Transcript downloads/larson 8 Hypothesis Testing

8 Hypothesis Testing-2 Sample
Elementary Statistics
Larson
Farber
Section 8.1
Testing the Difference
Between Two Means
(Large Independent
Samples)
Overview
To test the effect of an herbal treatment on improvement of
memory you randomly select two samples, one to receive the
treatment and one to receive a placebo. Results of a memory test
taken one month later are given.
Sample
1
Experimental Group
Treatment
Sample
2
Control Group
Placebo
The resulting test statistic is 77 – 73 = 4. Is this difference
significant or is it due to chance (sampling error)?
Independent Samples
When members of one sample are not related to
members of the other sample.
Person’s receiving herbal treatment were not related
or paired with those in the control group who took a
placebo.
x1 x1
x2
x2
x1
x2
x1
x1
x2
x2
x1 x1
Experimental Group
Control Group
Dependent Samples
Each member of one sample is paired with a member of the
other sample.
The test score for each person in the sample could be
recorded before and after taking the herbal treatment.
x1
x1
x2
x2
x1
x1
x1
x1
Score Before
The difference
x2
x2
x2
x2
Score After
can be calculated for each pair.
Application
To test the effect of an herbal treatment on
improvement of memory, you randomly select a
sample of 95 to receive the treatment and a
sample of 105 to receive a placebo. Both groups
take a test after one month. The mean score for
the experimental group is 77 with a standard
deviation of 15. For the control group, the mean
is 73 with a standard deviation of 12. Test the
claim that the herbal treatment improves memory
at = 0.01.
1. Write the null and alternative hypothesis.
Null Hypothesis H0 usually contains the equality condition.
(There is no difference between the parameters of two
populations.)
Alternative Hypothesis Ha is true when H0 is false.
Claim
2. State the level of significance.
= 0.01.
This is the probability that H0 is true, but you reject it.
3. Identify the sampling distribution.
The distribution for the sample statistic
is normal since both samples are large.
Rejection Region
z
0
z0
2.33
4. Find the critical value.
Critical Value z0
5. Find the
rejection region.
6. Find the test statistic.
When both samples are
large, you can use s1
and s2 in place of
and
.
7. Make your decision
0
2.33
z = 2.07 does not fall in the rejection region. Do not
reject the null hypothesis. The P-value is .019 >.01. Do
not reject H0.
8. Interpret your decision.
There is not enough evidence to support the claim that the herbal
treatment improves memory.
z
Section 8.2
Testing the Difference
Between Two Means
(Small Independent Samples)
Testing Difference Between
Means (Small Samples)
When you cannot collect samples of 30 or more, you can use a t-test,
provided both populations are normal. The sampling distribution
depends on whether or not the population variances are equal.
If the variances of the two populations are equal, you can combine or
“pool” information from both samples to form a pooled estimate of the
standard deviation.
The standard error is
d.f. = n1 + n2 - 2
If the variances are not equal, the standard error is:
And d.f. is the smaller of
n1 – 1 or n2 – 1.
Application
Crash tests at 5 miles per hour were performed on 5 small
pickups and 8 SUVs. For the small pickups the mean
bumper repair cost was $1520 and the standard deviation
was $403. For the SUVs the mean bumper repair cost was
$937 and the standard deviation was $382. At = 0.05 test
the claim that the bumper repair cost is greater for small
pickups than for SUVs. Assume equal variances.
n
Pickup
5
1520
SUV
8
937
s
403
382
1. Write the null and alternative hypothesis.
Claim
2. State the level of significance.
= 0.05.
3. Identify the sampling distribution.
Since the variances are equal, the distribution for the sample
statistic
is a t-distribution with d.f. = 5 + 8 – 2 = 11.
4. Find the critical value.
t
0
t0
5. Find the
rejection region
1.796
6. Find the test statistic.
When variances are equal
find the pooled value.
7. Make your decision.
t
0
1.796
t = 2.624 falls in the rejection region. Reject the
null hypothesis.
8. Interpret your decision.
There is enough evidence to support the claim
that bumper repair costs are greater for
pickups than for SUVs.
Application
A real estate agent claims there is no
difference between the mean household
incomes of two neighborhoods. The mean
income of 12 households from the first
neighborhood was $48,250 with a standard
deviation of $1200. In the second
neighborhood, 10 households has a mean
income of $50,375 with a standard deviation of
$3400. Assume the incomes are normally
distributed and the variances are not equal.
Test the claim at = 0.01.
1. Write the null and alternative hypothesis.
Claim
n
First
Second
12.000
10.000
48.250
50.375
s
1200.000 3400.000
2. State the level of significance.
.
3. Identify the sampling distribution.
Since the variances are not equal, the distribution for the sample
statistic
is a t-distribution with d.f. = 9. (The smaller
sample size is 10 and 10 - 1 = 9.)
4. Find the critical values.
–t0
t
–3.250
0
5. Find the
rejection regions.
t0
3.250
6. Find the test statistic.
7. Make your decision.
0
t
–3.250
3.250
t = –1.881 does not fall in the rejection region.
Do not reject the null hypothesis. (The P-value
is .087 > .01.)
8. Interpret your decision.
There is not enough evidence to reject the
claim that there is no difference in mean
household incomes in the two neighborhoods.
Section 8.3
Testing the Difference
Between Two Means
(Dependent Samples)
The Difference Between MeansDependent Samples
When each value from one sample is paired with a data value in
the second sample, the samples are dependent.
x1
x1
x2
x2
x1
x1
x1
x1
x2
x2
x2
x2
The difference, d = x1 – x2 is calculated for each data pair.
The sampling distribution for , the mean of the differences, is a
t-distribution with n – 1 degrees of freedom. (n is the number of pairs.)
Application
The table shows the heart rates (beats per minute) of 5
people before exercising and after. Is there enough
evidence to conclude that heart rate increases with
exercise? Use
.
Person
1
2
Before
65
72
After
127
135
d
62
63
3
4
5
85
78
93
140
136
150
55
58
57
The mean of the differences d is 59
The standard deviation of d is 3.39
1. Write the null and alternative hypothesis.
Claim
2. State the level of significance.
3. Identify the sampling distribution.
The distribution for the sample statistic
t-distribution with d.f. = 4.
is a
(Since there are 5 data pairs, d.f.= 5 – 1 = 4.)
4. Find the critical value.
t
0
t0
2.132
6. Find the test statistic.
5. Find the
rejection region.
7. Make your decision.
0
t0
t
2.132
t = 38.92 falls in the rejection region. Reject the
null hypothesis. The P-value is very close to 0.
8. Interpret your decision.
There is enough evidence to support the claim
that heart rate increases with exercise.
Using Minitab
The Minitab printout
Test of
= 0.00 vs
> 0.00
Variable N Mean StDev SE Mean
T
P
diff.
5 59.00 3.39 1.52
5
38.90 0.0000
The P-value is 0.0000. Since 0.0000 < 0.05, reject
the null hypothesis.
Section 8.4
Testing the Difference
Between
Two Proportions
The Difference Between Proportions
If independent samples are taken from each of two
populations and the samples are large enough you can test
for the difference between population proportions p1 – p2.
x1 and x2 represent the number of successes in 1st
and 2nd samples.
n1 and n2 represent the total number in the 1st and
2nd samples.
Sample proportions of successes.
Since the proportions will be assumed equal, an estimate for
the common value is:
Two Sample z-Test
If
are each at least 5,
the sampling distribution for
The mean is p1 – p2 = 0
and the standard error
The standardized test statistic is:
is normal.
Application
In a survey of 3420 college students attending private
schools, 917 said they smoked in the last 30 days. In a
survey of 5131 college students attending public
schools, 1503 said they had smoked in the last 30 days.
At
, can you support the claim that the
proportion of college students who said they had
smoked in the last 30 days in the private schools is less
than the proportion in public schools? Use
.
private
public
n1 = 3420
x1 = 917
n2 = 5131
x2 = 1503
1. Write the null and alternative hypothesis.
Claim
2. State the level of significance.
3. Identify the sampling distribution.
The distribution for the sample statistic
is normal since
are each at
least 5.
Rejection Region
Critical Value z0
-2.33
4. Find the critical value.
6. Find the test statistic.
0
5. Find the
rejection region.
z
7. Make your decision.
-2.33
0
z = –2.514 falls in the rejection region. Reject the
null hypothesis.
8. Interpret your decision.
There is enough evidence to support the claim that
there is a lower proportion of students who smoke
in private colleges than in public colleges.