The lifetime (in thousands of hours) X of a light bulb has pdf
g(x)= \lambda e^{-\lambda x}, \hspace{3mm} x\geq 0
for some unknown \lambda >0.
We collect {\color{blue}{n=33}} independent lightbulbs at random and record their lifetime X_1,\ldots ,X_ n, which are all independent copies of X. We find that {\color{blue}{\overline{X_ n}=42.6}} thousand hours.
We now want to test
\displaystyle \displaystyle H_0\, :\, \lambda =0.03 \displaystyle \text {vs} \displaystyle H_1\, :\, \lambda \neq 0.03
at (significance) level \alpha =5\%.
Write down the test statistic T_{n,\text {LR}} for the likelihood ratio test, in terms of \hat{\lambda }^{\text {MLE}}, n.
(Enter hatlambda for \hat{\lambda }^{\text {MLE}} and S_n for S_ n=\sum _{i=1}^ n X_ i.)
T_{n}^{\text {LR}}=\quad
unanswered
What is p-value p^{\text {LR}} of the likelihood ratio test?
(Enter an answer accurate to at least 3 decimal places.)
p^{\text {LR}}=\quad
What is the conclusion of the Likelihood Ratio test?
(Read the choices carefully, especially the subscripts.)
Reject H_0
Do not reject H_0.
Conclude H_0 is true.
Reject H_1
Do not reject H_1.
Conclude H_1 is true.
1 answer
T_{n}^{\text {LR}} = -2\ln\left(\frac{L(\lambda_0)}{L(\hat{\lambda}^{\text{MLE}})}\right)
where \lambda_0 is the null value of \lambda (0.03 in this case), L(\lambda) is the likelihood function, and \hat{\lambda}^{\text{MLE}} is the maximum likelihood estimator of \lambda.
In this case, the likelihood function is given by:
L(\lambda) = \prod_{i=1}^{n} \lambda e^{-\lambda x_i} = \lambda^n e^{-\lambda\sum_{i=1}^{n}x_i}
Given that \overline{X_n} = 42.6 thousand hours, we can estimate \lambda using the maximum likelihood estimator:
\hat{\lambda}^{\text{MLE}} = \frac{1}{\overline{X_n}}
Substituting the values, the test statistic is:
T_{n}^{\text{LR}} = -2\ln\left(\frac{L(0.03)}{L(\hat{\lambda}^{\text{MLE}})}\right) = -2\ln\left(\frac{(0.03)^n e^{-0.03\sum_{i=1}^{n}x_i}}{(\frac{1}{\overline{X_n}})^n e^{-(\frac{1}{\overline{X_n}})\sum_{i=1}^{n}x_i}}\right) = 2n\ln\left(\frac{1}{0.03\overline{X_n}}\right)
The p-value of the likelihood ratio test is the probability, assuming the null hypothesis is true, of obtaining a test statistic as or more extreme than the observed test statistic. It is given by:
p^{\text{LR}} = P(T_{n}^{\text{LR}} \geq k | H_0)
where k is the observed test statistic.
We do not have the observed test statistic, k, so we cannot calculate the p-value or make a conclusion about the test.