5 A Statistics practitioner selected a random sample of 50 o

5. A Statistics practitioner selected a random sample of 50 observation from a population(known) standard deviation equal to 25 and computed a sample mean equal to 100. a. Estimate the population mean with 95% confidence. b. Repeat Part a using a 99% level of confidence. c. Repeat Part a using a population standard deviation equal to 50. d. Repeat Part a using a sample size equal to 400.

Solution

a)
CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=100
Standard deviation( sd )=25
Sample Size(n)=50
Confidence Interval = [ 100 ± Z a/2 ( 25/ Sqrt ( 50) ) ]
= [ 100 - 1.96 * (3.536) , 100 + 1.96 * (3.536) ]
= [ 93.07,106.93 ]

b)
AT 0.01 LOS
= [ 100 - 2.58 * (3.536) , 100 + 2.58 * (3.536) ]
= [ 90.878,109.122 ]

c)

WHEN Standard deviation( sd )=50
Confidence Interval = [ 100 ± Z a/2 ( 50/ Sqrt ( 50) ) ]
= [ 100 - 2.58 * (7.071) , 100 + 2.58 * (7.071) ]
= [ 81.757,118.243 ]

d)WHEN Sample Size(n)=400
Confidence Interval = [ 100 ± Z a/2 ( 25/ Sqrt ( 400) ) ]
= [ 100 - 2.58 * (1.25) , 100 + 2.58 * (1.25) ]
= [ 96.775,103.225 ]