Twenty motors were put on test under a high-temperature setting. The lifetimes in hours of the motors under these conditions are
given below. Suppose we assume that the lifetime of a randomly selected motor under these conditions, X, has a Gamma (1, theta) distribution.
(a) Derive the maximum likelihood estimate of 9 and evaluate it numerically.
(b) Obtain the sample median of the data, which is an estimate of the median lifetime of a motor. What parameter is it estimating (i.e., determine the median of X)?
(c) Based on the mle, what is another estimate of the median of X?
(d) Obtain a large sample approximate 95% confidence interval for the mean lifetime of a motor.
(a) Derive the maximum likelihood estimate of 9 and evaluate it numerically.
(b) Obtain the sample median of the data, which is an estimate of the median lifetime of a motor. What parameter is it estimating (i.e., determine the median of X)?
(c) Based on the mle, what is another estimate of the median of X?
(d) Obtain a large sample approximate 95% confidence interval for the mean lifetime of a motor.
1 answer:
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The bases are both 2, so we would subtract the exponents. This is because the rule is
(a^b)/(a^c) = a^(b-c)
In this case,
a = 2
b = 3/4
c = 1/2
So this means
b - c = (3/4) - (1/2) = (3/4) - (2/4) = 1/4
After subtracting the exponents, the final exponent is 1/4
So the expression simplifies to 2^(1/4) which is the same as![\sqrt[4]{2}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B2%7D)
(fourth root of 2)
(a^b)/(a^c) = a^(b-c)
In this case,
a = 2
b = 3/4
c = 1/2
So this means
b - c = (3/4) - (1/2) = (3/4) - (2/4) = 1/4
After subtracting the exponents, the final exponent is 1/4
So the expression simplifies to 2^(1/4) which is the same as
