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Ramin Shamshiri STA 5328, HW #3 Due 07/20/09

STA 5328, Homework #3

Due July 20, 2009

Ramin Shamshiri

UFID # 90213353 Note: Problem numbers are according to the 6

th text edition.

8.2. Suppose that 𝐸 𝜃 1 = 𝐸 𝜃 2 = 𝜃,𝑉 𝜃 1 = 𝜎12, and 𝑉 𝜃 2 = 𝜎2

2 .

Consider the estimator 𝜃 3 = 𝑎𝜃 1 + 1 − 𝑎 𝜃 2.

a. Show that 𝜃 3is an unbiased estimator for 𝜃.

b. If 𝜃 1 and 𝜃 2 are independent, how should the constant a be chosen in order to minimize the variance of

𝜃 3?

Solution a:

𝜃 is an unbiased estimator if 𝐸(𝜃 ) = 𝜃. Here we have

𝐸(𝜃 3) = 𝐸[𝑎𝜃 1 + 1− 𝑎 𝜃 2]

= 𝑎𝐸(𝜃 1) + 1− 𝑎 𝐸(𝜃 2)

= 𝑎𝜃 + 1 − 𝑎 𝜃

= 𝒂𝜽+ 𝜽− 𝒂𝜽 = 𝜽

■

Solution b:

Recall that:

𝑉 𝑎𝑋 + 𝑏𝑌 = 𝑎2𝑉 𝑋 + 𝑏2𝑉(𝑌)

Therefore,

𝑉 𝜃 3 = 𝑉 𝑎𝜃 1 + 1− 𝑎 𝜃 2

= 𝑎2𝑉 𝜃 1 + 1 − 𝑎 2𝑉(𝜃 2)

𝑉 𝜃 3 = 𝑎2𝜎12 + 1− 𝑎 2𝜎2

2

In order to find the minimum value of 𝑉 𝜃 3 for a critical point a, we should set the first derivative of 𝑉 𝜃 3 with respect to a equal to zero to find a.

𝑑𝑉(𝜃 3)

𝑑𝑎= 0

2𝑎𝜎12 − 2 1− 𝑎 𝜎2

2 = 0

𝑎𝜎12 − 𝜎2

2 + 𝑎𝜎22 = 0

𝑎 𝜎12 + 𝜎2

2 = 𝜎22

𝒂 =𝝈𝟐𝟐

𝝈𝟏𝟐+𝝈𝟐

𝟐

In order to check that the value of a will minimize 𝑉 𝜃 3 , we can either substitute a in 𝑉 𝜃 3 or we can use the

second derivative test.

𝑑2𝑉(𝜃 3)

𝑑𝑎2= 2𝑎𝜎1

2 + 2𝑎𝜎22 ≥ 0

Using the second derivative test, it is confirmed that 𝐚 =𝝈𝟐𝟐

𝝈𝟏𝟐+𝝈𝟐

𝟐

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Ramin Shamshiri STA 5328, HW #3 Due 07/20/09

8.13. If Y has a binomial distribution with parameters n and p, then 𝑝 1 = 𝑌/𝑛 is an unbiased estimator of p.

Another estimator of p is 𝑝 2 = (𝑌 + 1)/(𝑛 + 2).

a. Derive the bias of 𝑝 2.

b. Derive 𝑀𝑆𝐸(𝑝 1) 𝑎𝑛𝑑 𝑀𝑆𝐸(𝑝 2).

c. For what values of p is 𝑀𝑆𝐸(𝑝 1) < 𝑀𝑆𝐸(𝑝 2)?

Solution a:

The bias of a point estimator 𝜃 is given by 𝐵 𝜃 = 𝐸 𝜃 − 𝜃

Therefore, 𝐵 𝑝 2 = 𝐸 𝑝 2 − 𝑝

= 𝐸 𝑌 + 1

𝑛 + 2 − 𝑝 =

1

𝑛 + 2𝐸 𝑌 + 1 − 𝑝

=1

𝑛 + 2𝐸 𝑌 +

1

𝑛 + 2𝐸 1 − 𝑝

=𝑛𝑝

𝑛 + 2+

1

𝑛 + 2− 𝑝 =

𝑛𝑝 + 1 − 𝑝(𝑛 + 2)

𝑛 + 2=𝟏 − 𝟐𝒑

𝒏+ 𝟐

■

Solution b:

𝑀𝑆𝐸(𝑝 1) = 𝑉 𝑝 1 + 𝐵 𝑝 1 2

𝑉 𝑝 1 =𝑝(1 − 𝑝)

𝑛 , 𝐵 𝑝 1 = 0

Therefore: 𝑴𝑺𝑬(𝒑 𝟏) = 𝑽 𝒑 𝟏 =𝒑(𝟏−𝒑)

𝒏

𝑀𝑆𝐸(𝑝 2) = 𝑉 𝑝 2 + 𝐵 𝑝 2 2

𝑉 𝑝 2 = 𝑉 𝑌+1

𝑛+2 , Recall that: 𝑉 𝑎𝑋 + 𝑏𝑌 = 𝑎2𝑉 𝑋 + 𝑏2𝑉(𝑌)

𝑉 𝑝 2 =1

𝑛 + 2 2𝑉 𝑌 +

1

𝑛 + 2 2𝑉 1 =

1

𝑛 + 2 2𝑉 𝑌 + 0 =

𝑛𝑝(1− 𝑝)

𝑛 + 2 2

𝑉 𝑝 2 =𝑛𝑝 1− 𝑝

𝑛 + 2 2 𝑎𝑛𝑑 𝐵 𝑝 2

2 = 1 − 2𝑝

𝑛 + 2

2

𝑀𝑆𝐸(𝑝 2) =𝑛𝑝 1−𝑝

𝑛+ 2 2 +

1 − 2𝑝

𝑛 + 2

2

=𝒏𝒑 𝟏 − 𝒑 + 𝟏 − 𝟐𝒑 𝟐

𝒏+ 𝟐 𝟐

■

Solution c:

𝑝(1 − 𝑝)

𝑛<𝑛𝑝 1 − 𝑝 + 1 − 2𝑝 2

𝑛 + 2 2

𝑛 + 2 2𝑝 1 − 𝑝 − 𝑛2𝑝 1− 𝑝 − 𝑛 1 − 𝑝2 < 0

𝑛 + 2 2 − 𝑛 𝑝 1 − 𝑝 − 𝑛 1 − 2𝑝 2 < 0

4𝑛𝑝 − 4𝑛𝑝2 + 4𝑝 − 4𝑝2 − 𝑛 + 4𝑛𝑝 − 4𝑛𝑝2 < 0

𝑝2 8𝑛 + 4 − 𝑝 4 + 8𝑛 + 𝑛 > 0

∆= 𝑏2 − 4𝑎𝑐 = 4 + 8𝑛 2 − 4 8𝑛 + 4 𝑛

𝑝1,𝑝2 = 4 + 8𝑛 ± 4 + 8𝑛 2 − 4 8𝑛 + 4 𝑛

2 8𝑛 + 4 =

1

2±

[ 4 + 8𝑛 2 − 4 8𝑛 + 4 𝑛]

4 8𝑛 + 4 2 =

𝟏

𝟐±

𝒏 + 𝟏

𝟖𝒏 + 𝟒

Therefore 𝒑 ≈𝟏

𝟐

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Ramin Shamshiri STA 5328, HW #3 Due 07/20/09

8.24. In a study of the relationship between birth order and college success, an investigator found that 126 in a

sample of 180 college graduates were first born or only children; in a sample of 100 nongraduates of comparable age and socioeconomic background, the number of firstborn or only children was 54. Estimate the

difference in the proportions of first-born or only children for the two populations from which these samples

were drawn. Give a bound for the error of estimation.

Solution:

𝑛1 = 180, 𝑌1 = 126

𝑛2 = 100, 𝑌2 = 54

𝑝 1 − 𝑝 2 =?

𝑝 1 =𝑌1

𝑛1=

126

180= 0.7

𝑝 2 =𝑌2

𝑛2=

54

100= 0.54

Estimating the difference in the proportions:

𝑝 1 − 𝑝 2 = 0.7− 54 = 𝟎.𝟏𝟔

The bound error estimation using probability of 95%:

2𝜎𝑝 1−𝑝 2 = 2 𝑝1(1− 𝑝1)

𝑛1+𝑝2(1− 𝑝2)

𝑛2= 2

0.7(0.3)

180+

0.54(0.44)

10= 2 0.6042 = 𝟎.𝟏𝟐𝟎𝟖𝟒

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Ramin Shamshiri STA 5328, HW #3 Due 07/20/09

8.76. Do SAT scores for high school students fifer depending on the student’s intended field of study? Fifteen

students who intended to major in engineering were compared with 15 students who intended to major language and literature. Given in the accompanying table are the means and standard deviations of the scores on the

verbal and mathematics portion of the SAT for the two groups of students.

Verbal Math

Engineering 𝑦 = 446 s=42 𝑦 = 548 s=57

Language/Literature 𝑦 = 534 s=45 𝑦 = 517 s=52

a. Construct a 95% confidence interval for the difference in average verbal scores of students majoring in engineering and of those majoring in language/literature.

b. Construct a 95% confidence interval for the difference in average math scores of students majoring in

engineering and of those majoring in language/literature. c. Interpret the results obtained in (a) and (b).

d. What assumptions are necessary for the methods used previously to be valid?

Solution a:

𝑛1 = 15 and 𝑛2 = 15, 𝑠1 = 42 and 𝑠2 = 45, 𝑌 1 = 446 and 𝑌 2 = 534

𝑠𝑝 = 𝑛1 − 1 𝑠1

2 + 𝑛2 − 1 𝑠22

𝑛1 + 𝑛2 − 2 =

14 42 2 + 14 45 2

28 = 43.525

95% confidence interval, (using t-table to find 𝑡(0.025 ,𝑑𝑓=28))

𝑌 1 − 𝑌 2 ± 𝑡(𝛼/2,𝑑𝑓=28)𝑠𝑝 1

𝑛1+

1

𝑛2 = 446 − 534 ± 2.048 43.525

1

15+

1

15= −𝟖𝟖± 𝟑𝟐.𝟓𝟒

= [−𝟏𝟐𝟎.𝟓𝟒,−𝟓𝟓.𝟒𝟔] Solution b:

𝑛1 = 15 and 𝑛2 = 15, 𝑠1 = 57 and 𝑠2 = 52, 𝑌 1 = 548 and 𝑌 2 = 517

𝑠𝑝 = 𝑛1 − 1 𝑠1

2 + 𝑛2 − 1 𝑠22

𝑛1 + 𝑛2 − 2 =

14 57 2 + 14 52 2

28 = 54.557

95% confidence interval, (using t-table to find 𝑡(0.025 ,𝑑𝑓=28))

𝑌 1 − 𝑌 2 ± 𝑡(𝛼/2,𝑑𝑓=28)𝑠𝑝 1

𝑛1+

1

𝑛2= 548 − 517 ± 2.048 54.557

1

15+

1

15= 𝟑𝟏± 𝟒𝟎.𝟕𝟗𝟕

= [−𝟗.𝟕𝟗𝟕,𝟕𝟏.𝟕𝟗𝟕] Solution c: It means that we are 95% confidence that the difference in average verbal scores of students

majoring in engineering and of those majoring in language/literature is in the interval = −120.54,−55.46 . Since this interval contains only negative values, we can claim that it appears to be difference in the two mean verbal scores achieved by engineering and language students.

Since the seconds interval, = [−9.797,71.797], contains both positive and negative values, we cannot claim

that there is difference in average math scores of engineering students and language students. In the other

words, we do not have 95% confidence to claim which group has a larger mean.

Solution d:

Samples are independent

Equal variance assumption (𝜎12 = 𝜎2

2)

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Ramin Shamshiri STA 5328, HW #3 Due 07/20/09

8.79. A factory operates with two machines of type A and one machine of type B. The weekly repair costs X for

type A machine are normally distributed with mean 𝜇1 and variance 𝜎2. The weekly repair costs Y for machines

of type B are also normally distributed but with mean 𝜇2 and variance 3𝜎2. The expected repair cost per week

for the factory is thus 2𝜇1 + 𝜇2. If you are given a random sample X1,X2,…,Xn on costs of type A machines and

an independent random sample Y1, Y2, …,Ym on costs for type B machines, show how you would construct a

95% confidence interval for 2𝜇1 + 𝜇2:

a. If 𝜎2 is known.

b. If 𝜎2 is not known.

Solution a:

Let 𝜃 = 2𝜇1 + 𝜇2 and 𝜃 = 2𝑋 + 𝑌 The 95% Confidence interval for 𝜃 = 2𝜇1 + 𝜇2 is:

𝜃 − 𝑍𝛼2

. (𝜎𝜃 ) < 𝜃 < 𝜃 + 𝑍𝛼2

. (𝜎𝜃 )

Where 𝜎𝜃 = 𝜎(2𝑋 +𝑌 ) = 4𝜎𝑋 + 𝜎𝑌 = 4𝜎2

𝑛+

3𝜎2

𝑚 and 𝑍𝛼

2= 𝑍0.25 = 1.96. Therefore:

2𝑋 + 𝑌 − 1.96. 4𝜎2

𝑛+

3𝜎2

𝑚 < 𝜃 < (2𝑋 + 𝑌 ) + 1.96. (

4𝜎2

𝑛+

3𝜎2

𝑚)

𝟐𝑿 + 𝒀 − 𝟏.𝟗𝟔𝝈 𝟒

𝒏+𝟑

𝒎< 𝟐𝝁𝟏 + 𝝁𝟐 < (2𝑿 + 𝒀 ) + 𝟏.𝟗𝟔𝝈

𝟒

𝒏+𝟑

𝒎

■

Solution b:

When 𝜎2 is not known, we should use pooled variance and t-distribution.

Pooled variance, 𝑆𝑝 is calculated as follow:

𝑆𝑝 = 𝑛 − 1 𝑠1

2 + 𝑚 − 1 𝑠22

𝑛 + 𝑚− 2

(2𝑋 + 𝑌 ) ± 𝑡𝛼/2. 𝑆𝑝 4

𝑛+

3

𝑚

In order to make an inference about the population variance 𝜎2 based on a random sample Y1, Y2, …,Ym from a

normal population, a good estimator of 𝜎2 is the sample variance.

𝑆12 =

1

𝑛 − 1 𝑋𝑖 − 𝑋

2

𝑛

𝑖=1

𝑆22 =

1𝑚− 1

𝑌𝑖 − 𝑌 2𝑛

𝑖=1

3

𝑆𝑝2 =

𝑋𝑖 − 𝑋 2𝑛

𝑖=1 + 𝑌𝑖 − 𝑌

2𝑛𝑖=1

3𝑛 + 𝑚− 2

𝟐𝑿 + 𝒀 ± 𝟏.𝟗𝟔 𝑿𝒊 −𝑿 𝟐

𝒏𝒊=𝟏 +

𝒀𝒊 − 𝒀 𝟐𝒎𝒊=𝟏

𝟑𝒏 +𝒎−𝟐

𝟒

𝒏+𝟑

𝒎

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Ramin Shamshiri STA 5328, HW #3 Due 07/20/09

6. Suppose on the average I will receive one phone call per evening. Show that under reasonable assumptions

that the probability y that I will receive exact one phone call tonight is e-1

Solution