Question 1 (a) Since X and Y be independent random variables both with the same mean µ 0. So E(X)=µ and E(Y)=µ Now W=aX + bY a,b are constants i. E(W)=E(aX + bY) E(W)=E(aX)+E(bY) ...

Question 1
(a) Since X and Y be independent random variables both with the same mean µ 0.
So
E(X)=µ and E(Y)=µ
Now
W=aX + bY a,b are constants
i. E(W)=E(aX + bY)
E(W)=E(aX)+E(bY) Linearity Property
E(W)=aE(X)+bE(Y) because E(cX)=cE(X)
E(W)=a*µ +b*µ
E(W)=(a+b)*µ
ii.
W is unbiased estimator of µ if
E(W)=µ
· E(W)=(a+b)*µ =µ
· (a+b)*µ =µ
· a+b=1
· b=1-a
So
W=a*X+b*Y=a*X+(1-a)*Y
W=a*X+(1-a)*Y
So W is unbiased estimator of µ
(b)
i. We know that value of Probability must lie between 0 and 1
So
Now
So
Multiply both sides by 4
Now
But
So
Thus
ii.
Now we know that Likelihood function of from given data

Now Let C== Constant    
iii.











iv. For Maximum Likelihood
So    
·
· So or or
XXXXXXXXXXBut We Know that
XXXXXXXXXXSo only value fulfills this condtions
XXXXXXXXXXNow for Fair unbiased die
XXXXXXXXXXSince so die is biased with higher probability of outcomes of 1 and 6
(C)
i. Let ‘X’ follows Geometric distribution with parameter p.
So
The Probability mass function is
Now Likelihood function is
Differentiate w.r.t P
For Maximum Likelihood
Now
So
ii. Now we know that
Question 2
(a)
Given data
The sample mean windscreen replacement time , 23515
Sample standard deviation windscreen replacement times was 5168 hours of flight i.e s=5168
n=84
i.
We will use t-test to find Confidence interval
Now Degree of Freedom dof=n-1=84-1=83
From t-table
t at 90% with 83 dof=
Now 90% confidence interval, CI
CI=
CI=[ XXXXXXXXXX, XXXXXXXXXX]
ii. Interpretation:- We are 90% confident that the mean replacement time of this type aircraft windscreen lies in in between XXXXXXXXXXto XXXXXXXXXXhours of flight.
(b)
i.
We have

Now

Standard E
or, SE=
Level of significance,
Critical value, z=1.96
So Required Confidence interval
CI= (
CI=[-0.0141 , XXXXXXXXXX]
Hence required 95% confidence interval is
[-0.0141,-0.0021]
ii. Confidence interval does not contain zero so there is no evidence to conclude that proportions of conventionally treated hypertension patients who later suffered a stroke and the proportion of hypertension patients treated with the alternative drug who later suffered a stroke. So there is no evidence to conclude that the proportions of hypertension patients who suffered a stroke would not depend on which treatment they were being given. The data is not consistent with that suggestion.
(c)
i.
Perform Descriptive analysis on Coin 1 and coin 4 using Descriptive Statistics Tool in Minitab.
We get
As we can see from above data standard deviation in Coin 1 is more than that of standard deviation in Coin 4.This means that Coin 1 has less Variability that of coin 4.
ii.
Now we have to obtain 90% two sample t-interval using minitab. We will use “2 sample t” tool from basics statistics. Then Select “Each sample is in its own column” option and Select “Coin 1” and “Coin 4” coloumn . Then set parameter like confidence interval 90%.Assume Equal Variances. Then click on OK.
We will get following output
So at 90% Confidence Interval true population means lies between interval (0.709,1.1551).
iii. The distribution used in constructing the confidence interval is T distribution (Two tailed).The value of its parameter is the difference i.e, 1.13 from the output. The t-value is a way to quantify the difference between the population means.
iv.
Do the process same as part two but select coin4 in sample 1 and coin 1 in sample 2.We will get following Output.
The sign of a t-value tells us the direction of the difference in sample mean. Negative t value means that it lies to the left of the mean.