Question

Jamie was working on a project for psychology class to determine whether students were sleep deprived during final week at her university. She interviewed 100 random students in the library and discovered that 70% of them claimed to be sleep deprived that week. Use this information to answer the questions below. (Round your answer to four decimal places.)

1- Find the point estimate as a proportion. (Round your answer to four decimal places.)

2- Find the appropriate z* to use when if you were to compute a 95% confidence interval for the data. (Round your answer to four decimal places.)

3- Compute the standard error estimate for the data. (Round your answer to four decimal places.)

4- Compute the margin of error for the data for a 95% confidence interval. (Round your answer to four decimal places.)

5- Compute the lower limit for the 95% confidence interval. (Round your answer to four decimal places.)

6- Compute the upper limit for the 95% confidence interval. (Round your answer to four decimal places.)

7- Which of the following is the correct interpretation of the confidence interval for the data?

Select one:

a) I am 95% confident this interval contains the population proportion of all students who were sleep deprived during final week at Jamie’s university.
b) This interval would contain 95% of the students who were sleep deprived in Jamie’s university.
c)  There is a 95% chance that the next student Jamie interviews will be sleep deprived.
d)  I am 95% confident, this interval contains the sample proportion of the 100 students in the survey.

Answer 1

1)

pt estiamte p̂ =x/n=

0.7000

2()

for 95 % CI value of z=

1.9600

from excel:normsinv((1+0.95)/2)

3)

se= √(p*(1-p)/n) =

0.0458

4)

margin of error E=z*std error   =

0.0898

5)

lower bound=p̂ -E                       =

0.6102

6)

Upper bound=p̂ +E                     =

0.7898

7)

a) I am 95% confident this interval contains the population proportion of all students who were sleep deprived during final week at Jamie’s university.