STATISTICS 401, Section 001
Homework for Chapter 7
SOLUTIONS
Question 1 Because σ is assumed known in this problem (which, by the way, is unrealistic), the appropriate confidence intervals are z -based.
(a)
1 .
96(3 /
√
With α =
25), or (60
.
.
05, we find from Table A.3 that
02 , 62 .
38).
z
α/ 2
(b)
1 .
96(3 /
√
With α = .
100), or (60
05, we find from Table A.3 that
.
61 , 61 .
79).
z
α/ 2
= 1
= 1 .
.
96, so the 95% CI is 61
96, so the 95% CI is 61 .
.
2
2
±
±
(c)
1 .
28(3 /
√
With α = .
20, we find from Table A.3 that
100), or (60 .
82 , 61 .
58).
z
α/ 2
= 1 .
28, so the 95% CI is 61 .
2 ±
(d) With α =
95% CI is 2(1 .
645)(3
.
10, we find from Table A.3 that n z
α/ 2
= 1 .
645, so the width of the
). Setting this expression equal to 1 and solving for n , we obtain n = 9 .
87 2 = 97 .
417. Since n must be an integer, the smallest n that works is n = 98.
Question 2 In this problem, I chose to use t -based confidence intervals due to the fact that
σ is unknown. However, if you chose to use z -based intervals because of the large sample size, that would also be fine. The two answers would be very close.
Both parts of this problem require the sample mean, which is given by x = 57385 .
7 / 150 =
382 .
6, and the sample standard deviation, which is given by s = s
22135771 −
149
57385 .
7 2
150
= 34 .
9 .
(a) With α = .
05 and 149 degrees of freedom, Table A.5 gives t
α/ 2 , 149
= 1 .
98 (even between 120 and ∞ anyway). Thus, the 95% CI is 382 .
6 ± 1 .
98(34 .
9 / 150), or (377 .
0 , 388 .
2).
(b) With α = .
05 and 149 degrees of freedom, the 95% upper confidence limit, or upper confidence bound, is found by using t
α, 149 instead of t
α/ 2 , 149
. From Table A.5, t
α, 149
= 1 .
66.
Thus, the 95% UCL is 382 .
6 + 1 .
66(34 .
9 / 150), or 387 .
3. The upper confidence limit is different from the upper limit in part (a) due to the fact that the confidence interval in part
(a) is constructed to have a lower limit, which thus excludes 2.5% of the distribution on the far left. The limit on part (b) does not have to exclude this extra 2.5%. One way to understand this distinction is to regard part (a) as a two-sided interval and part (b) as a one-sided interval.
Question 3 (a) Here, we have n = 487 and ˆ = .
072. Furthermore, α = .
01, which means that z
α/ 2
= 2 .
575. You may use either formula (7.10) or formula (7.11) to compute the confidence interval for the true proportion, p . Using (7.10) gives
.
072 +
2 .
575
2
2(487)
± 2 .
575 q
( .
072)( .
928)
487
+
2 .
575 2
4(487 2 )
1 +
2 .
575 2
487
= ( .
047 , .
108) , while using (7.11) gives
.
072 ± 2 .
575 r
( .
072)( .
928)
487
= ( .
042 , .
102) .
(b) Since the confidence interval is always widest when ˆ = .
5, this is the value we use in this case in order that the answer will work irrespective of ˆ . Again, you may use either formula (7.10) or formula (7.11) to compute the width of the confidence interval. Using
(7.10) gives that the width equals w = 2
2 .
575 q
( .
5)( .
5) n
1 +
+
2 .
575 2 n
2 .
575 2
4 n 2
, while the width according to (7.11) is w = 2 × 2 .
575 r
( .
5)( .
5) n
.

Setting w = .
01 and solving for n (then rounding up to the next integer) gives n = 66300 for the first equation and n = 66307 for the second.
Question 4 (a) With α = .
05 and 10 degrees of freedom, Table A.5 gives t
α/ 2 , 10
= 2 .
228.
(b) With α = .
05 and 15 degrees of freedom, Table A.5 gives t
α/ 2 , 15
(c) With α = .
01 and 15 degrees of freedom, Table A.5 gives t
α/ 2 , 15
= 2 .
131.
= 2 .
947.
(d) With α = .
01 and 4 degrees of freedom, Table A.5 gives t
α/ 2 , 4
(f ) With α = .
01 and 37 degrees of freedom, Table A.5 gives t
α/ 2 , 37
= 4
(e) With α = .
02 and 24 degrees of freedom, Table A.5 gives t
α/ 2 , 24
.
604.
= 2 .
492.
= 2 .
715 (I interpolated between 36 and 38; the last decimal place isn’t that important anyway).
Question 5 (a) This problem calls for a tolerance interval. You are free to interpret “a high degree of confidence” as either 95% confidence or 99% confidence. Table A.6 gives the tolerance critical values as 2.752 for 95% confidence and 3.168 for 99% confidence. Therefore, the correct interval could be given as .
0705 ± 2 .
752 × .
0062 or .
0705 ± 3 .
168 × .
0062, depending on which level of confidence you chose.
These intervals become ( .
053 , .
088) for 95% confidence and ( .
051 , .
090) for 99% confidence.
(b) The main reason that the tolerance interval is wider than the confidence interval is because a confidence interval is an estimate of the mean of a distribution, whereas a tolerance interval is an estimate of the distribution itself. It’s only natural that if we wish only the mean of that distribution. This is why we divide s by
TI.
n for the CI for not for the
Question 6 With n − 1 = 8 degrees of freedom and α = .
05, the required chi-squared critical values from Table A.7 are
χ
2
.
025 , 8
= 17 .
534 and χ
2
.
975 , 8
= 2 .
180 .
Thus, a 95% confidence interval for σ
2 is
8(2 .
81 2
17 .
534
)
,
8(2 .
81 2 )
2 .
18
= (3 .
60 , 28 .
98) .
Taking square roots gives the interval (1 .
90 , 5 .
38) for σ .