HW Set 6: Due 5/11 in quiz session.
HW16: hw_Z, hw_AA, hw_AB (lecture 18), 7.4(a), 7.11(a,b,c).
HW17: 7.15, 7.18, 7.22 (using messy eqn in lecture 19), 7.23 (using simple eqn in
lecture 19; typo in answer), 7.76 (using simple eqn in lect 19).
HW18: 7.27 (There is a typo in the statement of the problem: ln(pi1,pi2) should be ln(pi1/
pi2), 7.28 (Hint: what value of pi gives the largest CI?), 7.33 (typo in answer.)
HW19: 7.39, 7.40(a), 7.48 (use df=Welch's formula), 7.50 (use df=37), 7.53. In all of
these, any boxplots and qqplots should be done by computer, and the rest by hand.
Graded problems (and points): TBA.
Hw_AA.
(a)
μ
p
=
π
= .80.
σ
p
=
n
)
1
(
π
π

=
25
)
80
.
1
(
80
.

= .08
b
(b)
Since 20% do not
favor the proposed changes (so
π
= .20), the mean & standard
deviation of the sampling distribution of this proportion are
μ
p
=
π
= .20 and
σ
p
=
n
)
1
(
π
π

=
25
)
20
.
1
(
20
.

= .08.
(c)
For n = 1000 and
π
= .80,
μ
p
=
π
= .80.
σ
p
=
n
)
1
(
π
π

=
100
)
80
.
1
(
80
.

= .04.
Notice
that it was necessary to quadruple
the sample size (from n=25 to n=100) in order to
cut
σ
p
in half (from
σ
p
= .08 to
σ
p
= .04).
Hw_AB (a)
x has a binomial distribution with n = 5 and
π
= .05.
Writing
P(.05.01
≤
p
≤
.05+.01) in
terms of x,
we find P(.05.01
≤
x/5
≤
.05+.01) = P((.04)5
≤
x
≤
(.06)5) = P(.2
≤
x
≤
.3).
Because x can only have integer values, there is no x between .2 and .3, so the probability of
this event is 0.
(b)
For n= 25,
P(.04
≤
p
≤
.06) = P((.04)25
≤
x
≤
(.06)25) = P(1
≤
x
≤
1.5) =
P(x = 1) =
24
1
25
1
)
95
(.
)
05
(.
=
0.36498.
(c)
For n= 1005,
P(.04
≤
p
≤
.06) = P((.04)100
≤
x
≤
(.06)100) = P(4
≤
x
≤
6).
Using
the normal approximation to the binomial, with
μ
=
n
π
= 100(.05) = 5 and
σ
2
= n
π
(1
π
) = 100(.05)(.95) = 4.75 and
σ
= 2.7195:
P(4
≤
x
≤
6)
≈
P(
1795
.
2
5
6
7195
.
2
5
4

≤
≤

z
) =
P(.46
≤
z
≤
.46) = .3544.
Using the
continuity correction
makes a substantial
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difference in this problem because the interval from 4 to 6 contains the mean of the
distribution (and hence, a large amount of the probability): P(4
≤
x
≤
6)
≈
P(
1795
.
2
5
5
.
6
7195
.
2
5
5
.
3

≤
≤

z
) = P(.69
≤
z
≤
.69) = .5098.
7.4
(a)
Recall:
(
29
n
p
p
π
π
σ
π
μ

=
=
1
and
Even though
π
is unknown, we can still set an upper bound on
p
σ
.
When
p
σ
π
,
5
.
=
is maximized.
So,
n
p
2
1
=
σ
.
In our case,
15811
.
10
=
⇒
=
p
n
σ
(
29
(
29
(
29
4714
.
63
.
63
.
15811
.
10
.
15811
.
10
.
10
.
10
.
So,
=
<
<

=
<
<

=
<

<

z
P
z
P
p
P
π
7.11
(a)
Decreasing the confidence level from 95% to 90% will decrease the associated z
value and therefore make the 90% interval narrower than the 95% interval.
(Note:
see the answer to Exercise 9 above)
(b)
The statement is not correct.
Once a particular confidence interval has been created/
calculated, then the true mean is either in the interval or not.
The 95% refers to the
process of creating confidence intervals; i.e., it means that 95% of all the possible
confidence intervals you could create (each based on a new random sample of size
n) will contain the population mean (and 5% will not).
(c)
The statement is not correct.
A confidence interval states where plausible values of
the population mean are, not where the individual data values lie.
In statistical
inference, there are three types of intervals:
confidence intervals
(which estimate
where a population mean is),
prediction intervals
(which estimate where a single
value in a population is likely to be), and
tolerance intervals
(which estimate the
likely range of values of the items in a population.
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 Spring '08
 Normal Distribution, Student's tdistribution

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