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A study has been conducted to investigate the association between type II diabetes and incidence of chronic kidney disease (CKD). 100 people with type II

A study has been conducted to investigate the association between type II diabetes and incidence of chronic kidney disease (CKD). 100 people with type II diabetes and 100 people with no diabetes were recruited into the study and followed up for 10 year. The participants had normal kidney function at recruitment and the primary outcome was whether the patients develop chronic kidney disease during follow-up.

The dataset contains 200rows (one row per patient) and three columns-Subject: subject number (not to be used in analysis)-Diabetes coded with 1 if the patient had type 2 diabetes at recruitment and 0 if the patient had no diabetes at recruitment-CKD coded with1 if the participant has developed CKD during follow-up and zero otherwise.

(i)At the 5% level of significance test for an association between Diabetes and CKD incidence.

(ii)Calculate a 95% confidence interval for the difference in proportions of patients with CKD in the diabetic vs non-diabetic group.

(iii)Compare the results under points (i) and (ii) and formulate the conclusion.

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Problem 1: 10 points A small hair dresser shop operates with three hair stylists. It has two waiting seats so that when a new customer arrives and all stylists are busy and there is no vacant seat, the customer leaves the shop. The state space S of this \"birth-and-death\" process is S={0,1,2,3,4,5}, where a: E 3 indicates the number of customers either waiting or having the ongoing service. Consider this as a process with \"birth rates\" and \"death rates\" as follows: birth rates: At = 3 per hour, for k = 0,1,. . . , 4, death rates: at) =0, pk = k- p for k=1,2, 3, and p4 =p5 = 311, where p: = 4 per hour. Obviously, pk = 0 for k > 5 and A1,, = 0 for k 2 5. 1. Derive the stationary distribution of the number of customers in the shop. 2. Determine the limiting probability of the idle state, a: = 0, that is: rm P [X (t) = D] iroo 3. Determine the limiting probability of losing customers because the shop is full, that is: h'm P [X (t) = 5] lDOO \fA random walk is expressed as X] =Z, X, = X +Z, t=2.3..... where Z, - WN (Hz: 52 ). That is, E (Z, ) = /z: Var(Z, ) = o,, and Cov( Z,, Z; )=0 for t # s. Let {X,) be the random walk. (Z, ) is iid white noise with mean /, and variance o . And Z, is uncorrelated with X, . From the given information, X, is not stationary. So, mean and variance may vary with t. Thus, E [ X] ] = E [Z] ] = / E[X,] = E[X]] + E[Z,] and so on. = /4 + /2 That implies, E[X, ] = EM, The random walk is nonstationary in the mean when E [X] ] = E [X,] =... =[X,] =0 for all t. That is / =0, then the random walk is nonstationary in the mean. But in the I statement /2 0, so it is wrong. Therefore, statement I is false.For the simple symmetric random walk pengno,1,2... with $0 0, show that Pr$4 05 Pr$3 1s (i) with using the probability mass functions of 54 and $3 from class, (ii) without using the probability mass functions of $4 and $3 from class. Hint: In this case, use the fact that Pr93 is Pr93 is due to symmetrie of the symmetric random walk Exercise 5.1 For the simple symmetric random walk (S,)-0,1,2.... with So - 0, show that P[S - 0] - P[4 - 1] (i) with using the probability mass functions of S, and S; from class, (ii) without using the probability mass functions of S, and Sa from class. Hint: In this case, use the fact that PJS, = 1] = P[S, = -1] due to symmetric of the symmetric random walk (see also Problem 5.2)

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