252solnA4 1/19/01 (Open this document in 'Page Layout' view!)
A. Parameter Estimation
1. Review of the Normal Distribution
A1
2. Point and Interval Estimation
3. A Confidence Interval for the Mean when the Population Variance is Known.
4. A Confidence Interval for the Mean when the Population Variance is not Known.
A2, text 8.20, 8.50 [8.21, 8.50] (8.21, 8.50) – Answers from both editions will be provided for 8.21. 8.95 on CD (8.93)
Graded Assignment 1 (Will be posted)
5. Deciding on Sample Size when working with a Mean
A3, 8.38 [8.36] (8.36)
6. A Confidence Interval for a Proportion.
Text 8.24, 8.25, 8.26, 8.58, 8.94 on CD [8.22, 8.23, 8.24, 8.58, 8.93a,c on CD] (8.22, 8.23, 8.25, 8.58, 8.91a,c)
7. A Confidence Interval for a Variance.
Text 12.1-12.2 [9.72] (9.67), A4
8. (A Confidence Interval for a Median.)
Optional - A5 -- solution is posted.
----------------------------------------------------------------------------------------------------------------------------- ---Problem A5 is in this document.
Problem involving a Confidence Interval for a Median
PROBLEM A5: a. Find the confidence level for an interval for the median using binomial tables, if from a
sample of 12 we take the third observation from both ends.
b. Do the same for the 19th observation from both ends in a sample of 50.
c. Do the same for an interval using the 10th observation from both ends in a sample of 40, using the
Normal approximation to the binomial distribution.
d. In part c, try to find a 95% confidence interval for the median.
Solution: a) If we take the third number from both the bottom and the top of the data, we get the interval
x3    x10 from the ordered numbers x1, x2 , x3 , x4 , x5 , x6 , x7 , x8 , x9 , x10 , x11, x12 . For example, if the
numbers in order are 13.2, 17.1,18.5, 21.3, 21.4, 22.0, 27.1, 27.7, 28.9, 29.2, 35.4, 35.9, we would say that
the interval is 18 .5    29 .2.
To find the confidence level, first find the significance level  , the probability that the interval is
wrong. The interval will be wrong if (i) x3 through x10 are all below the median or (ii) x3 through x10
are all above the median. The probabilities of these two events are both the same, so that we can figure out
the probability that x3 through x10 are all below the median and double it. The probability of any given
number being below (or above) the median is 0.5, and the probability that x3 through x10 are all below the
median is the probability that the first 10 or more numbers are all below the median, and is the same as the
probability of getting ten or more heads in twelve flips of a coin.
From the binomial table for n  12 and p  .5 , we find that Px  10   1  Px  9  1  .98071
 .01929 . But note that the binomial distribution with p  .5 is symmetrical so that Px  10   Px  2 .
252solnA4 1/19/01
Also remember that we stated in the previous paragraph that to get the significance level, we must double
this probability, so that   2.01929   .03858 . Thus the confidence level is 1    1  2.01929   .96142 .
More generally, if k is the index of the number at the bottom of the confidence interval, (in the case we
just did k  3 ) the confidence level is 1    1  2Px  k  1 .
b) If we take a sample of n  50 , put it in order, and then pick the 19th number k  19  from both
the top and the bottom, so that the confidence interval is x19    x32  , the confidence level is
1    1  2Px  k  1  1  2Px  19  1  1  2Px  18   1  2.03245   .93510 . This can also be done
using the Normal Distribution. If we ignore the continuity correction, and recall that for the binomial
distribution with p  .5 and q  1  p  .5 ,   np  .5n and  2  npq  n.5.5  .25n ,
k  1     P z  k  1  .5n   P z  18  .550    Pz  1.98   .5  .4761  .0239

Px  k  1  P z 








.25 n 
.5 50 


.and 1    1  2Px  k  1  1  2.0239   .9522 . This looks way off, so try the same problem with a

18  .5  .550  
continuity correction Px  k  1  Px  18   P z 
  Pz  1.84   .5  .4671  .0329
.5 50


and the confidence level is 1    1  2Px  k  1  1  2.0329   .9342 . (Note: The continuity correction
widens any interval by .5 on both sides, and should be used when we replace a discrete distribution by a
continuous distribution. Here we replaced 18 by 18.5, but did not make a change in the bottom of the
interval because there was no bottom.)
c) If n  40 and k  10 we have no binomial table, so use the normal approximation to the
binomial distribution with a continuity correction.
k  1  .5     P z  k  1  .5  .5n   P z  9  .5  .540    Pz  3.32   .5  .4995  .0005

Px  k  1  P z 








.25 n
.5 40




and the confidence level is 1    1  2Px  k  1  1  2.0005   .9990 .
d) If we want a 95% confidence interval and n  40 , we require that 1    1  2Px  k  1  1 2.025 
k  1  .5     P z  k  1  .5  .5n   .025 . But since

 .95 . This means that Px  k  1  P z 




.25 n




z .025  1.960 , we know that
Px  k  1  Pz  1.960   .025 . So we can say that
k  1  .5  .5n  1.960 . Solve this with n  40 , or note that k  1  .5  .5n  1.960 .25n and, solving
.25 n
for
k
k  .5  .5n  1.960 .25n . If we substitute
, find
n  40 ,
k  .5  .540   1.960 .2540 
 20.5  10  20.5  6.26  14.30 . We could also follow the formula in the outline that says
k
n  1  z

n

40  1  z

40
 14 .30 . Obviously k must be a whole number and the more
2
2
conservative choice would be to round it down, so that the interval is x14    x 27 .
2
2