Question

.As a political scientist, you are interested in evaluating the average number of campaign stops
that Senate candidates make per week during the general election campaign. Previous research
on the subject suggests that the average number of campaign stops per weeks is equal to 26
(conventional wisdom). You are interested in testing this against the possibility that this is not
the parameter value (two-tailed test). You collect a sample of n = 20 candidates, and find a
sample mean of 32 with a sample standard deviation of 10. Use significance level α = 0:05, and
clearly state the hypotheses, the test statistic, the p-value, and your conclusion (in complete, plain
language sentences). Confirm your results by constructing a 95% confidence interval, and
explain how this confidence interval offers confirmation of your hypothesis test.

Answer 1

Solution :

Given that ,

= 26

= 32

s = 10

n = 20

The null and alternative hypothesis is ,

H₀ :   = 26

H_a :    26

This is the two tailed test .

Test statistic = z

= ( - ) / / n

= ( 32 - 26) / 10 / 20

= 2.68

The test statistic = 2.68

df = n - 1 = 20 - 1 = 19

P - value = 2 * 1 - ( t_2.68,19 )

= 2 * 1 - 0.9926

= 2 * 0.0074

= 0.0148

P-value = 0.0148

= 0.05

0.0148 < 0.05

P-value <

Reject the null hypothesis .

conclusion : - There is sufficient evidence to test the claim , that the average number of campaign stops per weeks is equal to 26 at α = 0:05 significance level .

Given that,

Point estimate = sample mean = = 32

sample standard deviation = s = 10

sample size = n = 20

Degrees of freedom = df = n - 1 = 20 - 1 = 19

At 95% confidence level

= 1 - 95%

=1 - 0.955= 0.05

/2 = 0.025

t/2,df = t_{0.05 . 19} = 2.093

Margin of error = E = t_/2,df * (s /n)

= 2.093 * (10 / 20)

Margin of error = E = 4.68

The 95% confidence interval estimate of the population mean is,

- E < < + E

32 - 4.68 < < 32 + 4.68

27.32 < < 36.68

( 27.32 , 36.68 )

The 95% confidence interval estimate of the population mean is :  ( 27.32 , 36.68 )

2.68 < ( 27.32 , 36.68 )

Test statistic < Confidence interval

Reject the null hypothesis .