Question

Q. As sample of n=60 had mean = 47.6 and a standard deviation = 12.6

Develop a 90% confidence interval for the population mean.

Q. The critical value for the confidence interval

Q. With 90% confidence, what is the margin of error?(provide three decimals)

Q. Lower limit and Upper limit? (Provide 2 decimals)

Q. What sample size is required to ensure the 99% confidence interval has a width no greater than 20 when sampling from a population with σ= 30?

Answer 1

Solution :

1) Given that,

Point estimate = sample mean = = 47.6

sample standard deviation = s = 12.6

sample size = n = 60

Degrees of freedom = df = n - 1 = 60 - 1 = 59

At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2 = 0.05

t/2,df = t0.05,59 = 1.671

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

= 1.671 * (12.6 / 60)

Margin of error = E = 2.718

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

  ± E  

= 47.6  ± 2.718

= ( 44.88, 50.32 )

lower limit = 44.88

upper limit = 50.32

2) Given that,

Population standard deviation = = 30

Margin of error = E = 20

At 99% confidence level the z is,

= 1 - 99%

= 1 - 0.99 = 0.01

/2 = 0.005

Z/2 = Z0.005 = 2.576

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

n = [2.576 * 30 / 20 ]2

n = 14.93

Sample size = n = 15