Exam
Name___________________________________

SHORT ANSWER.  Write the word or phrase that best completes each statement or answers the question.
A statistics professor asked her students whether or not they were registered to vote. In a sample of 50 of her students (randomly sampled from her 700 students), 35 said they were registered to vote.
1) Find a 95% confidence interval for the true proportion of the professor's students who were registered to vote. (Make sure to check any necessary conditions and to state a conclusion in the context of the problem.)	1)  _____________

2) Explain what 95% confidence means in this context.	2)  _____________

3) What is the probability that the true proportion of the professor's students who were registered to vote is in your confidence interval?	3)  _____________

4) According to a September 2004 Gallup poll, about 73% of 18- to 29-year-olds said that they were registered to vote. Does the 73% figure from Gallup seem reasonable for the professor's class? Explain.	4)  _____________

5) If the professor only knew the information from the September 2004 Gallup poll and wanted to estimate the percentage of her students who were registered to vote to within ±4% with 95% confidence, how many students should she sample?	5)  _____________

The countries of Europe report that 46% of the labor force is female. The United Nations wonders if the percentage of females in the labor force is the same in the United States. Representatives from the United States Department of Labor plan to check a random sample of over 10,000 employment records on file to estimate a percentage of females in the United States labor force.
6) The representatives from the Department of Labor want to estimate a percentage of females in the United States labor force to within ±5%, with 90% confidence. How many employment records should they  sample?	6)  _____________

7) They actually select a random sample of 525 employment records, and find that 229 of the people are females. Create the confidence interval.	7)  _____________

8) Interpret the confidence interval in this context.	8)  _____________

9) Explain what 90% confidence means in this context.	9)  _____________

10) Should the representatives from the Department of Labor conclude that the percentage of females in their labor force is lower than Europe's rate of 46%? Explain.	10)  _____________

A state's Department of Education reports that 12% of the high school students in that state attend private high schools. The State University wonders if the percentage is the same in their applicant pool. Admissions officers plan to check a random sample of the over 10,000 applications on file to estimate the percentage of students applying for admission who attend private schools.
11) The admissions officers want to estimate the true percentage of private school applicants to within ±4%, with 90% confidence. How many applications should they sample?	11)  _____________

12) They actually select a random sample of 450 applications, and find that 46 of those students attend private schools. Create the confidence interval.	12)  _____________

13) Interpret the confidence interval in this context.	13)  _____________

14) Explain what 90% confidence means in this context.	14)  _____________

15) Should the admissions officers conclude that the percentage of private school students in their applicant pool is lower than the statewide enrollment rate of 12%? Explain.	15)  _____________

Pew Research reports that 63% of the U.S. adult cell phone owners use their phone to go online. A company wants to target 16- to 24-year olds for advertising and they wonder if that age group has a similar pattern of phone use.
16) The company wants to estimate the true percentage of 16- to 24-year old cell phone owners who use their phone to go online to within ±7.5%, with 95% confidence. How many cell phone owners in this age group should they sample?	16)  _____________

17) They ignore your advice in Question 1 and just select a random sample of 300 cell phone users aged 16 to 24, and find that 206 of those surveyed use their phone to go online. Create the confidence interval.	17)  _____________

18) Interpret the confidence interval in this context.	18)  _____________

19) Explain what 95% confidence means in this context.	19)  _____________

20) Should the company conclude that the percentage of cell phone owners in this age group who use their phone to go online is different from 63%? Explain.	20)  _____________

1) We have a random sample of less than 10% of the professor's students, with 35 expected successes (registered) and 15 expected failures (not registered), both of which are at least ten, so a Normal model applies.
n = 50,  =  = 0.70,  = 1 -  = 0.30, so SE() =  =  = 0.065
Our 95% confidence interval is
  ± z*SE() = 0.70 ± 1.96(0.065) = 0.70 ± 0.127 = 0.573 to 0.827

We are 95% confident that between 57.3% and 82.7% of the professor's students are registered to vote.
2) If many random samples were taken, 95% of the confidence intervals produced would contain the actual percentage of the professor's students who are registered to vote.
3) There is no probability involved-once the interval is constructed, the true proportion of the professor's students who were registered to vote is in the interval or it is not.
4) The 73% figure from Gallup seems reasonable since 73% lies in our confidence interval.
5) ME = 
0.04 = 1.96
n =   = 473.24 ≈ 474
Note Since there are only 700 students in the professor's class, she cannot sample this many students without violating the 10% condition!
6) ME = z*  
0.05 = 1.645 
 = 
n = 270.6025 ≈ 271
They should sample at least 271 employment records.
7) We have a random sample of less than 10% of the employment records, with 229 successes (females) and 296 failures (males), so a Normal model applies.
n = 525,   = 0.436 and  = 0.564, so SE() =  =  = 0.022
margin of error ME = z* × SE() =(1.645)(0.022) = 0.0362.

Confidence interval  ± ME = 0.436 ± 0.0362 or (0.3998, 0.4722)
8) We are 90% confident that between 40.0% and 47.2% of the employment records from the United States labor force are for females.
9) If many random samples were taken, 90% of the confidence intervals produced would contain the actual percentage of all female employment records in the United States labor force.
10) No. Since 46% lies in the confidence interval, (0.3998, 0.4722), it is possible that the percentage of females in the labor force matches Europe's rate of 46% females in the labor force.
11) ME = 
0.04 = 1.645
 = 
n = 178.60 ≈ 179
They should sample at least 179 applicants. (423 if  = 0.5 is used)
12) We have a random sample of less than 10% of the applicants, with 46 successes and 404 failures, both of which are at least ten, so a Normal model applies.  The confidence interval is (0.079, 0.126).
13) We are 90% confident that between 7.9% and 12.6% of the applicants attend private high schools.
14) If many random samples were taken, 90% of the confidence intervals produced would contain the actual percentage of all applicants who attend private schools.
15) No. Since 12% lies in the confidence interval it's possible that the percentage of private school students in the applicant pool matches the statewide enrollment rate.
16) ME = z*
0.075 = 1.96
 = 
n = 159.196 ≈ 160
They should sample at least 160 applicants.
17) We have a random sample of less than 10% of the number of cell phone users in this age group, with 206 successes and 944 failures, both of which are at least ten, so a Normal model applies. The confidence interval is (0.634, 0.739).
18) We are 95% confident that between 63.4% and 73.9% of the all cell phone owners aged 16 to 24 use their phone to get online.
19) If many random samples were taken, about 95% of the confidence intervals produced would contain the actual percentage of all cell phone owners aged 16 to 24 who use their phone to get online.
20) Yes. Since 63% does not lie in the confidence interval it's unlikely that the percentage of all cell phone owners aged 16 to 24 who use their phone to get online is 63%.