Question

A sample of 260 observations is selected from a normal population with a population standard deviation of 26. The sample mean is 15.

1. Determine the standard error of the mean.
2. Determine the 98% confidence interval for the population mean. (Use z Distribution Table.) (Round your answers to 3 decimal places.)

Answer 1

TRADITIONAL METHOD
given that,
standard deviation, σ =26
sample mean, x =15
population size (n)=260
I.
standard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
standard error = ( 26/ sqrt ( 260) )
= 1.612
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.02
from standard normal table, two tailed z α/2 =2.326
since our test is two-tailed
value of z table is 2.326
margin of error = 2.326 * 1.612
= 3.751
III.
CI = x ± margin of error
confidence interval = [ 15 ± 3.751 ]
= [ 11.249,18.751 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, σ =26
sample mean, x =15
population size (n)=260
level of significance, α = 0.02
from standard normal table, two tailed z α/2 =2.326
since our test is two-tailed
value of z table is 2.326
we use 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
confidence interval = [ 15 ± Z a/2 ( 26/ Sqrt ( 260) ) ]
= [ 15 - 2.326 * (1.612) , 15 + 2.326 * (1.612) ]
= [ 11.249,18.751 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 98% sure that the interval [11.249 , 18.751 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 98% of these intervals will contains the true population mean
Answer:
a.
standard error =1.612
b.
confidence interval =11.249,18.751