Question

Marc wants to know if the mean age of the prison population in his city is more than 34 years. He obtains a random sample of 25 prisoners, and finds a mean age of 35.2 years. The population standard deviation is 4.3 years. Use a significance level of 0.05.

use the first picture to answer 13 and 14

Answer 1

Given that,
population mean(u)=34
standard deviation, σ =4.3
sample mean, x =35.2
number (n)=25
null, Ho: μ=34
alternate, H1: μ>34
level of significance, α = 0.05
from standard normal table,right tailed z α/2 =1.645
since our test is right-tailed
reject Ho, if zo > 1.645
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 35.2-34/(4.3/sqrt(25)
zo = 1.395
| zo | = 1.395
critical value
the value of |z α| at los 5% is 1.645
we got |zo| =1.395 & | z α | = 1.645
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value : right tail - ha : ( p > 1.395 ) = 0.081
hence value of p0.05 < 0.081, here we do not reject Ho
ANSWERS
---------------
null, Ho: μ=34
alternate, H1: μ>34
test statistic: 1.395
critical value: 1.645
13.
decision: do not reject Ho
option:A
p-value: 0.081
14.
we do not have enough evidence to support the claim that if the mean age of the prison population in the city is more than 34 years.
20.
Given that,
mean(x)=189.1
standard deviation , s.d1=38.7
number(n1)=85
y(mean)=178.7
standard deviation, s.d2 =37.2
number(n2)=79
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, α = 0.01
from standard normal table,right tailed t α/2 =2.35
since our test is right-tailed
reject Ho, if to > 2.35
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (84*1497.69 + 78*1383.84) / (164- 2 )
s^2 = 1442.873
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=189.1-178.7/sqrt((1442.873( 1 /85+ 1/79 ))
to=10.4/5.936
to=1.752
| to | =1.752
critical value
the value of |t α| with (n1+n2-2) i.e 162 d.f is 2.35
we got |to| = 1.752 & | t α | = 2.35
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value: right tail -ha : ( p > 1.7519 ) = 0.04084
hence value of p0.01 < 0.04084,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 > u2
test statistic: 1.752
critical value: 2.35
decision: do not reject Ho
p-value: 0.04084
we do not have enough evidence to support the claim that treatment population 1 is larger than control population2.