Question

2) According to the Social Security Administration, in 2018 the average wages and earnings in the US was $50,000, the median was $32,838 and σ=$10,200 based on 167 million wage earners. a) From the above data, are earnings normally distributed? Explain your reasoning. b) Regardless of you answer above, assume wages and earnings are normally distributed. Construct a 95% confidence interval for the mean using the above data. c) Regardless of you answer above, assume wages and earnings are normally distributed. Change the confidence level to 99% and construct a confidence interval.

Answer 1

2.

a.
assume wages and earnings are normally distributed
b.
TRADITIONAL METHOD
given that,
standard deviation, σ =10200
sample mean, x =50000
population size (n)=167
I.
standard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
standard error = ( 10200/ sqrt ( 167) )
= 789.3
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
value of z table is 1.96
margin of error = 1.96 * 789.3
= 1547.027
III.
CI = x ± margin of error
confidence interval = [ 50000 ± 1547.027 ]
= [ 48452.9729999999,51547.027 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, σ =10200
sample mean, x =50000
population size (n)=167
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
value of z table is 1.96
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 = [ 50000 ± Z a/2 ( 10200/ Sqrt ( 167) ) ]
= [ 50000 - 1.96 * (789.3) , 50000 + 1.96 * (789.3) ]
= [ 48452.9729999999,51547.027 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [48452.9729999999 , 51547.027 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean
c.
TRADITIONAL METHOD
given that,
standard deviation, σ =10200
sample mean, x =50000
population size (n)=167
I.
standard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
standard error = ( 10200/ sqrt ( 167) )
= 789.3
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.01
from standard normal table, two tailed z α/2 =2.576
since our test is two-tailed
value of z table is 2.576
margin of error = 2.576 * 789.3
= 2033.236
III.
CI = x ± margin of error
confidence interval = [ 50000 ± 2033.236 ]
= [ 47966.764,52033.236 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, σ =10200
sample mean, x =50000
population size (n)=167
level of significance, α = 0.01
from standard normal table, two tailed z α/2 =2.576
since our test is two-tailed
value of z table is 2.576
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 = [ 50000 ± Z a/2 ( 10200/ Sqrt ( 167) ) ]
= [ 50000 - 2.576 * (789.3) , 50000 + 2.576 * (789.3) ]
= [ 47966.764,52033.236 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 99% sure that the interval [47966.764 , 52033.236 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contains the true population mean