Problem 3 Suppose Xi, i = 1, , n are i.id. with density fo(x) = -291(z 0), where
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Problem 3 Suppose Xi, i = 1, , n are i.id. with density fo(x) = -291(z 0), where > 0 is an unknown parameter. I. Find the likelihood for , L(9) 2. Find the maximum likelihood estimator (MLE) of , -arg max L(9) Prove that the suggested estimator is indeed the MLE 3. Using the result in 3.2., find the MLE of = exp(9). Prove that it is indeed the MLE. Hint: you should not need any calculus to solve this once you solve d 3.2. 4. Prove that the MLE of is consistent. Prove that the MLE of is consistent Problem 4 It is thought that ACT scores are normally distributed with a mean of 22. A particular teacher claims their teaching method produces a mean result higher than 22. 35 of their students take the test, and obtain an average score of 23.3 and standard deviation of 6.1. 1. Check their claim by using an -005 level hypothesis test. Clearly state the null hypothesis, alternative hypothesis, test statistic, and rejection region, as wel as the result of the test. [You may assume the sample size is large enough for the standardized test statistic to be approximately normal.] 2. If it is known that 37, does the result of the test in part (a) change?Explanation / Answer
ANSWER 1.
Given that,
population mean(u)=22
sample mean, x =23.3
standard deviation, s =6.1
number (n)=35
null, Ho: =22
alternate, H1: >22
level of significance, = 0.05
from standard normal table,right tailed t /2 =1.691
since our test is right-tailed
reject Ho, if to > 1.691
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =23.3-22/(6.1/sqrt(35))
to =1.2608
| to | =1.2608
critical value
the value of |t | with n-1 = 34 d.f is 1.691
we got |to| =1.2608 & | t | =1.691
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :right tail - Ha : ( p > 1.2608 ) = 0.10798
hence value of p0.05 < 0.10798,here we do not reject Ho
ANSWERS
---------------
null, Ho: =22
alternate, H1: >22
test statistic: 1.2608
critical value: 1.691
decision: do not reject Ho
p-value: 0.10798
ANSWER 2.
conidering population standard deviation, =3.7
level of significance, = 0.05
from standard normal table,right tailed z /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 = 23.3-22/(3.7/sqrt(35)
zo = 2.0786
| zo | = 2.0786
critical value
the value of |z | at los 5% is 1.645
we got |zo| =2.0786 & | z | = 1.645
make decision
hence value of | zo | > | z | and here we reject Ho
p-value : right tail - ha : ( p > 2.0786 ) = 0.0188
hence value of p0.05 > 0.0188, here we reject Ho
ANSWERS
---------------
null, Ho: =22
alternate, H1: >22
test statistic: 2.0786
critical value: 1.645
decision: reject Ho
p-value: 0.0188
yes, the result is diffrent with population standard
deviation data..
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