Download Homework 5 Solutions 3.84) a) High bias (center of

Homework 5 Solutions
3.84) a) High bias (center of the distribution is far from the population parameter)
High variability (large range, variance, and IQR)
b) Low bias (center of the distribution close to the population parameter)
Low variability (small range, variance, and IQR)
c) Low bias, High variability
d) High bias, Low variability
3.86) Margin of error is roughly twice the standard deviation of the sampling
distribution of a sample proportion or mean. As the sample size increases, the
sampling distribution becomes more concentration around the true value.
However, this has nothing to do with population size. Thus, as long as sample
sizes are comparable, the margin of error should be comparable since the
population is sufficiently large.
4.90) Consider the probability function of cash intake for the individual policies:
Age at Death
25
26
27
28
29
30
Cash Intake
-$99,825
-$99,650
-$99,475 -$99,300 -$99,125
$857
Probability
.00039
.00044
.00051
.00057
.00060
.99749
The mean of the distribution is very close to $875 since that is where most of the
probability mass is. The mean of this distribution is $623. But as the sample size
grows large, the sampling average gets more and more concentrated around the
population mean of $623, thus the negative intakes for a single sample are
counteracted with the thousands of positive intakes.
5.12) a) No: There must be only two possible outcomes.
b) Yes: Binomial(20, p) is reasonable assuming that the number of shoes
produced is much higher than the sample size of 20
c) Yes: Binomial(500, 1/12) is reasonable assuming that the sample size is small
relative to the population size.
5.14) a) Let X be the number of males in the sample of 15 who have visited an online
auction site in the past month. It is reasonable to assume X is Binomial(15, .5)
b) P(X 8) = P(X = 8, 9, 10, 11, 12, 13, 14, or 15)
= P(X = 8) + P(X = 9) +
+ P(X = 15) = 0.5 (See table C for probabilities)
5.16) a) X can be approximated by a Binomial(15, 0.5) distribution.
So the mean of X is np = 15*0.5 = 7.5
Standard deviation of X is np (1 p ) = 15 * 0.5 * 0.5 = 1.94
p (1 p )
= 0.13
n
b) As n increases, the mean of X increases and the standard deviation decreases,
but for the estimate of p , it does not change. The mean of X is 75 for n = 150,
and 750 for n = 1500. The mean of p remains 0.5.
For p = X/n, its mean is 0.5, and standard deviation
5.18) a) Since the sample size is large relative to the population size that the probability
of success will change significantly with every draw. The population is only 2.5
times the size of the sample, much smaller than the rule of thumb where the
population size should be at least 20 times the sample size.
b) np and n(1 p) must be at least 10 for the normal approximation to work so
that the edge effect will not be strongly present when making inferences and
calculations. The value of np = 1000(.0002) = 2, which is smaller than 10.
5.23) Recall that for sufficiently large n, p is approximately normal with mean p, and
variance np(1-p).
.28 .30
.32 .30
a) For p = .30, = .01441, P(.28 < p < .32) = P
Z
.01441
.01441
= P(-1.39 < Z < 1.39) = .8348
.04 .06
.08 .06
b) For p = .06, = .00747, P(.04 < p < .08) = P
Z
.00747
.00747
= P(-2.68 < Z < -2.68) = .9926
c) P(-.2 < p p < .02) increases to 1 as p gets closer to 0 because gets closer to
0 so that .02/ gets bigger.
5.26) Since p = np (1 p ) , .0042 = (.49)(.51)/n => n = 15,618.75. Round this number
up so that the standard deviation is slightly less than .004. So n = 15,619.
5.28) a)
= (1200)(.75) = 900,
b) P(X
951) = P(Z
=
(1200)(.75)(.25) = 15
951 900 ) = P(Z
15
3.4) = .0003
c) When n = 1300,
= (1300)(.75) = 975 and
P(X
951 975 ) = P(Z
15.6
951) = P(Z
= 1300(.75)(.25) = 15.6
-1.54) = .9382
Note that numerical answers may vary depending on the continuity correction as
the binomial distribution is discrete while the normal distribution is continuous.
P(X 950.5) and P(X 951) are using the binomial distribution, but will slightly
differ when using normal approximation.
5.77) a) Since npM = 400(.88) = 352 and npF = 400(.82) = 328 are far enough from 0
and 400 that a normal approximation is reasonable for pM and pF.
So p M ~N(.88, 400(.88)(.12) = .01625)
and p F ~N(.82, 400(.82)(.18) = .01921)
b) The difference of two independent normal distributions is normal.
So approximately, p M - p F ~N(.88 - .82 = .06, .01625 2 .019212 = .02516)
c) P( p F > p M ) = P( p M - p F < 0) = P(Z < .06 ) = P(Z < -2.38) = .0087
.02516
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