1. Cans of soda marked 12oz should average at least 12 ounces. In testing H0: = 12
versus H1: < 12, at level  = .05, the following sample quantities are recorded.
N = 67,
x ( with a line over it) = 11.960,
s = .136
(a) Write the test statistic and determine the rejection region
(b) Calculate the test statistic and draw a conclusion
solution
(a)
N (  x )
has student t-distribution with N-1 = 66 degrees of
s
freedom. P(T > t) =  = 0.05 => t = 1.668
The rejection region is (t, ) = (1.668, )
The test statistic T 
(b)
67 (12  11.96)
 2.4075 > 1.668 => H0 is rejected
0.136
Conclusion: with 95% confidence, the cans of soda average at less than 12 ounces
T
2. A zoologist wishes to estimate the mean blood sugar level of a species of animal
when injected with a specified dosage of adrenaline. A sample of 55 animals of a
common breed are injected with adrenaline, and their blood sugar measurements
are recorded in units of milligrams per 100 milliliters of blood. The mean and
standard deviation of these measurements are found to be 126.9 and 10.5,
respectively
(a) Give a point estimate of the population mean and find a 95.4% error
margin
(b) Determine a 90% confidence interval for the population mean.
solution
(a)
N = 55, x = 126.9, s = 10.5,  = 1 – 0.954 = 0.046
N (  x )
T 
has student t-distribution with N-1 = 54 degrees of freedom
s
P(-t < T < t) = 0.954 => t = 2.042
ts
2.042  10.5

 2.8911
The 95.4% error margin is E 
N
55
The point estimate is x = 126.9
(b)
 = 1 – 0.9 = 0.1
P(-t < T < t) = 0.9 => t = 1.674
ts
1.674  10.5
The 90% error margin is E 

 2.3701
N
55
The 90% confidence interval is ( x  E , x  E )  (124.53, 129.27)
3. Computing from a large sample, one finds the 95% confidence interval for  to be
(6.8, 14.2). Based on this information alone, determine the 90% confidence
interval for .
solution
6.8  14.2
 10.5 E  14.2  10.5  3.7   0.05
2
N (  x )
T
has approximately normal distribution (since N is large)
s
P(-t < T < t) = 0.95 => t = 1.96
N
t 1.96
 
 0.5297
s
E 3.7
P(-t1 < T < t1) = 0.90 => t1 = 1.645
ts
1.645
 3.1055
90% error margin is E1  1 
N 0.5297
The 90% confidence interval is ( x  E1 , x  E1 )  (7.39, 13.61)
x
4. Measurements of the amount of suspended solids in river water, on 14 Monday
mornings, yield x ( with a line over it) = 47 and s= 9.4. Obtain a 95% confidence
interval for the mean amount of suspended solids. State any assumption you
make about the population.
solution
I assume that the population is normally distributed.
x  47, s  9.4,   0.05, N  14
N (  x )
has t-distribution with N-1 = 13 degrees of freedom
s
P(-t < T < t) = 0.95 => t = 2.16
ts
2.16  9.4
E

 5.4265
N
14
The 95% confidence interval is ( x  E , x  E )  (41.57, 52.43)
T
5. The following measurements of the diameter ( in feet ) of Indian mounds in
southern Wisconsin were gathered by examining reports in Wisconsin
Archeologist.
22, 24, 24, 30, 22, 20, 28, 30, 24, 34, 36, 15, 37
Do these data substantiate the conjecture that the population mean diameter is
larger than 21 feet? Test at  = .01
Notation for question 5: Use H0 :  = 21 and H1 : >21
solution
N = 13
1
x   x k = 26.6154
N
1
s
 ( xk  x ) 2  6.5643
N 1
N (  x )
T
has t-distribution with N-1 = 12 degrees of freedom
s
P(T < t) = 0.01 => t = -2.68
13 (21  26.6154)
T
 3.0844  2.68 => H0 is rejected
6.5643
Conclusion: the data shows that, with 99% confidence, the population mean diameter is
larger than 21 feet