Stat 322 – HW 1
Due Friday, Jan.
12
1) Complete the Stat 322 Questionnaire in Blackboard (under the Assignments button).
2)
Find 2 uses of statistics (e.g., in the news, on the web, from other
courses). For each, write a paragraph
summarizing the use of statistics (what did they do) and then asking a few
questions about it, e.g., what additional information do you want to know? Be sure to cite the sources and dates of the
articles.
3)
(p. 127) problem 54. If you use Minitab, include a well-labeled copy of your
output.
Binomial with n = 10 and p =
0.01
x P( X <= x )
2 0.999886
Binomial with n = 10 and p =
0.05
x P( X <= x )
2 0.988496
Binomial with n = 10 and p =
0.1
x P( X <= x )
2 0.929809
Binomial with n = 10 and p =
0.2
x P( X <= x )
2 0.677800
Binomial with n = 10 and p =
0.25
x P( X <= x )
2 0.525593
Binomial with n = 10 and p =
0.01
x P( X <= x )
1 0.995734
Binomial with n = 10 and p =
0.05
x P( X <= x )
1 0.913862
Binomial with n = 10 and p =
0.1
x P( X <= x )
1 0.736099
Binomial with n = 10 and p =
0.2
x P( X <= x )
1 0.375810
Binomial with n = 10 and p =
0.25
x P( X <= x )
1 0.244025
(d)
Binomial with n = 15 and p =
0.01
x P( X <= x )
2 0.999584
Binomial with n = 15 and p =
0.05
x P( X <= x )
2 0.963800
Binomial with n = 15 and p =
0.1
x P( X <= x )
2 0.815939
Binomial with n = 15 and p =
0.2
x P( X <= x )
2 0.398023
Binomial with n = 15 and p =
0.25
x P( X <= x )
2 0.236088
(e)
n=15
curve does the best job of being high when p is low and low when p
is high. The P(X 
1, n = 10) has
high probabilities throughout, but we don’t want to accept the batch when p
is large.
4)
(p. 172) problem 34.
.6827
.9987
Second one wins hands down!
4)
(p. 242) problem 52, make sure you justify your steps. Hint: You might
want to consider calculating the probability in terms of the average lifetime
for the package instead.
We
want to find t such that P(T > t) < .05 where the
random variable T represents the total lifetime of all batteries. We can instead find 
such that P(
> ) < .05
and convert.
Since
the lifetimes follow a normal distribution with m = 10 and s
= 1, then follows a normal
distribution with mean 10 and standard deviation equal to s/sqrt(4)
= .5.
To find the 95th percentile, this corresponds to z = 1.645. So our observation needs to be 1.645 standard deviations above the mean. So we would need to observe a sample mean of 10 + 1.645(.5) = 10.8 hours. This corresponds to a total lifetime of 43.2 hours.
5)
(p. 298) problem 20 using formula 7.11 (p. 295) and/or Minitab (including your
output in the latter case)
n =4722 
= .15 z =
2.576 for 99% confidence
Our sample size is quite large so we can use the “wald procedure”:
± z 
= .15 ± 2.576sqrt(.15(.85)/4722) = .15 ± .0134 = (.1366, .1634)
Minitab:
We are 99% confident that the proportion of all American youngsters who are seriously overweight is between .137 and .163.