Description

Ravish Kamath: 213893664
Question 1
a. Find the appropriate allocation of the sample to the two strata.
b. Find the appropriate sample size.
Solution
Part A
We are given 2 strata which are 20 small plants and 26 large plants. Hence our Nh is 20 and 26. Furthermore we are given our range for each strata which are 0 to 100 for small plants and 10 to 200 for large plants. We are unfortunately not given the cost of each observation. Recall the formula for the the sample allocation fraction:
NhSh ah = H
Ph=1NhSh
We have already stated what Nh but need to figure out Sh which is range 4 .
Nh = c(20,26) range = c(100-0, 200-10) Sh = range/4
Let us now input the these values into our sample allocation fraction eqaution.
ah = round((Nh*Sh)/sum(Nh*Sh),4) ah
## [1] 0.2882 0.7118
Part B
To find the appropriate sample size we will use this formula:
(PHh=1NhSh)2

N2D + PHh=1NhSh2
Furthermore, we are given a bound of 100 days.
B = 100
NsqrdD = Bˆ2/4
Solution
n = ceiling((sum(Nh*Sh)ˆ2) / (NsqrdD + sum(Nh*(Shˆ2)))) n
## [1] 41
nh = ceiling(n*ah) nh
## [1] 12 30
We are given a full sample size of 41, with an allocation of 12 to small plants and 30 to the large plants. However we come accross an issue since there are only 26 large plants. Hence we will use the full 26 large plants in our sample and the rest of the small plants will be used in the sample which would be 15.
Question 2
Wage earners in a large firm are stratified into management and clerical classes, the first having 300 and the second having 500 employees. To assess attitude on sick-leave policy, independent random samples of 100 workers each were selected, one sample from each of the classes. After the sample data were collected, the responses were divided according to gender. In the table of results, a = Number who like the policy; b = Number who dislike the policy; and c = Number who have no opinion on the policy.
Find an estimate and an estimated variance of that estimate for each parameter listed: a. Proportion of managers who like the policy
b. Proportion of wage earners who like the policy
c. Total number of female wage earners who dislike the policy
d. Difference between the proportion of male managers who like the policy and the proportion of femalemanagers who like the policy
e. Difference between the proportion of managers who like the policy and the proportion of managers whodislike the policy
Solution
Let us first recreate the table given in the question.
## Mang a Mang b Mang c Cler a Cler b Cler c
## Male 60 15 5 24 4 2
## Female 10 7 3 42 30 8
Further more we are given our populations and sample strata sizes which gives us our information on overall population and sample size. Finally we can also figure out the sampling fraction denoted by fh.
nh = c(100,100) n = sum(nh)
Nh = c(300,500)
N = sum(Nh) Wh = Nh/N fh = nh/Nh
Part A
Our estimated proportion, pˆh, of managers who like the policy is 0.7.
hat_pMang = (df[1,1] + df[2,1])/100
Our estimated variance, Vˆ(pˆ), is 0.0014.
ssh = (nh[1]/(nh[1]-1))*(hat_pMang*(1-hat_pMang)) varhat_pMang = (1-fh[1])*(ssh/(nh[1]-1))
Finally our bound on error, B = 0.07558904:
B = 2*sqrt(varhat_pMang) hat_pMang + B hat_pMang – B
We have a confidence interval of (0.624411, 0.775589).
Part B
Our estimated proportion, pˆst of wage earners who like the policy is 0.675:
Solution
hat_pCler = (df[1,4] + df[2,4])/100 hat_ph = c(hat_pMang, hat_pCler) hat_pSt = sum(Wh*hat_ph)
Our estimated variance, Vˆ(pˆst), is 0.000907197.
hat_pq = (hat_ph*(1-hat_ph))
hat_var_pSt = sum((Whˆ2)*(1-fh)*hat_pq/(nh-1))
Finally our bound on error, B = 0.06023942:
B = 2*sqrt(hat_var_pSt) hat_pSt + B hat_pSt – B
We have a confidence interval of (0.6147606, 0.735294).
Part C
The estimated total, τˆst, number of female wage earners who dislike the policy is 171.
hat_pMang = df[2,2]/nh[1] hat_pCler = df[2,5]/nh[2] hat_ph = c(hat_pMang, hat_pCler) hat_pSt = sum(Wh*hat_ph) hat_tauSt = N*hat_pSt
Our estimated total variance, Vˆ(τˆst), is 463.697:
hat_pq = (hat_ph*(1-hat_ph)) hat_var_pSt = sum((Whˆ2)*(1-fh)*hat_pq/(nh-1)) hat_var_tauSt = Nˆ2 * hat_var_pSt
Finally our bound on error, B = 43.06725:
B = 2*sqrt(hat_var_tauSt) hat_tauSt + B hat_tauSt – B
We have a confidence interval of (127.9328, 214.0672).
Part D
Our estimated difference between the proportion of male managers who like the policy and the proportion of female managers who like the policy is 0.5.
hat_pMangM = df[1,1]/nh[1] hat_pMangF = df[2,1]/nh[1]
We have the estimated variance for Male Managers to be 0.0002272727 and for Female Managers to be 8.522727e-05.
sh_sqrd = (nh[1]/(nh[1]-1))*(hat_pMangM*(1-hat_pMangM)) varhat_pMangM = Wh[1]ˆ2*(1-fh[1])*sh_sqrd/nh[1]
sh_sqrd2 = (nh[1]/(nh[1]-1))*(hat_pMangF*(1-hat_pMangF)) varhat_pMangF = Wh[1]ˆ2*(1-fh[1])*sh_sqrd2/nh[1]

We have the bound on error, B = 0.03535534.
B = 2*sqrt(varhat_pMangM + varhat_pMangF)
(hat_pMangM – hat_pMangF) + B
(hat_pMangM – hat_pMangF) – B
We have a confidence interval of (0.4646447, 0.5353553).
Part E
Our estimated difference between the proportion of managers who like the policy and the proportion of managers who dislike the policy is 0.48.
hat_pMang_a = (df[1,1] + df[2,1])/100 hat_pMang_b = (df[1,2] + df[2,2])/100
Here we have the estimated variance for managers who like the policy to be 0.0001988 and for managers who dislike the policy to be 0.0001625.
sh_sqrd= (nh[1]/(nh[1]-1))*(hat_pMang_a*(1-hat_pMang_a)) varhat_pMang_a = Wh[1]ˆ2*(1-fh[1])*sh_sqrd/nh[1]
sh_sqrd2 = (nh[1]/(nh[1]-1))*(hat_pMang_b*(1-hat_pMang_b)) varhat_pMang_b = Wh[1]ˆ2*(1-fh[1])*sh_sqrd2/nh[1]
We have a bound of error, B = 0.03801913.
B = 2*sqrt(varhat_pMang_a + varhat_pMang_b)
(hat_pMang_a – hat_pMang_b) – B
(hat_pMang_a – hat_pMang_b) + B
We have a confidence interval of (0.4419809, 0.5180191).
Question 3
a. Estimate the population proportion of those who think they have worked beyond a safe limit.
Calculate a bound on the error of estimation.
b. Do anesthesiologists differ significantly from residents in this matter?
c. Do anesthesiologists differ significantly from nurse anesthetists in this matter?
Solution
Part A
We are given 3 strata which are anesthesiologist, anesthesiology resident and nurse anesthetist. We need to assume Sh2 are equal to each strata, hence ah = Wh. Furtheremore, we can assume the population strata size, Nh, is quite large since as the overall population is for the United States, we can ignore fpc to calculate the variance.
nh = c(913 + 417, 136 + 29, 860 + 240) n = sum(nh) ah = nh/n
yes_ph = c(.687,.824,.782) no_ph = c(.314,.176,.218)
hat_pSt = sum(ah*yes_ph) hat_pSt
## [1] 0.7359807
hat_pq = yes_ph*no_ph hat_var_pSt = sum( (ahˆ2) * (hat_pq/(nh – 1)) ) hat_var_pSt
## [1] 7.408495e-05
B = 2*sqrt(hat_var_pSt)
B
## [1] 0.01721452
hat_pSt + B hat_pSt – B
As shown above we are given the estimated population proportion to be 0.736 with a bound of error of B =
0.017. This gives us a confidence of interval (0.7187662, 0.7531953).
Part B
(yes_ph[1] – yes_ph[2])
## [1] -0.137
ssh = (nh/(nh – 1))*(yes_ph*no_ph) hat_var_ph = ssh/(nh-1)
B = 2*sqrt(hat_var_ph[1] + hat_var_ph[2])
B
## [1] 0.06487289
As we can see from above the difference between anesthesiologists from residents is, -0.137, with a bound of error 0.6487289, and a confidence interval (-0.07212711, -0.2018729). We can deem this as not significant.
Part C
(yes_ph[1] – yes_ph[3])
## [1] -0.095
ssh = (nh/(nh – 1))*(yes_ph*no_ph) hat_var_ph = ssh/(nh-1) B = 2*sqrt(hat_var_ph[1] + hat_var_ph[3])
B
## [1] 0.0356482
As we can see from above the difference between anesthesiologists from nurse anesthestists is, -0.095, with a bound of error 0.0356482, and a confidence interval (-0.1306482,-0.0593518). We can deem this as not significant.
Question 4
In the same survey discussed in Exercise 5.31, the respondents were asked for the longest continuous time (in hours) of administering anesthesia without a break over the last six months. A summary of the results is as follows:
a. Estimate the mean time for the population of those giving anesthesia, with an estimated bound on theerror.
b. Do residents have a significantly higher average than the other groups? Justify your answer statistically.
Solution
Part A
nh = c(1347, 163, 1095) n = sum(nh) ah = nh/n sd = c(0.15, 0.35, 0.11)
ybar_h = c(7.63, 7.74, 6.55) nh = c(1347, 163, 1095) ybarSt = sum(ah*ybar_h) ybarSt
## [1] 7.18291
hat_var_ybarSt = sum((ahˆ2)*((sdˆ2)/nh)) hat_var_ybarSt
## [1] 9.361077e-06
B = 2*sqrt(hat_var_ybarSt) B
ybarSt + B ybarSt – B
The estimated mean time for the population of those giving anesthesia is 7.18291 with a bound on error of
0.006119175. Furthermore we have a confidence interval of (7.176791, 7.189029).
Part B
This is for comparing the significance between the resident and the anesthesiologist.
ybar_h[2] – ybar_h[1]
## [1] 0.11
B = 2*sqrt(sd[1]/nh[1] + sd[2]/nh[2])
B
## [1] 0.09504942
ybar_h[2] – ybar_h[1] – B
## [1] 0.01495058
ybar_h[2] – ybar_h[1] + B
## [1] 0.2050494
As shown above, we have a difference 0.11 with a bound on error of 0.09504942, with a confidence interval, (0.01495058, 0.2050494). As we can see we do see some difference hence we can say there is a significantly higher average.
This is for comparing the significance between the resident and the nurse anesthetist.
ybar_h[2] – ybar_h[3]
## [1] 1.19
B = 2*sqrt(sd[3]/nh[3] + sd[2]/nh[2])
B
## [1] 0.09481974
ybar_h[2] – ybar_h[3] – B
## [1] 1.09518
ybar_h[2] – ybar_h[3] + B
## [1] 1.28482
As shown above, we have a difference 1.19 with a bound on error of 0.09481974, with a confidence interval, (0.09481974, 1.28482). As we can see we do see some difference hence we can say there is a significantly higher average.