1. Hypothesis Testing: You randomly sample N individuals from a population of ba
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Question
1. Hypothesis Testing: You randomly sample N individuals from a population of basketball players and estimate each sampled individual's height. The average height (u) of the population of basketball players is unknown, but the standard deviation (a) is known and equals 4.0 inches. You calculate the sample average (X) and obtain X-73 inches. You would like to test the hypothesis Ho: = 72 inches. Carry out a two-sided hypothesis test for =0.05 and -001. For each hypothesis test, state the rejection region and the p-value that you obtain. Assume the following sample sizes (N) were used to compute (R). a. ii. iii. N=25 N-100 N=1000Explanation / Answer
PART A.
when N=25
AT 0.05 LOS
Given that,
population mean(u)=72
standard deviation, =4
sample mean, x =73
number (n)=25
null, Ho: =72
alternate, H1: !=72
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 = 73-72/(4/sqrt(25)
zo = 1.25
| zo | = 1.25
critical value
the value of |z | at los 5% is 1.96
we got |zo| =1.25 & | 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 != 1.25 ) = 0.21
hence value of p0.05 < 0.21, here we do not reject Ho
ANSWERS
---------------
null, Ho: =72
alternate, H1: !=72
test statistic: 1.25
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.21
AT 0.01 LOS
critical value
the value of |z | at los 1% is 2.576
we got |zo| =1.25 & | z | = 2.576
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != 1.25 ) = 0.21
hence value of p0.01 < 0.21, here we do not reject Ho
------------------------------------------------------------------------------------------
when N=100
AT 0.05 LOS
Given that,
population mean(u)=72
standard deviation, =4
sample mean, x =73
number (n)=100
null, Ho: =72
alternate, H1: !=72
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 = 73-72/(4/sqrt(100)
zo = 2.5
| zo | = 2.5
critical value
the value of |z | at los 5% is 1.96
we got |zo| =2.5 & | z | = 1.96
make decision
hence value of | zo | > | z | and here we reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != 2.5 ) = 0.01
hence value of p0.05 > 0.01, here we reject Ho
ANSWERS
---------------
null, Ho: =72
alternate, H1: !=72
test statistic: 2.5
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0.01
AT 0.01 LOS
critical value
the value of |z | at los 1% is 2.576
we got |zo| =2.5 & | z | = 2.576
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != 2.5 ) = 0.01
hence value of p0.01 < 0.01, here we do not reject Ho
------------------------------------------------------------------------------------------
PART C.
AT 0.05 LOS
when N=1000
Given that,
population mean(u)=72
standard deviation, =4
sample mean, x =73
number (n)=1000
null, Ho: =72
alternate, H1: !=72
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 = 73-72/(4/sqrt(1000)
zo = 7.91
| zo | = 7.91
critical value
the value of |z | at los 5% is 1.96
we got |zo| =7.91 & | z | = 1.96
make decision
hence value of | zo | > | z | and here we reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != 7.91 ) = 0
hence value of p0.05 > 0, here we reject Ho
ANSWERS
---------------
null, Ho: =72
alternate, H1: !=72
test statistic: 7.91
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0
AT 0.01 LOS
critical value
the value of |z | at los 1% is 2.576
we got |zo| =7.91 & | z | = 2.576
make decision
hence value of | zo | > | z | and here we reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != 7.91 ) = 0
hence value of p0.01 > 0, here we reject Ho
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