Question

Suppose for a random sample of 49 CMU students, the mean amount of time spent eating or drinking per day is 1.58 hours with a S.D. of 0.62 hours!

a. What is the Standard Error of the estimate of the population mean?

b. What is the Margin of Error at a C.L. of 90%?

c. What is the right boundary of the C.I. at a C.L. of 90%?

d. At a 98% C.L., if the researcher wants to limit the error within 0.2 hours range, what size of the sample is needed?

Answer 1

Solution :

Given that,

mean = = 1.58

standard deviation = = 0.62

n = 49

a)

= = 1.58

= / n = 0.62/ 49 = 0.0886

The Standard Error of the estimate of the population mean is 0.0886

b)

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

Margin of error = E = Z_/2* ( /n)

= 1.645 * ( 0.62/ 49 )

= 0.15

The Margin of Error at a C.L. of 90% is 0.15

c)

At 90% confidence interval estimate of the population mean is,

- E < < + E

1.58 - 0.15 < < 1.58 + 0.15

1.43 < < 1.73

(1.43 , 1.73)

The right boundary of the C.I. at a C.L. of 90% is (1.43 , 1.73)

d)

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

Z_/2 = Z_0.01 = 2.326

sample size = n = (Z_/2* / E) ²

n = (2.326 *0.62 / 0.2)²

n = 51.99

n = 52

The size of the sample is 52