Transcript Statistics 303

Statistics 303
Chapter 7
Inference for Means
Inference for Means
• To this point, when examining the mean of a population we
have always assumed that the population standard
deviation (s) was known.
• In practice this is seldom the case.
• We usually must estimate the population standard
deviation with the sample standard deviation s (for a
review of s, see pp. 49-50 of the book).
• When we do this, the sampling distribution of the sample
mean is no longer normally distributed, because of the
adjustment for estimating s with s.
• Thus, instead of using the Z, the standard normal
distribution, we must use the appropriate t-distribution.
Inference for Means
• The t-distribution
– Although there is only one Z-distribution, there
are many, many t-distributions.
– In fact, there is a different t-distribution for
each sample size used.
– The shape of each t-distribution is very similar
to the Z-distribution, but is slightly flatter.
– The larger the sample size, the closer the tdistribution is to the Z-distribution.
Inference for Means
• The t-distribution
– The way we distinguish between various t-distributions
is by finding the degrees of freedom (df) that
correspond to the sample size.
– When we are looking at only one sample, the degrees
of freedom are the sample size minus one: df = n – 1.
– We say that the one-sample t-statistic:
tn 1
x 

s
n
has the t distribution with n – 1
degrees of freedom.
Inference for Means
• The t-distribution
– A table of t distribution critical values can be found in Table D (the
last page of the book).
– Note that these values are areas to the right, not areas to the left as
in the Z-table.
– In Table D, the degrees of freedom are listed in the left column.
– The probabilities are on top (these probabilities are inside for the
Z-table)
– The individual t-values are inside the table.
– Make sure to get acquainted with this table and how it differs from
the Z-table.
Inference for Means
• The t-distribution
– In the book, p.452, we see an example of how the
distributions compare:
Inference for Means
• The t-distribution
– With the change from s to s, and the change from z* to
t*, the steps in producing confidence intervals and
hypothesis tests are the same as we have seen
previously.
– In Chapter 1, p. 50, we find that s is calculated from the
data using the formula:
s
1 n
2
xi  x 

n  1 i 1
This formula is very cumbersome.
Ideally, a computer is used to
calculate s, particularly for large data
sets.
Confidence Interval for  with
Unknown s
• The formula for a confidence interval for 
with unknown s is
Calculated
from the data.
x t
*
s
n
Calculated
from the data.
Sample size
t* is found in table D at the back of the book. It must correspond
to the appropriate df = n – 1. It is easiest to find the confidence
level at the bottom of the table and go up to the correct df.
Confidence Interval for  with
Unknown s
• Confidence Interval Example
– An economist wants to determine the average
amount a family of four in the United States
spends on housing annually. He randomly
selects 85 families of size four and finds the
amount they spent on housing the previous
year.
– The economist wishes to estimate the mean
with 99% confidence.
Confidence Interval for  with
Unknown s
• Confidence Interval Example
– Information given:
Sample size: n = 85.
Data: $6,789, $8,233, $4,784, …, $5,974 (85 numbers)
x  $6,219
s  $1,978
df = n – 1 = 85 – 1 = 84
Calculated from the data.
Confidence Interval for  with
Unknown s
• Confidence Interval Example
x t
*
s
n
1,978
 6,219 566.18
 6,219 2.639
85
 (5652.82,6785.18)
t* is found in table D. We first
go to the 99% confidence level
at the bottom. Then we go up
to 80 df (always round down).
Thus, t* = 2.639.
This is a 99% confidence interval
for the true average amount a
family of four in the United States
spends on housing annually.
Hypothesis Test for  with
Unknown s
• The steps for a hypothesis test are the same
as those seen previously, namely,
–
–
–
–
1. State the null hypothesis.
2. State the alternative hypothesis.
3. State the level of significance (i.e., a = 0.05).
4. Calculate the test statistic (note change):
x  0
t
s
n
Hypothesis Test for  with
Unknown s
– 5. Find the P-value:
• For a two-sided test:
P - value  PrT  t or T   t   2PrT  t 
• For a one-sided test:
P - value  PrT  t 
• For a one-sided test:
P - value  PrT  t 
Because of the limited number of t-values given in Table D, it is more
common to find a range for the P-value, rather than the exact value (as
will be seen in the example). Computers can be used to obtain exact
values.
Hypothesis Test for  with
Unknown s
– 6. Reject or fail to reject H0 based on the P-value.
• If the P-value is less than or equal to a, reject H0.
• It the P-value is greater than a, fail to reject H0.
– 7. State your conclusion.
• If H0 is rejected, “There is significant statistical evidence that
the population mean is different than 0.”
• If H0 is not rejected, “There is not significant statistical
evidence that the population mean is different than 0.”
Notice that these last two steps are exactly the
same as for the case where s is known.
Hypothesis Test for  with Unknown
s
• T.V. Example
– Suppose that the data collected from our class
survey is a random sample from the entire
university (which it obviously is not). We wish
to see if there is evidence that the average
amount of television watched for students here
is more than 7 hours per week.
Hypothesis Test for  with Unknown
s
3
4
3
10
2
5
20
10
5
10
20
3
6
10
2
3
1
3
x  8.05
9
5
1
4
5
30
s  7.46
1
10
30
10
4
10
6
3
10
0
15
21
3
9
• T.V. Example
– Information given:
Sample size: n = 38.
df  n  1  38  1  37
Hypothesis Test for  with Unknown
s
• T.V. Example
– 1. State the null hypothesis:
H0 :   7
or H0 :   7
– 2. State the alternative hypothesis:
Ha :   7
– 3. State the level of significance
Assume a = 0.05
from “is more than”
Hypothesis Test for  with Unknown
s
• T.V. Example
– 4. Calculate the test statistic.
x  0
t
s
n
1.05
8.05  7

 0.87

7.46
1.21
38
– 5. Find the P-value.
P - value  PrT  t   PrT  0.87
Remember the table gives probabilities
to the right so we do not use the
technique of subtracting from 1.
 between 0.15 and 0.20
Use df = 30
(rounding down)
Hypothesis Test for  with Unknown
s
• T.V. Example
– 6. Do we reject or fail to reject H0 based on the Pvalue? P-value = between 0.15 and 0.20 is greater
than a = 0.05.
Therefore, we fail to reject H0
– 7. State the conclusion.
“There is not significant statistical evidence that
the average amount of television watched is
more than 7 hours per week at the 0.05 level of
significance.”
Matched Pairs t-test
• To this point we have only looked at tests for single samples.
• Soon we will look at confidence intervals and hypothesis tests for
comparing two groups.
• When each individual can be given both treatments, we can reduce the
two samples to a single sample using a matched pairs design.
• Examples:
– Students are each given a pre-test and a post-test to determine the amount
of material learned in a given time interval.
– To examine the effect of a new drug, a large group of identical twins is
identified. One twin is given a treatment and the other a placebo.
– A ophthalmologist is examining the importance of the dominant eye in
reading. A large group of subjects is asked to read a passage with
dominant eye covered and again with the non-dominant eye covered.
– It can be seen in each of these examples that something pairs the two
responses.
Matched Pairs t-test
• To analyze matched pairs data, we first reduce the data
from two samples to one sample and then analyze the data
using one-sample techniques.
• The data is reduced from two samples to one by
subtracting one of the responses from the other.
– We could subtract each pre-test score from each post-test score.
– We could subtract each placebo response from each treatment
response.
– We could subtract the time taken to read the passage with the nondominant eye from the time taken to read the passage with the
dominant eye.
Matched Pairs t-test
• Example: Keyboards
– “Suppose we want to compare two brands of computer keyboards,
which we will denote as keyboard 1 and keyboard 2. Keyboard 1
is a standard keyboard, while keyboard 2 is specially designed so
that the keys need very little pressure to make them respond. The
manufacturer of keyboard 2 would like to claim that typing can be
done faster using keyboard 2…A simple random sample of n = 30
teachers was selected from a population of high-school teachers
attending a national conference. Each teacher typed the same page
of text once using keyboard 1 and once using keyboard 2. For
each teacher the order in which the keyboards were used was
determined by the toss of a coin. For each teacher the variable
measured was the time (in seconds) to correctly type the page of
text…” (from Graybill, Iyer and Burdick, Applied Statistics, 1998).
Matched Pairs t-test
• Example: Keyboards
– Information given:
Sample size: n = 30.
xdiff  3.53
sdiff  8.56
df  n  1  30  1  29
Subject
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Keyboard 1 Keyboard 2
348
350
435
442
369
356
357
360
376
373
412
405
396
376
317
314
366
366
340
337
347
352
315
303
349
338
330
328
335
322
345
351
374
361
374
370
380
375
319
318
387
382
313
317
303
310
404
393
355
362
364
364
348
355
361
368
301
291
348
323
Reduction to
one sample
Difference = K2 - K1
2
7
-13
3
-3
-7
-20
-3
0
-3
5
-12
-11
-2
-13
6
-13
-4
-5
-1
-5
4
7
-11
7
0
7
7
-10
-25
Matched Pairs t-test
• Example: Keyboards
– 1. State the null hypothesis:
H0 :   0
or H0 :   0
– 2. State the alternative hypothesis:
Ha :   0
– 3. State the level of significance
Assume a = 0.05
from carefully reading
Matched Pairs t-test
• Example: Keyboards
– 4. Calculate the test statistic.
x  0
 3.53  0  3.53
t

 2.26

s
8.56
1.56
n
30
– 5. Find the P-value.
P - value  PrT  t   PrT  2.26
Remember the table gives probabilities
to the right.
 PrT  2.26
 between 0.01 and 0.02
Use df = 29
Matched Pairs t-test
• Example: Keyboards
– 6. Do we reject or fail to reject H0 based on the Pvalue? P-value = between 0.01 and 0.02 is less
than a = 0.05.
Therefore, we reject H0
– 7. State the conclusion.
“There is significant statistical evidence that the
average amount of time needed to type the
passage is lower for keyboard 2 than keyboard 1
at the 0.05 level of significance.”
Matched Pairs Confidence Interval
• After reducing the data to a single sample, we use
the same formula as for a confidence interval for 
with unknown s, namely,
x t
*
s
n
using the mean and standard deviation of the differences.
Matched Pairs Confidence Interval
• Example: Golf Balls
– “In the manufacture of golf balls two procedures are
used. Method I utilizes a liquid center and method II, a
solid center. To compare the distance obtained using
both types of balls, 12 golfers are allowed to drive a ball
of each type, and the length of the drive (in yards) is
measured.” (from Milton, McTeer, and Corbet,
Introduction to Statistics, 1997)
– The manufacturer wants to estimate the mean difference
with 90% confidence.
Matched Pairs Confidence Interval
• Example: Golf Balls
– Information given:
Sample size: n = 12.
xdiff  9.52
sdiff  3.12
df = n – 1 = 12 – 1 = 11
Golfer
liquid
1
2
3
4
5
6
7
8
9
10
11
12
solid
difference (liquid - solid)
180
172.7
7.3
215.8
202.5
13.3
140.6
128.1
12.5
182.7
173.9
8.8
193.8
180.7
13.1
100.2
88.7
11.5
195.2
188.9
6.3
117.6
108.8
8.8
199
186.5
12.5
179.5
175.9
3.6
122.3
112.7
9.6
106.7
99.8
6.9
Matched Pairs Confidence Interval
• Example: Golf Balls
x t
*
s
n
3.12
 9.52  1.796
 9.52  1.62
12
 (7.90,11.14)
t* is found in table D. We first
go to the 90% confidence level
at the bottom. Then we go up to
11 df. Thus, t* = 1.796.
This is a 90% confidence interval
for the true average difference for
the distance traveled for the two
types of golf balls.
Comparing Two Means
• We use the same basic principles for comparing two
population means as those used for examining one
population mean.
• If the standard deviations s1 and s2 for each of the two
populations are known, the two-sample z-statistic is then

x1  x2   1   2 
z
s 12
n1

s 22
n2
But it is very rare that both population standard deviations are
known. We will examine the situation in which they are not known.
Comparing Two Means
• When we are interested in comparing two population
means and we are estimating the population standard
deviations s1 and s2 with s1 and s2, the two-sample tstatistic is then

x1  x2   1   2 
t
2
1
2
2
s
s

n1 n2
with degrees of freedom equal to the smaller of n1-1 and n2-1
(or an appropriate estimate using computer software).
Comparing Two Means
• The null hypothesis can be any of the following:
H0 : 1  2
or H0 : 1  2
or H0 : 1  2
• The alternative hypothesis can be any of the following
(depending on the question being asked):
Ha : 1  2
or Ha : 1  2
or Ha : 1  2
The other steps are the same as those used for the
tests we have looked at previously.
Comparing Two Means
• Example: Tomatoes
– “There has been some discussion among amateur gardeners
about the virtues of black plastic versus newspapers as weed
inhibitors for growing tomatoes. To compare the two,
several rows of tomatoes are planted. Black plastic is used
around nine randomly selected plants and newspaper around
the remaining ten. All plants start at virtually the same
height and receive the same care. The response of interest is
the height in feet after a month’s growth.” (from Milton,
McTeer, and Corbet, Introduction to Statistics, 1997).
– Perform a test to see if there is any difference between the
average heights with significance level 0.10.
Comparing Two Means
• Example: Tomatoes
– Information given:
Sample sizes: n1 = 9, n2 = 10.
x1  1.87
x2  1.49
s1  0.63
s2  0.43
black
plastic
1.8
1.29
1.13
2.92
2.2
1.25
2.61
1.6
2.06
newspaper
2.57
1.59
1.78
1.37
1.22
1.34
1.43
1.06
1.44
1.12
df  n1 1  9 1  8 because n1 is smaller than n2
Comparing Two Means
• Example: Tomatoes
– 1. State the null hypothesis:
H0 : 1  2
– 2. State the alternative hypothesis:
Ha : 1  2 from “any difference between”
– 3. State the level of significance
a = 0.10
Comparing Two Means
• Example: Tomatoes
– 4. Calculate the test statistic.
t
x1  x2   1   2 
2
1
2
2

s
s

n1 n2
1.87  1.49  0
2
2
0.63
0.43

9
10

0.38
0.25
 1.52
– 5. Find the P-value.
P - value  2 * PrT | t |  2 * PrT  1.52
Remember the table gives probabilities
to the right.
 2 * (between 0.05and 0.10)
Use df = 8
 between 0.10 and 0.20
Comparing Two Means
• Example: Tomatoes
– 6. Do we reject or fail to reject H0 based on the Pvalue?
P-value = between 0.10 and 0.20 is greater
than a = 0.10.
Therefore, we fail to reject H0
– 7. State the conclusion.
“There is not significant statistical evidence that
the average tomato plant heights are different for
the two types of weed inhibitors at the 0.10 level
of significance.”
Comparing Two Means
• The confidence interval for the difference of two
population means (1- 2) is
x1  x2   t
*
2
1
2
2
s
s

n1 n2
Where t* comes from Table D and corresponds to the
confidence level desired and df = smaller of n1-1 and n2-1 .
Comparing Two Means
• Example: Commercials
– “There is some concern that TV commercial breaks are
becoming longer. The observations on the following
slide are obtained on the length in minutes of
commercial breaks for the 1984 viewing season and the
current season.” (from Milton, McTeer, and Corbet,
Introduction to Statistics, 1997)
– Find a 95% confidence interval for the difference
between the true averages of the two seasons.
Comparing Two Means
• Example: Commercials
– Information given:
Sample sizes: n1 = 16, n2 = 16.
x1  2.01
x2  2.36
s1  0.49
s2  0.19
1984 current
2.42
2.28
2
2.36
1.17
2.05
1.18
2.45
2.32
2.64
1.84
2.62
2.16
2.39
2.35
2.63
2.4
2.29
1.47
2.39
2.82
2.11
2.04
2.04
2.23
2.25
1.95
2.31
1.38
2.44
2.42
2.57
df  16 1  15 because n1 and n2 are thesame.
Comparing Two Means
• Example: Commercials
x1  x2   t *
0.492 0.192
s12 s22
 2.01 2.36  2.131


16
16
n1 n2
 0.35  0.28
t* is found in table D. We first
go to the 95% confidence level
at the bottom. Then we go up to
15 df. Thus, t* = 2.131.
 (0.63,0.07)
This is a 95% confidence interval
for the true difference of average
length in minutes for commercials
between 1984 and the present.
Pooled t test: Comparing Two Means
• The null hypothesis can be any of the following:
H0 : 1  2
or H0 : 1  2
or H0 : 1  2
• The alternative hypothesis can be any of the following
(depending on the question being asked):
Ha : 1  2 or Ha : 1  2
or Ha : 1  2
Pooled Estimator
• Previously, we discussed two-sample t procedures from
two populations with two unknown standard deviations.
We then used the sample standard deviations to estimate
the population standard deviations. But what about when
the two populations have the same standard deviation.
This estimate is called the pooled estimator of σ2 because it
combines the information in both samples.
2
2
(
n

1
)
s

(
n

1
)
s
2
1
2
2
sp  1
n1  n2  2
Test Statistic
• Suppose that an SRS of size n1 is drawn from a normal
population with unknown mean μ1 and that an independent SRS
of size n2 is drawn from another normal population with
unknown mean μ2. Suppose also that the two populations have
the SAME standard deviation. Thus, the two-sample t statistic is
x1  x2
t
1
1
sp

n1
n2
• With degrees of freedom equal to n1 + n2 – 2
Confidence Interval
• A level C confidence interval for μ1 – μ2 is
( x1  x2 )  t * s p
1 1

n1 n2
• Where t* comes from Table D and corresponds to the
confidence level desired and df = n1 + n2 – 2