Practice Problems for Midterm Exam 2 - Statistical Methods | STAT 51100, Exams of Data Analysis & Statistical Methods

Material Type: Exam; Professor: Levine; Class: Statistical Methods; Subject: STAT-Statistics; University: Purdue University - Main Campus; Term: Fall 2006;

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Statistics 511: Statistical Methods
Dr. Levine
Purdue University
Fall 2006
Midterm2: Practice Problems
Aug, 2006
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Purdue University

Midterm2: Practice Problems

Aug, 2006

Purdue University

By how much must the sample size

n

be increased if the width

of the CI

¯x

z

α/

2 σ/

n

is to be halved? If the sample size is

width of the interval? Justify your assertions.increased by a factor of 25, what effect will this have on the

The width of the interval is

L

z

α/

2 σ/

n

; therefore, if we

increase the sample size by the factor of

we have the new

width

L

=

z

α/

2 σ/

n

L/

.

If the new sample size is

n

= 25

n

, the new confidence interval

would have the width of

L

=

L/

.

Aug, 2006

Purdue University

with parameterteachers absent on any given day has a Poisson distributioncourse in probability and statistics, believes that the number ofThe superintendent of a large school district, having once had a

λ

. Use the accompanying data on absences for

days to derive a large-sample CI for

λ

.

λ Hint: The mean and variance of a Poisson variable both equal

, so

Z

X

λ

λ/n

has approximately a standard normal

p distribution. Now proceed as in the derivation of the interval for

by making a probability statement (with probability

λ

)

and solving the resulting inequalities for

λ

.

Absences

0 1 2 3 4 5 6 7 8 9

10

Frequency

1

6

8

10

8 7 5 3 2 1 1

Aug, 2006

Purdue University

The parameter we estimate is

θ

λ

; therefore,

θˆ

X

and

σ

θˆ

n λ^

(^). We will estimate the latter quantity by

n ¯x

(^). Hence,

the large sample confidence interval will be

¯x

z

α/

2

n ¯x

Computations give us

x

i

= 205

and

¯x

. The final

answer is

Aug, 2006

Purdue University

sample ofThe amount of lateral expansion (mils) was determined for a

n

pulsed-power gas metal arc welds used in

deviation wasLNG ship containment tanks. The resulting sample standard

s

mils. Assuming normality, derive a 95%

CI for

σ

2

and for

σ

.

The number of df is

n

; respective critical values are

χ

0 2 . 025

, 8

and

χ

0 2 . 975

, 8

.

The resulting confidence interval for

σ

2

is

2

2

The confidence interval for

σ

is

Aug, 2006