Let X1, ...., Xn be a random sample from a normal population given by the normal density f(x) = 1 σ √2 2π e −(x−µ) 2 2σ2 , −∞ < x < ∞1. Construct for σ known the 90 percent confidence interval for the unknown mean µ by using the standard normal distributed random variable Z = √2 n(X − µ) σ with X = 1 n Pn i=1 Xi . (Hint:you may use that for a standard normal distribution P(Z ≤ 1.65) ≈ 0.95)2. Consider the random variable Z = √2 n(X − µ) S with S 2 = 1 n − 1 Xn i=1 (Xi − X) 2 the so-called sample variance and determine the cdf of this random variable. Construct now for σ unknown the 90 percent confidence interval for the unknown mean µ by using this random variable Z (Hint:you may use all the knowledge you know about normal distributions. In particular, you may use that the random variable (n−1)S 2 σ2 has a chisquare distribution with n − 1 degrees of freedom and for the random variable Z there exist some number tn−1,0.05 satisfying P(Z ≤ tn−1,0.05) = 0.953. If we consider two independent samples of size n1 and size n2 both from a normal population with unknown parameters µ1, σ1 and µ2, σ2 having the same unknown variance σ1 = σ2 = σ determine the distribution of the statistic Z = (X − Y ) − (µ1 − µ2) Sp 2 q 1 n1 + 1 n2 with S 2 p = (n1 − 1)S 2 1 + (n2 − 1)2S 2 2 n1 + n2 − 2 and S 2 i the sample variance of population i, i = 1, 2. (hint:You may use all the knowledge you know about chi-square distributions and normal populations without proof) 4. Construct a 90 percent confidence interval for the difference µ1 −µ2 by using the statistic defined in part 3.5. Toyota manufacturers a variety of cars over a wide range of prices. In particular they sell the Corolla and the Yaris. The view of the managers is that in general buyers of a Corolla are older then buyers of a Yaris. The ages and the model of each buyer were recorded and these statistics are given by the following Type Total sells sample mean age standard deviation S Corolla 30 42 11 Yaris 22 48 10 Construct now the 90 percent confidence interval for the mean age of the buyers of a corolla and of a Yaris and a 90 percent confidence interval for the difference of these ages. (since no calculators are allowed you can leave all your products and sums without evaluation.)6. In part 5 we applied the result of the previous three parts for a normal population. Is that realistic in this particular example? Explain! with(n1 – 1)S? + (n2 – 1)²Sпi + n2 — 2and S? the sample variance of population i, i = 1, 2. (hint:You may use all the knowledgeyou know about chi-square distributions and normal populations without proof)84.Construct a 90 percent confidence interval for the difference u1 – µz by using thestatistic defined in part 3.5.Toyota manufacturers a variety of cars over a wide range of prices. In particularthey sell the Corolla and the Yaris. The view of the managers is that in general buyers ofa Corolla are older then buyers of a Yaris. The ages and the model of each buyer wererecorded and these statistics are given by the followingТуреTotal sells sample mean agestandard deviation SCorolla304211Yaris224810Construct now the 90 percent confidence interval for the mean age of the buyers of a corollaand of a Yaris and a 90 percent confidence interval for the difference of these ages. (sinceno calculators are allowed you can leave all your products and sums without evaluation.)6.In part 5 we applied the result of the previous three parts for a normal population.Is that realistic in this particular example? Explain! Let X1, .., Xn be a random sample from a normal population given by the normal density1f(x):o27e202-0 < x <∞1.Construct for o known the 90 percent confidence interval for the unknown meanu by using the standard normal distributed random variableVn(X – u)with X = ;D- X;. (Hint:you may use that for a standard normal distribution P(Z <1.65) - 0.95)2.Consider the random variableZ =S(1 – X)uAwithn1(X; – X)?1i=1the so-called sample variance and determine the cdf of this random variable. Construct nowfor o unknown the 90 percent confidence interval for the unknown mean u by using this ran-dom variable Z (Hint:you may use all the knowledge you know about normal distributions.In particular, you may use that the random variable (n-YS-n – 1 degrees of freedom and for the random variable Z there exist some number tn-1.0.05satisfyinghas a chisquare distribution withP(Z < tn-1,0.05)= 0.953.If we consider two independent samples of sizen1and sizen2both from a normalpopulation with unknown parameters µ1, 0i and µ2,02 having the same unknown variance01 = 02 = 0 determine the distribution of the statistic(X – Y) – (µ1 – µz)-%3DSp +d,nin2with1 a?112 a?

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MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

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Let X1, ...., Xn be a random sample from a normal population given by the normal density f(x) = 1 σ √2 2π e −(x−µ) 2 2σ2 , −∞ < x < ∞

1. Construct for σ known the 90 percent confidence interval for the unknown mean µ by using the standard normal distributed random variable Z = √2 n(X − µ) σ with X = 1 n Pn i=1 Xi . (Hint:you may use that for a standard normal distribution P(Z ≤ 1.65) ≈ 0.95)

2. Consider the random variable Z = √2 n(X − µ) S with S 2 = 1 n − 1 Xn i=1 (Xi − X) 2 the so-called sample variance and determine the cdf of this random variable. Construct now for σ unknown the 90 percent confidence interval for the unknown mean µ by using this random variable Z (Hint:you may use all the knowledge you know about normal distributions. In particular, you may use that the random variable (n−1)S 2 σ2 has a chisquare distribution with n − 1 degrees of freedom and for the random variable Z there exist some number tn−1,0.05 satisfying P(Z ≤ tn−1,0.05) = 0.95

3. If we consider two independent samples of size n1 and size n2 both from a normal population with unknown parameters µ1, σ1 and µ2, σ2 having the same unknown variance σ1 = σ2 = σ determine the distribution of the statistic Z = (X − Y ) − (µ1 − µ2) Sp 2 q 1 n1 + 1 n2 with S 2 p = (n1 − 1)S 2 1 + (n2 − 1)2S 2 2 n1 + n2 − 2 and S 2 i the sample variance of population i, i = 1, 2. (hint:You may use all the knowledge you know about chi-square distributions and normal populations without proof)

4. Construct a 90 percent confidence interval for the difference µ1 −µ2 by using the statistic defined in part 3.

5. Toyota manufacturers a variety of cars over a wide range of prices. In particular they sell the Corolla and the Yaris. The view of the managers is that in general buyers of a Corolla are older then buyers of a Yaris. The ages and the model of each buyer were recorded and these statistics are given by the following Type Total sells sample mean age standard deviation S Corolla 30 42 11 Yaris 22 48 10 Construct now the 90 percent confidence interval for the mean age of the buyers of a corolla and of a Yaris and a 90 percent confidence interval for the difference of these ages. (since no calculators are allowed you can leave all your products and sums without evaluation.)

6. In part 5 we applied the result of the previous three parts for a normal population. Is that realistic in this particular example? Explain!

with
(n1 – 1)S? + (n2 – 1)²S
пi + n2 — 2
and S? the sample variance of population i, i = 1, 2. (hint:You may use all the knowledge
you know about chi-square distributions and normal populations without proof)
8
4.
Construct a 90 percent confidence interval for the difference u1 – µz by using the
statistic defined in part 3.
5.
Toyota manufacturers a variety of cars over a wide range of prices. In particular
they sell the Corolla and the Yaris. The view of the managers is that in general buyers of
a Corolla are older then buyers of a Yaris. The ages and the model of each buyer were
recorded and these statistics are given by the following
Туре
Total sells sample mean age
standard deviation S
Corolla
30
42
11
Yaris
22
48
10
Construct now the 90 percent confidence interval for the mean age of the buyers of a corolla
and of a Yaris and a 90 percent confidence interval for the difference of these ages. (since
no calculators are allowed you can leave all your products and sums without evaluation.)
6.
In part 5 we applied the result of the previous three parts for a normal population.
Is that realistic in this particular example? Explain!

Let X1, .., Xn be a random sample from a normal population given by the normal density
1
f(x):
o27
e
202
-0 < x <∞
1.
Construct for o known the 90 percent confidence interval for the unknown mean
u by using the standard normal distributed random variable
Vn(X – u)
with X = ;D- X;. (Hint:you may use that for a standard normal distribution P(Z <
1.65) - 0.95)
2.
Consider the random variable
Z =
S
(1 – X)uA
with
n
1
(X; – X)?
1
i=1
the so-called sample variance and determine the cdf of this random variable. Construct now
for o unknown the 90 percent confidence interval for the unknown mean u by using this ran-
dom variable Z (Hint:you may use all the knowledge you know about normal distributions.
In particular, you may use that the random variable (n-YS-
n – 1 degrees of freedom and for the random variable Z there exist some number tn-1.0.05
satisfying
has a chisquare distribution with
P(Z < tn-1,0.05)
= 0.95
3.
If we consider two independent samples of size
n1
and size
n2
both from a normal
population with unknown parameters µ1, 0i and µ2,02 having the same unknown variance
01 = 02 = 0 determine the distribution of the statistic
(X – Y) – (µ1 – µz)
-
%3D
Sp +
d,
ni
n2
with
1 a?
112 a?

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