A random sample of n 64 observations has a mean x 278 and

A random sample of n = 64 observations has a mean x = 27.8 and a standard deviation s = 3.1.

(a) Give the point estimate of the population mean ?.

_______

Find the 95% margin of error for your estimate. (Round your answer to four decimal places.)

_______

(b) Find a 90% confidence interval for ?. (Round your answers to three decimal places.)

(c) Find a 90% lower confidence bound for the population mean ?. (Round your answer to two decimal places.)

(d) How many observations do you need to estimate ? to within 0.5, with probability equal to 0.95? (Round your answer up to the nearest whole number.)
observations

Solution

a)
Point estimate = Mean(x)=27.8
b)
Margin of Error = t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value

Standard deviation( sd )=3.1
Sample Size(n)=64
Margin of Error = t a/2 * 3.1/ Sqrt ( 64)
= 1.998 * (0.3875)
= 0.7742
c)
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x)=27.8
Standard deviation( sd )=3.1
Sample Size(n)=64
Confidence Interval = [ 27.8 ± t a/2 ( 3.1/ Sqrt ( 64) ) ]
= [ 27.8 - 1.998 * (0.3875) , 27.8 + 1.998 * (0.3875) ]
= [ 27.0258,28.5742 ]
d)
Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Z/2 at 0.05% LOS is = 1.998 ( From Standard Normal Table )
Standard Deviation ( S.D) = 3.1
ME =0.5
n = ( 1.998*3.1/0.5) ^2
= (6.1938/0.5 ) ^2
= 153.4526 ~ 154