Statistics
PBLM SET- Hypothesis Test and CI Practice #2
KEY
1.
Name:____________________________________
Date:_____________________________________
Measures of the left- and right-hand gripping strengths of 10 left-handed writers are recorded:
Left
Hand
Right
Hand
1
140
2
90
3
125
4
129
5
95
Person
6
121
7
85
8
97
9
131
10
110
138
87
110
131
96
120
86
90
129
100
Test the claim that left-handed people have greater gripping strength in their left-hands as opposed to their right
hands. Construct a 99% confidence interval for the mean difference. What does your confidence interval
suggest about the equality of the two population means.
SOLUTION:
*Think of as Matched Pairs*
Input " left" in L1
let : L3  L1  L2
Input " right " in L2
Use stat calc 2: select data and let: “List:” be L3
Let  = 0.05
H 1 : 1   2
H 0 : 1   2
P-value= 0.0334
P  value    reject H 0
There is sufficient evidence to reject the claim that there is no difference in population mean gripping
strengths between left-handed individuals’ left and right hands at the 0.05 significance level. Therefore,
the claim of left-handed people have greater gripping strength in their left-hands as opposed to their
right hands is true.
CI= 99%
Use stat test #8
(-2.012, 9.2122)
We are 99% confident that the difference in mean gripping strengths between the left and right hands of
lefties is between -2.012 and 9.2122.
The above confidence interval supports equality between the two means because 0 is in the set. When 0 is in a
confidence interval it shows that the difference between the two means is likely 0 which implies equality.
2.
Suppose that a production process is considered to be out of control if more than 3% of the items
produced are defective. Inspection of a random sample of 500 items produced in a particular week revealed that
25 were defective. Use a 0.05 significance level to test the claim that the process was out of control during that
week. Construct a 95% confidence interval estimate of the population proportion of defects during any given
week.
SOLUTION:
n  500
x  25
  0.05
H 1 : p  0.03
H 0 : p  0.03
Use stat test #5
P-value= 0.00438
P  value    reject H 0
There is sufficient evidence to reject the claim that the population proportion of defects is 3% at the 0.05
significance level. Therefore, the claim that it is greater is true (process is out of control).
CI= 95%
Use stat test #A
(0.0309, 0.0691)
We are 95% confident that the population proportion of defects is between 3.09% and 6.91%.
3.
Recorded here are the germination times (number of days) for seven seeds of a new strain of snap bean.
12
16
15
20
17
11
18
Test the claim that the population mean number of germination days is more than 15 at the 0.01 significance
level. Construct a 95% confidence interval estimate for the true mean germination time for this strain.
SOLUTION:
Input data into L1
Use stat test #2 (select “data”)
P-value= 0..327
P  value    Don ' t reject H 0
H 1 :   15
H 0 :   15
There is insufficient evidence to reject the claim that the population mean number of days for germination is 15
at the 0.01 significance level. Therefore, the claim that it is greater is false.
CI= 95%
Use stat test #8
(12.605, 18.537)
We are 95% confident that the population mean germination time is between 12.605 and 18.537 days.
4.
Cann’s Classes I:
Class A has 13 students. Class B has 36 students. Both classes took identical
tests on “Descriptive Statistics.” The mean and standard deviation for class A was 60 and 26.1 respectively.
The mean and standard deviation for class B was 77 and 26.0 respectively. Use a 0.10 significance level to test
the claim that there is no significant difference in population means between the two classes. Construct a 90%
confidence interval estimate of the true mean difference.
SOLUTION:
B
A
x  77
x  60
s  26.1
s  26
n  13
n  36
P-value= 0.0568
H 1 : 1   2
H 0 : 1   2
Use stat test #4
  0.10
P  value    reject H 0
There is sufficient evidence to reject the claim that there is no difference in population mean scores between
class A and B at the 0.10 significance level. Therefore, the claim of a significant difference is true.
CI= 90%
Use stat test #0
(-31.51, -2.489)
We are 90% confident that the difference in population mean scores is between -31.51 & -2.489 pts.
5.
Cann’s Classes II: Class A has 13 students. Class B has 36 students. Both classes took identical
tests on “Probability.” There were 5 passing papers in Class A and 23 passing papers in Class B. Use a 0.10
significance level to test the claim that there is no significant difference in population proportions between the
two classes. Construct a 90% confidence interval estimate of the true population proportion difference.
SOLUTION:
A
B
n  13
x5
n  36
x  23
P-value= 0.112
H 1 : p1  p 2
H 0 : p1  p 2
Use stat test #6
  0.10
P  value    Don ' t reject H 0
There is insufficient evidence to reject the claim that there is no significant difference in population proportion
of passing papers between classes A and B at the 0.10 significance level.
CI= 90%
Use stat test #B
(-.5123, 0.00379)
We are 90% confident that the difference in population proportion of passing papers between the two classes
is between -51.23% & .379%.