12.1 – Tests about a Population Mean
What do we do if we don’t know ?
T-distribution!
T-statistic:
estimate  hypothesized value
test statistic =
standard deviation of the estimate
t=
x  O
s
n
Calculator Tip: TTest
Stat – Tests - TTest
Calculator Tip: Tprobability
2nd – Dist – tcdf( lower, upper, df)
Conditions:
SRS
Normality
Independence
Normal?
• If n < 15, graph it. Normal?
• If 15 <n< 30, only moderate skewness in graph
allowed to approximate normality
• If n  30, CLT rocks!
Example #1
A new treatment for hepatitis is being tested for effectiveness. The
standard treatment takes an average of 10 days from the
beginning of treatment until the patient’s health is improved. A
random sample of 11 patients with hepatitis is selected and given
the new treatment. The number of days until improvement of the
patient’s health is recorded:
4
12
4
5
3
3
8
8
5
6
7
Assume the population is normally distributed and test the claim
that the new treatment is better than the standard treatment at
level  = 0.05.
P: True # of days until improvement in health
H:
H o :   10
H A:   10
A:
SRS (says so)
Normality (n=11, and only moderate skewness in graph)
Independence
(Safe to assume more than 110 hepatitis patients)
N: TTest
=?
T:
x  O
5.909  10 4.091
 5.02
t


2.700
s
0.8141
11
n
O:
-5.02
P(t < -5.02) =
df = 11 – 1 = 10
?
O:
-5.02
P(t < -5.02) < 0.0005
df = 11 – 1 = 10
OR:
P(t < -5.02) = 0.000259
M:
<
p ____

0.000259
Reject the Null
0.05
S:
There is enough evidence to claim that the true # of
days until improvement in health is less than 10 days
with treatment.
Example #2
A theory of Charles Darwin’s was that plants that are the
progeny of sexual reproduction are taller than those that are the
progeny of self-fertilization. Darwin took random pairs of
seedlings of the same age: one produced by cross-fertilization
and the other by self-fertilization and grew them under nearly
identical conditions. The data below are the final heights of the
plants after a certain period of time. Assume the population is
approx. normal. Test Darwin’s theory at level 0.10.
Pair
Crossfertilized
Selffertilized
Difference
1
2
3
4
5
23.5
19.0
21.0
22.0
19.1
17.4
20.4
20.0
20.0
18.4
6.1
-1.4
1
2
0.7
P:
μC = Cross-fertilized height
μS = self-fertilized height
μD = true difference in height from cross-fertilized
and self-fertilized
H:
Ho : D  0
H A:  D  0
A:
SRS (says so)
Normality (n=5, and graph looks normal)
Independence
(Safe to assume more than 50 plants)
N: TTest
=?
T:
xD   D
1.68  0
t


2.764
sD
5
n
1.68
 1.3594
1.236
O:
1.3594
P(t > 1.3594) =
df = 5 – 1 = 4
O:
-5.02
0.10 < P(t >1.3594) < 0.15 OR: P(t >1.3594) = 0.1228
df = 5 – 1 = 4
M:
>
p ____

0.1228
0.10
Accept the Null
S:
There is enough evidence to claim that the true
difference in height from cross-fertilized and selffertilized plants is zero, or there is no difference.
12.2 – Tests about a Population Proportion
T-statistic:
estimate  hypothesized value
test statistic =
standard deviation of the estimate
Z=
p̂  pO
po (1  po )
n
Calculator Tip: 1-Proporiton Z-test
Stat – Tests – 1-PropZTest
Conditions:
SRS
Normality
Independence
np0  10
n(1  p0 )  10
Example #3
According to the Energy Information Administration, 53% of
households nationwide used natural gas for heating in 2006.
Recently a survey of 3600 randomly selected households showed
that 54% used natural gas. Use a 0.05 significance level to test the
claim that the 53% national rate has changed.
P: True proportion of households nationwide using
natural gas for heating in 2006
H:
H o : p  0.53
H A: p  0.53
A:
SRS (says so)
n(1  p0 )  10
np0  10
(3600)(0.53)  10 3600(1  .53)  10
1908  10
1692  10
Normality
Independence
(Safe to assume more than 36,000 households)
N: 1-Proportion ZTest
T:
Z
pˆ  pO
pO (1  pO )
n

0.54  0.53
 1.202
0.53(1  0.53)
3600
O:
1.202
2[P(Z < -1.202)] =
1.202
O:
1.202
2[P(Z < -1.202)] = 2[0.1151] = 0.2302
OR:
P = 0.229299
1.202
M:
>
p ____

0.229299
Accept the Null
0.05
S:
There isn’t enough evidence to say the national rate of natural
gas use for heating has changed.