.96 and z_crit = -1.96

- A one-sided z-test with z_crit = -1.645

C) Calculate the appropriate test statistic, showing all your work. Enter the standard error you calculated in the box below (without units).

D) Calculate the appropriate test statistic, showing your work. Enter the test statistic you calculated in the box below (without units).

E) Calculate the corresponding p-value from the appropriate table.

F) What conclusions can you draw from the hypothesis test? Be sure to comment on evidence from both the test statistic and p-value.

G) Construct a 95% confidence interval around the sample proportion estimate. Enter the lower bound of the interval you calculated in the box below. (In this case, be sure to use the standard error you calculated when determining the test statistic that uses information about the population proportion.)

H) Construct a 95% confidence interval around the sample proportion estimate. Enter the upper bound of the interval you calculated in the box below. (In this case, be sure to use the standard error you calculated when determining the test statistic that uses information about the population proportion.)

I)  How would you interpret the confidence interval?

J) What connections can you draw between the confidence interval and the hypothesis test?

Answer 1

n = 837

x = 456

p̄ = x/n = 0.5448

α = 0.05

a)

Null and Alternative hypothesis:

Ho : p ≤ 0.50

H1 : p > 0.50

b)

Critical value :

Right tailed critical value, z_crit = ABS(NORM.S.INV(0.05)) = 1.645

Reject Ho if z > 1.645

- A one-sided z-test with z_crit = 1.645

c)

Standard error, se = √( p *(1- p )/n) = 0.0173

d)

Test statistic:

z = (p̄ -p)/se = (0.5448 - 0.5)/0.0173 = 2.59

e)

p-value = 1- NORM.S.DIST(2.5924, 1) = 0.0048

f) Conclusion:

As z < z_crit, so reject the null hypothesis.

or As p-value < α, Reject the null hypothesis

g)

95% Confidence interval :

At α = 0.05, two tailed critical value, z_c = NORM.S.INV(0.05/2) = 1.960

Lower Bound = p̄ - z_c*se = 0.5448 - 1.96 *0.0173 = 0.5109

h)

Upper Bound = p̄ + z_c*se = 0.5448 + 1.96 *0.0173 = 0.5787

i)

As the confidence interval do not contain 50, so we reject the null hypothesis.