Question

1. Assume that a sample is used to estimate a population mean μμ. Find the margin of error M.E. that corresponds to a sample of size 21 with a mean of 58.7 and a standard deviation of 17.9 at a confidence level of 80%.

Report ME accurate to one decimal place because the sample statistics are presented with this accuracy.
M.E. = __________________

Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places.

2. Express the confidence interval (174.8,273.6)(174.8,273.6) in the form of ¯x ± ME

-x ± ME= ______________ ± _____________

Answer 1

Solution :

Given that,

1) Point estimate = sample mean = = 58.7

sample standard deviation = s = 17.9

sample size = n = 21

Degrees of freedom = df = n - 1 = 21-1 = 20

At 90% confidence level

= 1-0.90% =1-0.9 =0.10

/2 =0.10/ 2= 0.05

t/2,df = t_{0.05,20 = 1.33}

t _/2,df = 1.33

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

= 1.33 * ( 17.9/ 21)

Margin of error = E = 5.177

2)

confidence interval (174.8,273.6)

  = (Lower confidence interval + Upper confidence interval ) / 2

Sample mean = = (174.8 + 273.6) / 2 = 224.2

Margin of error = E = Upper confidence interval -   = 273.6 - 224.2 = 49.4

± ME

224.2 ± 49.4