The US Bureau of Labor Statistics publishes a variety of une

The U.S. Bureau of Labor Statistics publishes a variety of unemployment statistics, including the number of individuals who are unemployed and the mean length of weeks that the individuals have been unemployed. For December 2014, the Bureau of Labor Statistics reported the mean length of unemployment was 40.8 weeks.

A study on the status of unemployment in City of Chicago with a sample size of 31 unemployed residents shows the average of unemployed length is 44.7 weeks and the sample standard deviation is 5.4 weeks.  

Find the confidence intervals of the mean length of unemployment for the City of Chicago at 95%.

Setup a hypothesis to test whether the length of time in Chicago is longer than national average, and explain the result.

Solution

a)
Confidence Interval
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)=44.7
Standard deviation( sd )=5.4
Sample Size(n)=31
Confidence Interval = [ 44.7 ± t a/2 ( 5.4/ Sqrt ( 31) ) ]
= [ 44.7 - 2.0423 * (0.97) , 44.7 + 2.0423 * (0.97) ]
= [ 42.72,46.68 ]

b)
Set Up Hypothesis
Null Hypothesis H0: U=<40.8
Alternate Hypothesis H1: U>40.8
Test Statistic
Population Mean(U)=40.8
Sample X(Mean)=44.7
Standard Deviation(S.D)=5.4
Number (n)=31
we use Test Statistic (t) = x-U/(s.d/Sqrt(n))
to =44.7-40.8/(5.4/Sqrt(30))
to =4.021
| to | =4.021
Critical Value
The Value of |t | with n-1 = 30 d.f is 1.697
We got |to| =4.021 & | t | =1.697
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho
P-Value :Right Tail - Ha : ( P > 4.0212 ) = 0.00018
Hence Value of P0.05 > 0.00018,Here we Reject Ho
Yes, it is longer than National Average