Question

We want to find the proportion of students at Saint Michael's Middle School who carpool or ride the bus to school.  

a. A quick survey of 30 randomly selected students at Saint Michael's found 7 either rode the bus or carpooled to school. What is the point estimate for the population proportion for students at Saint Michael's who carpool or ride the bus to school?  (a) The point estimate = 7/30 = 0.233

b. Would you say that this would be a good estimate of the proportion of students who carpool or ride the bus to school?  

(b) Yes, because this is a random sample.

c. How large a sample needs to be taken to determine the proportion of students who carpool or ride the bus to school within an error of 2% at a 99% level of confidence? (c) n = (2.576/0.02)^2*0.233*(1 - 0.233) = 2968

d. Assuming the student population at Saint Michael's is 8000 students, what is the largest sample size you could use and still consider the subjects independent?  (d) n = 8000*0.05 = 400

e. Is the sample size found in (c) acceptable according to your answer to (d)?  

f. If we insist on a 99% level of confidence and using your point estimate from (a) what margin of error would we expect if we used the largest possible sample size based on the student population at Saint Michael's?  

g. Using your point estimate from (a), what is the largest level of confidence we could use for a 2% margin of error if we used the largest possible sample size based on the student population at Saint Michael's?

Answer 1

We want to find the proportion of students at Saint Michael's Middle School who carpool or ride the bus to school.  

a. A quick survey of 30 randomly selected students at Saint Michael's found 7 either rode the bus or carpooled to school. What is the point estimate for the population proportion for students at Saint Michael's who carpool or ride the bus to school?

This is a binomial experiment where the outcomes are either riding the bus or/and carpooling or none

x: No. of people riding or carpooling

n: Total surveryed

= x /n

(a) The point estimate = 7/30 = 0.233

b. Would you say that this would be a good estimate of the proportion of students who carpool or ride the bus to school?  

(b) Yes, because this is a random sample.

Since the sample n > 30, we can use the normal approximation.

c. How large a sample needs to be taken to determine the proportion of students who carpool or ride the bus to school within an error of 2% at a 99% level of confidence?

(c) n = (2.576/0.02)^2*0.233*(1 - 0.233) = 2968

Margin of error = Critical value * SE

= *      

alpha = 1 - 0.99 = 0.01

C.V. = Z_0.005

₌ 2.5758.............using normal percentage tables with p = 0.005

Subsituting in the MOE formula

0.02 = 2.5758 *

n = 2967.27

=2968

d. Assuming the student population at Saint Michael's is 8000 students, what is the largest sample size you could use and still consider the subjects independent?  (d) n = 8000*0.05 = 400

The sample size should be at most 0.05 of the population

Population = 8000

maximum n = 0.05 * 8000

= 400

e. Is the sample size found in (c) acceptable according to your answer to (d)?  

Since 400 < 2968

No the sample size is beyond the acceptnce level.

f. If we insist on a 99% level of confidence and using your point estimate from (a) what margin of error would we expect if we used the largest possible sample size based on the student population at Saint Michael's?  

MOE = Critical value * SE

= *      

The level of confidence and 'p are same so C.V. is same and n = 400

MOE = 2.5758 *

MOE = 0.0545

g. Using your point estimate from (a), what is the largest level of confidence we could use for a 2% margin of error if we used the largest possible sample size based on the student population at Saint Michael's?

To have a 2% margin of level we have

MOE = *      

0.02 = *

= 0.9457

We check the normal probability tables for it. where z -score = 0.95

p = 0.8289

But this is 1 - = 0.8289

This is at level of significance = 0.172

= 0.3443

Therefore the confidence llevel = 1 -

C = 0.6557

65.57%