| << Chapter < Page | Chapter >> Page > |
The confidence interval can be used only if the number of successes np′ and the number of failures nq′ are both greater than five.
For the normal distribution of proportions, the z -score formula is as follows.
If then the z -score formula is
Suppose that a market research firm is hired to estimate the percent of adults living in a large city who have cell phones. Five hundred randomly selected adult residents in this city are surveyed to determine whether they have cell phones. Of the 500 people surveyed, 421 responded yes - they own cell phones. Using a 95% confidence level, compute a confidence interval estimate for the true proportion of adult residents of this city who have cell phones.
Press
STAT and arrow over to
TESTS .
Arrow down to
A:1-PropZint . Press
ENTER .
Arrow down to
and enter 421.
Arrow down to
and enter 500.
Arrow down to
C-Level and enter .95.
Arrow down to
Calculate and press
ENTER .
The confidence interval is (0.81003, 0.87397).
Suppose 250 randomly selected people are surveyed to determine if they own a tablet. Of the 250 surveyed, 98 reported owning a tablet. Using a 95% confidence level, compute a confidence interval estimate for the true proportion of people who own tablets.
(0.3315, 0.4525)
For a class project, a political science student at a large university wants to estimate the percent of students who are registered voters. He surveys 500 students and finds that 300 are registered voters. Compute a 90% confidence interval for the true percent of students who are registered voters, and interpret the confidence interval.
Press
STAT and arrow over to
TESTS .
Arrow down to
A:1-PropZint . Press
ENTER .
Arrow down to
and enter 300.
Arrow down to
and enter 500.
Arrow down to
C-Level and enter 0.90.
Arrow down to
Calculate and press
ENTER .
The confidence interval is (0.564, 0.636).
A student polls his school to see if students in the school district are for or against the new legislation regarding school uniforms. She surveys 600 students and finds that 480 are against the new legislation.
a. Compute a 90% confidence interval for the true percent of students who are against the new legislation, and interpret the confidence interval.
(0.7731, 0.8269); We estimate with 90% confidence that the true percent of all students in the district who are against the new legislation is between 77.31% and 82.69%.
b. In a sample of 300 students, 68% said they own an iPod and a smart phone. Compute a 97% confidence interval for the true percent of students who own an iPod and a smartphone.
Press STAT and arrow over to TESTS.
Arrow down to A:1-PropZint. Press ENTER.
Arrow down to x and enter 300*0.68.
Arrow down to n and enter 300.
Arrow down to C-Level and enter 0.97.
Arrow down to Calculate and press ENTER.
The confidence interval is (0.6531, 0.7069).
There is a certain amount of error introduced into the process of calculating a confidence interval for a proportion. Because we do not know the true proportion for the population, we are forced to use point estimates to calculate the appropriate standard deviation of the sampling distribution. Studies have shown that the resulting estimation of the standard deviation can be flawed.
Notification Switch
Would you like to follow the 'Introductory statistics' conversation and receive update notifications?
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|