1. Hypothesis Testing: You randomly sample N individuals from a population of ba

Question

1. Hypothesis Testing: You randomly sample N individuals from a population of basketball players and estimate each sampled individuals height. The average height (u) of the population of basketball players is unknown, but the standard deviation () is known and equals 4.0 inches. You calculate the sample average (X) and obtain 73 inches. You would like to test the hypothesis Ho: = 72 inches. Carry out a two-sided hypothesis test for qs0.05 and =0.01. 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 (X) a. N-25 N-100 N-1000 Suppose that a random sample of heights is obtained from a population having =72 inches and =4.0 inches. The sample average (X) is computed for 500 independent samples of size N. This data is provided in the Excel file labeled Basketball Heights for i) N-25 ii) N=100 iii) N=1000 b. c. For i)-ili), compute a test statistic for the hypothesis test: Ho: H-71 inches. Compute the percentage d. For i)-iii), compute a test statistic for the hypothesis test: Ho: 70 inches. Compute the percentage e, what can you conclude about the Type l and Type error probabilities of your hypothesis test? For i)-iii), compute a test statistic for the hypothesis test: Ho: = 72 inches. Compute the percentage of times that the test statistic falls outside the rejection region for =0.05, =0.01 of times that the test statistic falls outside the rejection region for =0.05, =0.01 of times that the test statistic falls outside the rejection region for =0.05, =0.01

Explanation / 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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