Review / Sample test ST 540
Chapter 6 and chapter 7
o Point estimation, Interval estimation and Testing of hypothesis based on a single
sample:
o Answer all the review questions printed in green at the end of each section.
o Write the formal statement of Central Limit Theorem.
o What is the difference between normal distribution and t-distribution, when used
as sampling distribution?
o Compare them with respect to shape, parameters, estimation of those
parameters, asymptotic behavior ( as sample size becomes large).
o Write a short note on interval estimation based on following points
o What width is preferable?
o What issues affect width favorable/ unfavorably
o How do we get the best interval estimate.
o How do we interprete the interval estimate.
o How do u connect the interval estimate with testing of hypothesis
o What is the meaning of confidence level.
o Chi-square is a sampling distribution, write a short note of chi square distribution,
its range, shape, parameters asymptotic behavior, its application in different
scenario, underlying assumptions.
o Know about one sided confidence intervals.
o Solve example 6.62 in the textbook.
o Solution to the problem 6.5 and 6.6 of text book (page 217)
o
05
.
 0.079 for normal men and
40
limitation.
o
It means that the distribution of mean triceps skin-fold thickness from repeated samples
of size 40 drawn from the population of normal men can be considered to be normal with
 2 s2 0. 52
mean  and variance


 0.0063 . A similar statement holds for men with
n
n
40
chronic airflow limitation.
sem 
0.4
 0.071 for men with chronic airflow
32
o Solution to the problems 6.13-6.21
13. We have that x 
x
14
215
 8.6 days. Therefore, a 95% confidence interval for  is given by
25
t24, .975s
n
2.0645.72
25
 8.6  2.36  6.24, 10.96
 8.6 
s
. We have that x  784
. ,
n
s  321
. . There-fore, we have the following 95% confidence interval
The 95% confidence interval is computed from x  tn 1,.975
7.84  t24,.975 
3.21
3.21
 7.84  2.064 
5
25
 7.84  133
.  6.51, 917
. 
15
A 90% confidence interval is given by
x  tn 1,.95
16
17
s
3.21
 7.84  t24,.95 
n
25
3.21
 7.84  1.711 
5
 7.84  110
.  6.74, 8.94
The 90% confidence interval should be shorter than the 95% confidence interval, since
we are requiring less confidence. This is indeed the case.
We refer to Table 6. The lower 2.5th percentile is 0.0506 and is denoted by  22,.025. The
upper 2.5th percentile is 7.38 and is denoted by  22,.975 .
18
19
20
11
.44 . Because p̂ is unbiased and has minimum
25
variance among all estimators of p.

pq
.44.56
The standard error =

.099 .
n
25
   25.44. 56  616
.  5 , we can use the normal theory method. Therefore, a
Since npq
95% confidence interval for the percentage of males discharged from Pennsylvania
hospitals is given by
Our best estimate is given by p 
p  1.96

pq
.44  1.96.099
n
.44.195
 .25, .63
21 There are 16 people after excluding women of childbearing age. Of these 16 people, 4
received a bacterial culture while in the hospital. Thus, the best point estimate
 p 
4
.25 .
16
o Solution to 6.41. .42
2
41 We assume that x1 , x 2 ,..., x 25 ~ N (  ,  ) , where  ,  2 are unknown, and find that
x  7.0 , s2  4.0 . Thus, a two-sided 95% confidence interval for the mean is given by
2.064  2 
2.064  2  
s
s  

, x  t24,.975
, 7.0 

 x  t24,.975
  7.0 
5
5
n
n 


  6.17, 7.83
42 A two-sided 99% confidence interval for the unknown variance  2 is given by
  n  1 s 2  n  1 s 2
 2
, 2
  24,.995
 24,.005

  24  4  24  4  

,

  45.56 9.89 

  2.11, 9.71
o In chapter 7 you should be able to write a short note on type one and type
errors describing their definitions, their interrelationship, effect of
different issues in testing on type one an type two error. Their behavior
with respect to values in null and alternative hypothesis. How do we
control them in our decision making process.
o Solutions to 7.39-7.46
.39 We wish to test the hypothesis H0:    0 versus H1:    0 , where   true mean daily iron
intake for 911-year-old boys below the poverty level and  0  true mean daily iron
intake for 911-year-old boys in the general population.
.40 We must use a one-sample t test. We reject H0 if t  tn1, / 2 , or t  tn 1,1 / 2 , where
x  0
, and accept H0 otherwise. We have that  0  14.44 , n  51 ,  .05 , x  1250
. ,
s/ n
s  475
. . Therefore,
t
12.50  14.44
4.75 / 51
194
.

 2.917
0.665
t
The critical values are t50,.025 and t50,.975 . Since t  t40,.025  2.021  t50,.025 , it follows that we
reject H0 at the 5% level. We conclude that 911-year-old boys below the poverty level have a
significantly lower mean iron intake than comparably aged boys in the general population.
41
To obtain the p-value, we must compute 2x (t0 >.2917). Since t40,.995  2.704 , t40,.9995  3551
.
2.704  2917
.
 3551
. , if we had 40 df, then 2  1.9995  p  2 1.995 or
and 2660
, if we
.
.001  p .01 . Similarly, since t60,.995  2.660 , t60,.9995  3460
.
 2917
.
 3460
.
and
had 60 df, it would also follow that .001  p .01 . Therefore, since we actually have 50 df, and
we reach the same conclusion with either 40 or 60 df, it follows that .001  p .01 .
42 The hypotheses to be tested are H0: 2   20 versus H1: 2   20 where  2  underlying
variance in low-income population,  20  underlying variance in the general population.
43 We use a one-sample chi-square test. We reject H0 if
X2 
n  1s2
 20
  2n 1, / 2
or
X 2   2n 1,1 / 2 .
We have  20  556
. 2  3091
. , n  51 ,  .05 , s2  4.752  22. 56 . Thus,
X2 

n  1s2
 20
504.752
2
 36.49 ~  50
under H0
556
. 2
2
2
.
. . Since 3236
The critical values are  50
it
.  3649
.  7142,
,.025  32.36 and  50,.975  7142
follows that we accept H0 at the 5% level and conclude that there is no significant difference
between the variance of iron intake for the low-income population and the general population.
2
2
 50
 50
,.05  34.76 ,
,.10  37.69
2.05  p  2.10 or .10  p .20 .
.44 Since
45
and
34.76  3649
.  37.69 ,
A 95% confidence interval for the underlying variance
it
follows
that
 2 is given by
(n  1) s 2 (n  1) s 2
, 2
 (15.80,34.86 )
2

n 1,. 975

n 1,. 025
Since this confidence interval contains  20  556
. 2  3091
. , we conclude that the underlying
variance of the low-income population is not significantly different from that of the general
population.
46
The inferences made with the hypothesis-testing approach in Problems 7.43 and 7.44 are the
same as those made with the CI approach in Problem 7.45, viz. there is no significant difference
between the variance of iron intake for the low income population and the variance of the
general population.
o