Several years ago, the reported mean age of an inmate on death row was 38.5 year

Question

Several years ago, the reported mean age of an inmate on death row was 38.5 years. A sociologist wondered whether the mean age of a death-row inmate has changed since then. He randomly selects 32 death-row inmates and finds that their mean age is 37.9 with a standard deviation of 9.6. Construct a 95% confidence interval about the mean age of death row inmates. What does the interval imply?

Ho:__ ___ H1:__ ___

Construct a 95% confidence level about the mean age. (____,_____) Use ascending order. Round to one decimal point as needed.

What does the interval imply?

Explanation / Answer

a.
Given that,
population mean(u)=38.5
standard deviation, =9.6
sample mean, x =37.9
number (n)=32
null, Ho: =38.5
alternate, H1: !=38.5
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 37.9-38.5/(9.6/sqrt(32)
zo = -0.35
| zo | = 0.35
critical value
the value of |z | at los 5% is 1.96
we got |zo| =0.35 & | z | = 1.96
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != -0.35 ) = 0.72
hence value of p0.05 < 0.72, here we do not reject Ho
ANSWERS
---------------
null, Ho: =38.5
alternate, H1: !=38.5
test statistic: -0.35
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.72
we do not have enough evidence to support the claim that the mean age of death row inmates

b.
TRADITIONAL METHOD
given that,
standard deviation, =9.6
sample mean, x =37.9
population size (n)=32
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 9.6/ sqrt ( 32) )
= 1.7
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
value of z table is 1.96
margin of error = 1.96 * 1.7
= 3.3
III.
CI = x ± margin of error
confidence interval = [ 37.9 ± 3.3 ]
= [ 34.6,41.2 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, =9.6
sample mean, x =37.9
population size (n)=32
level of significance, = 0.05
from standard normal table, two tailed z /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 = [ 37.9 ± Z a/2 ( 9.6/ Sqrt ( 32) ) ]
= [ 37.9 - 1.96 * (1.7) , 37.9 + 1.96 * (1.7) ]
= [ 34.6,41.2 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [34.6 , 41.2 ] 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
[ANSWERS]
best point of estimate = mean = 37.9
standard error =1.7
z table value = 1.96
margin of error = 3.3
confidence interval = [ 34.6 , 41.2 ]

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