Question

a)A sample of 30 patients in a doctor's office showed that they had to wait an average of 35 minutes before they could see the doctor. The sample standard deviation is 9 minutes. Assume the population of waiting times is normally distributed.At 90% confidence, compute the margin of error.

b)Consider the population of electric usage per month for houses. The standard deviation of this population is 119 kilowatt-hours. What is the smallest sample size to provide a 90% confidence interval for the population mean with a margin of error of 32 or less? (Enter an integer number.)

Answer 1

solution

a.

Given that,

= 35

s =30

n = 9

Degrees of freedom = df = n - 1 = 9 - 1 = 8

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

t /2,df = t0.05,8 = 1.860 ( using student t table)

Margin of error = E = t/2,df * (s /n)

= 1.860* ( 30/ 9) = 18.6

b.

Solution :

Given that,

standard deviation =s =   =119

Margin of error = E = 32

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table ( see the 0.05 value in standard normal (z) table corresponding z value is 1.645 )  

sample size = n = [Z/2* / E] 2

n = ( 1.645 * 119 / 32 )2

n =37.42189

Sample size = n =38