I have done part(a), I would like to know how to do part(b) and part(c), includi

Question

I have done part(a), I would like to know how to do part(b) and part(c), including R codes.

Let X1, , Xn be a random sample from the inverse Gaussian distribution, IG(, ), whose pdf is 1S: 1/2 r > 0. 2Tc (a) Show that the MLE of and are X and (b) Schwartz and Samanta (1991) proved that n/X"X2-1. Use this result to derive a 100 , (1-a)% confidence interval for (c) Kumagai and Matsunaga (1995) gave the following dataset of cholorbenzene ex- posure concentrations in ppm (15 min weighted time average) for a worker in a particular industr 13.7, 10.2, 9.9, 4.3, 5.6, 45.6, 42.0, 14.1, 3.8, 9.3, 10.6, 91.3, 91.3, 2.2, 3.8, 6.0, 17.8, 131.8, 31.0, 4.2 1.5, 7.5, 2.5, 2.4, 51.9, 2.6, 27.6, 1.7, 7.0, 2.1, 12.9, 12.3 i. Assuming an inverse Gaussian distribution is an appropriate model for these data, compute maximum likelihood estimates for and give a 95% confidence interval ii. From the data above, construct a QQ plot comparing quantiles of the fitted in- verse Gaussian distribution IG(A, X) with the empirical quantiles. Is the inverse normal model appropriate for these data? Hint: Quantiles for the IG distribution may be computed using the function qinvgauss in the R package statmod.]

Explanation / Answer

b) We are given the result that n/^ ~2n-1 distribution. Then we have to derive a 100(1-)% confidence interval for .

Here we derive a two sided confidence interval.

P(21-/2,n-1 < n/^ < 2/2, n-1) =100(1-)%

Rearranging this we get:

P[ (^21-/2,n-1)/n < < (^2/2, n-1)/n ] = 100(1-)%

So, the required two sided confidence interval is:

[ (^21-/2,n-1)/n , (^2/2, n-1)/n ]

(c) (i)So we have the data provided to us. To get the maximum likelihood estimates we put the data values in the MLE Estimator obtained in (a) and the expression for the confidence interval obtained in (b).

So, we have

n = 32 and Xbar=21.265625 and (X-1i – Xbar-1) = 4.423362

^ = 32/4.423362 = 7.234316341

Thus, the MLE estimate of is 7.234316341.

And to get the value of the 95% confidence interval we put the value in the Expression for the confidence interval.

From the tables we have:

20.975 =17.53873858   20.025,31=48.23188959

Thus, the value of the confidence interval is (2.320,10.9038).

          maxit=200L, tol=1e-14, trace=FALSE).

Or try putting qinvgauss(x, mu, lambda=1) here x will be your vector of the values.

You should get the required plot using this command.

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