Download There is a conjecture that Japanese

Solutions
1. a)
Histogram of IBI
Boxplot of IBI
(with 90% t-confidence interval for the mean)
(with 90% t-confidence interval for the mean)
10.0
Frequency
7.5
5.0
2.5
_
X
_
X
0.0
36
48
60
IBI
72
84
30
40
50
60
IBI
70
80
90
b)
Descriptive Statistics: IBI
Variable
IBI
N
49
N*
0
Variable
IBI
Maximum
91.00
Mean
65.35
SE Mean
2.70
Range
62.00
StDev
18.90
Minimum
29.00
Q1
53.50
Median
71.00
Q3
82.00
IQR
28.50
c) The distribution is left skewed. (The boxplot and histogram both suggest this. Notice that
mode (about 84) > median (71) > mean (65.4).)
d) 60.82 < µ < 69.88: We’re 90% confident that the mean IBI for all streams is between 60.82
and 69.88. Here’s the Minitab:
One-Sample T: IBI Sort
Variable
IBI Sort
N
49
Mean
65.35
StDev
18.90
SE Mean
2.70
90% CI
(60.82, 69.88)
The point estimate is 65.35; the error margin is 4.53.
e) The 5th percentile is 30 (31.4 is OK); The 95th percentile is 89 (88.6).
30 < X NEW < 89
We’re 90% confident that the IBI for a future randomly sampled stream is between 30 and 89.
Some Minitab guidance is on page 3 of this assignment, available in the online version.
f) The confidence interval for the mean of all streams (the population mean) has smallest error
margin (4.53); the prediction interval for a single stream has largest. (While this interval is
computed differently, you could look at it as 59.5  29.5 – so the error margin is effectively 29.5.
g) (1.96*18.90/2.5)2 = 219.56. So sample at least 220 streams.
2. The “answer” depends on what you would reasonably think the standard deviation might be.
Since most nickels are between 0 and 40 years old, a guess of 10 would be OK. (Or you might
actually collect some data and compute the standard deviation to get a decent start.) For σ = 10,
 2.576  10 
you’d get n  
 = 2564.3, so sample at least 2565 nickels.
0.5 

2
Actually: In class last year we discovered that this standard deviation is probably closer to 15.
Two different samples of 40 nickels yielded standard deviations of 15.5 and 14.4. Using 15:
 2.576  15 
n
 = 5972.2, so sample at least 5973 nickels.
0.5 

2
 2.576  20 
A guess of 20 gives n  
  10617.24, so sample 10618 nickels.
0.5


2
Somewhere between 2565 and 10618 nickels (about 6000 seems appropriate).
Both of these are mere guidelines. But: It will take a lot of nickels!
3.
a) You should have pretty close to 0% confidence in that interval. The student is interested in TV
viewing for children in Oswego, yet only uses data from a highly nonrandom sample of students,
many of whom have parents that are highly educated, or in college. Further: Even if the student
were interested in children of this sort, bias is almost certain to creep in, because parents will
inevitably understate their kids’ TV viewing times.
b) I would not have 95% confidence in this interval. In order to have the stated confidence, the
requirements must be met. One requirement is: If not a Normal population, then a large sample
size. Survival for this type of situation is not Normal; the sample here is rather small. The T
interval should not be applied here. In this situation there are ways to obtain a confidence
interval that you can have 95% confidence in, but they are not covered in detail in “Intro”
courses.
Some Minitab guidance is on page 3 of this assignment, available in the online version.