Question

The CEO of a company wants to estimate the percent of employees that use company computers to go on Facebook during work hours with 95% confidence. He selects a random sample of 150 of the employees and finds that 53 of them logged onto Facebook that day. Compute the 95% confidence interval for the population proportion.

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b. Image

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The CEO of a company wants to estimate the percent of employees that use company computers to go on Facebook during work hours with 95% confidence. He selects a random sample of 150 of the employees and finds that 53 of them logged onto Facebook that day. What is the point estimate of the proportion of the population that logged onto Facebook that day?

a. 0.35

b. 0.53

c. 0.25

d. 0.65

A statistics teacher started class one day by drawing the names of 10 students out of a hat and asked them to do as many pushups as they could. The 10 randomly selected students averaged 15 pushups per person with a standard deviation of 9 pushups. Suppose the distribution of the population of number of pushups that can be done is approximately normal. What is the standard error of the mean?

a. 4.743

b. 3.061

c. 0.900

d. 2.876

Answer 1

1)

sample success x =		53
sample size          n=		150
sample proportion p̂ =x/n=		0.3533
std error se= √(p*(1-p)/n) =		0.0390
for 95 % CI value of z=		1.960
margin of error E=z*std error   =		0.0765
lower bound=p̂ -E                       =		0.2768
Upper bound=p̂ +E                     =		0.4298
from above 95% confidence interval for population proportion =(0.277,0.43)

2)

estimate of the proportion of the population that logged onto Facebook that day =53/150=0.35

3)

standard error of the mean =9/sqrt(10)=2.846