Question 1077133: Previously, 7.8% of workers had a travel time to work of more than 60 minutes. An urban economist believes that the percentage has increased since then. She randomly selects 120 workers and finds that 18 of them have a travel time to work that is more than 60 minutes. Test the economist's belief at theα=0.05 level of significance
what is the p-value? three decimal place
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the assumed population proportion is .078 (7.8% / 100 = .078)
the sample proportion is 18/120 = .15.
pp is the population proportion
sp is the sample proportion
the sample size is 120.
n is the sample size.
the standard error of the sample is equal to sqrt(pp * pq / n) = sqrt(.078 * .922 / 120) = .024481 rounded to 6 dp.
sp is equal to the sample proportion.
pp is equal to the assumed population proportion
sq is = to 1 - sp
pq is equal to 1 - pp
n is the sample size.
dp = decimal places.
the test is to see if the proportion has increased, therefore the test is for a one tail distribution.
the alpha is .05.
the critical z-score is 1.645 rounded to 3 dp.
the z-score of the sample is equal to (.15 - .078) / .024481 = 2.941 rounded to 3 dp.
this is based on the general formula that z = (x-m)/s
z is the z-score
x is the sample mean proportion
m is the assumed population mean proportion
s is the standard error.
since the test z-score of 2.941 is much higher than the critical z-score of 1.645, you can reject the null hypothesis indicating that the percentage has, in fact, increased since then.
this involves hypothesis testing of a proportion.
a reference is shown below:
https://onlinecourses.science.psu.edu/stat200/node/53
|
|
|
