Transcript PPT 10 - halsnarr

Sampling Distribution of x1  x2
If x1 and x2 are normally distributed and samples 1 and 2 are
independent, their difference is
2
1
n1
1 -2

 22
n2
Sampling Distribution of x1  x2
If x1 and x2 are normally distributed and samples 1 and 2 are
independent, their difference is
2
s1
s22

n1
n2
1 -2
The normal distribution is not used if 1 and 2 are unknown. The
t-distribution is used instead with
2
 s12 s22 
  
n1 n2 

df 
2 2
2 2
1  s1 
1  s2 

 
 
n1  1  n1  n2  1  n2 
Sampling Distribution of p1  p2
If p1 and p2 are normally distributed and samples 1 and 2 are
independent, their difference is
2
1
n1
n1p1 > 5 n1(1  p1) > 5
n2p2 > 5 n2(1  p2) > 5
p1 -p2
 12  p1 (1  p1 )
 22  p2 (1  p2 )
The sample proportions are pooled (averaged) when testing
the equality of proportions.

 22
n2
Sampling Distribution of p1  p2
If p1 and p2 are normally distributed and samples 1 and 2 are
independent, their difference is
2
1
n1

p1 -p2
 12  p (1  p )
 22  p (1  p )
The sample proportions are pooled (averaged) when testing
whether the proportions are equal or not.
n1 p1  n2 p2
p
n1  n2
 22
n2
Interval estimate of 1  2
Example 1
Par, Inc. is a manufacturer of golf equipment and
has developed a new golf ball that has been designed
to provide “extra distance.”
In a test of driving distance using a mechanical
driving device, a sample of Par golf balls was
compared with a sample of golf balls made by Rap,
Ltd., a competitor. The sample statistics appear on the
next slide.
Interval estimate of 1  2
Example 1
Sample Size
Sample Mean
Sample #1
Par, Inc.
120 balls
275 yards
Sample #2
Rap, Ltd.
80 balls
258 yards
Based on data from previous driving distance tests, the two
population standard deviations are 15 and 20 yards, respectively.
Develop a 95% confidence interval estimate of the difference
between the mean driving distances of the two brands of golf
balls.
x1  x2 = 275 – 258 = 17
Interval estimate of 1  2
Example 1
x1  x2  z / 2
275  258  z.0250
 12  22

n1 n2
(15) 2 (20) 2

120
80
17  1.96 6.875
17 + 5.14 yards
11.86 yards to 22.14 yards
We are 95% confident that the difference between the mean driving
distances of Par and Rap golf balls is between 11.86 to 22.14 yards.
Interval estimate of 1  2
Example 2
Specific Motors of Detroit has developed the M car.
24 M cars and 28 J cars (from Japan) were road tested to
compare miles-per-gallon (MPG) performance. The
sample statistics are given below. Develop a 90%
confidence interval estimate of the difference between
the MPG performances of the two models of automobile.
M Cars
J Cars
24 cars
28 cars
n
29.8 mpg
27.3 mpg
x
s
2.56 mpg
1.81 mpg
Point estimate of 1  2 = x1 – x2
= 29.8 – 27.3
= 2.5 MPG
Interval estimate of 1  2
Example 2
The degrees of freedom are:
2
 (2.56) (1.81) 
2


24 
28 
.2731 .1170
 40.585  402
df 
2
2
2
.2731
1 (2.56.1170
) 2   1  (1.81) 2 






24 231  24 27
28

1
2
8



2
With 1 = .9000
2
 = .1000
/2 = .0500 (the column of t-table)
df = 40 (the row of t-table)
-t.0500 = -1.684
t.0500 = 1.684
Interval estimate of 1  2
Example 2
( x1  x2 )  t / 2
s12 s12

n1 n2
(2.56)2 (1.81)2
29.8  27.3  1.684

24
28
2.5 + 1.051
1.448 to 3.552 mpg
We are 90% confident that the difference between the
MPG performances of M cars and J cars is
1.448 to 3.552 mpg
Interval estimate of p1  p2
Example 3
Market Research Associates is conducting research to
evaluate the effectiveness of a client’s new advertising
campaign. Before the new campaign began, a telephone
survey of 150 households in the test market area showed 60
households “aware” of the client’s product.
The new campaign has been initiated with TV and
newspaper advertisements running for three weeks. A survey
conducted immediately after the new campaign showed 120
of 250 households “aware” of the client’s product. Develop
a 95% confidence interval estimate of the difference between
the proportion of households that are aware of the client’s
product.
60
p2 
 .40
150
120
p1 
 .48
250
p1  p2  .08
Interval estimate of p1  p2
Example 3
1 = .9500
 = .0500
p1  p2  za / 2
/2 = .0250
z.0250 = 1.96
p1 (1  p1 ) p2 (1  p2 )

n1
n2
.48(1  .48) .40(1  .40)
.48  .40  1.96

250
150
.08 + .10
We are 95% confident that the “change in awareness”
due to the campaign is between -0.02 and 0.18…
Hypothesis Tests About 1  2
Example 1 (continued)
Can we conclude, using a 5% level of significance, that
the mean driving distance of Par, Inc. golf balls is
greater than the mean driving distance of Rap, Ltd. golf
balls?
1. Develop the hypotheses.
1   2
1   2   2   2  0
1   2  0
H0: 1 – 2 < 0
Ha: 1 – 2 > 0
Hypothesis Tests About 1  2
2. Determine the critical value
 = .0500
 / 1  .0500
z.0500 = 1.645
3. Compute the value of the test statistic.
z-stat 
( x1  x2 )  D0
 12
n1


 22
n2
( 275  258)  0
(15)2 (20)2

120
80
 6.49
Hypothesis Tests About 1  2
4. Reject or do not reject the null hypothesis
Do Not Reject H0
Reject H0
.050
0
z
1.645 6.49
z-stat
At the 5% level of significance, the sample evidence indicates
the mean driving distance of Par, Inc. golf balls is greater
than the mean driving distance of Rap, Ltd. golf balls.
Hypothesis Tests About 1  2
Example 2 (continued)
Can we conclude, using a 5% level of significance,
that the miles-per-gallon (MPG) performance of M cars
is greater than the MPG performance of J cars?
Recall
n
x
s
M Cars
24 cars
29.8 mpg
2.56 mpg
1. Develop the hypotheses.
J Cars
28 cars
27.3 mpg
1.81 mpg
H0: 1 - 2 < 0
Ha: 1 - 2 > 0
Hypothesis Tests About 1  2
2. Determine the critical value.
 /1 = .050
(column)
(row)
df = 40
3. Compute the value of the test statistic.
t -stat 

( x1  x2 )  D0
s12 s22

n1 n2
( 29.8  27.3)  0
( 2.56) 2 (1.81) 2

24
28
 4.003
t.050 = 1.684
Hypothesis Tests About 1  2
4. Reject or do not reject the null hypothesis
At 5% significance,
fuel economy of M
cars is greater than
the mean fuel
economy of
J cars.
Do Not Reject H0
Reject H0
.050
0
1.684 4.003
t050 t-stat
t
Hypothesis Tests About p1  p2
Example 3 (continued)
Can we conclude, using a 5% level of significance,
that the proportion of households aware of the client’s
product increased after the new advertising campaign?
120
p1 
 .48
250
1. Develop the hypotheses.
p2 
60
 .40
150
H0: p1  p2 < 0
Ha: p1  p2 > 0
2. Determine the critical value.  = .0500
3. Compute the value of the test statistic.
.08
(.48  .40)  0
 1.57

z-stat 
.0510
.0510
z.05 = 1.645
Hypothesis Tests About p1  p2
4. Reject or do not reject the null hypothesis
Do Not Reject H0
Reject H0
.050
0
1.57
z-stat
1.645
z.050
z
We cannot conclude that the proportion of households aware
of the client’s product increased after the new campaign.
Hypothesis Tests About p1  p2
Example 4
John is a political candidate who wants to know if female
support for his candidacy is the same as his male support using
a 5% level of significance. A survey conducted shows 310 of
620 females surveyed said they will vote for John while the
same survey said 362 of 680 males said they will vote for John.
p1 
310
 .5
620
1. Develop the hypotheses.
p2 
362
 .532
680
H0: p1  p2 = 0
Ha: p1  p2 ≠ 0
n1 p1  n2 p2 620  0.5  680  0.532 310  362
 .5169


p
620  680
620  680
n1  n2
Hypothesis Tests About p1  p2
s p1  p2 
.5169(1  .5169) .5169(1  .5169)

620
680
 .02775
2. Determine the critical value.
 = .0500
 / 2  .0250
± z.0250 = ± 1.96
3. Compute the value of the test statistic.
(.5  .532 )  0 .032
 1.15

z-stat 
.02775
.02775
Hypothesis Tests About p1  p2
4. Reject or do not reject the null hypothesis
Reject H0
1 –Not
a =Reject
.9700H0
Do
.0250
Reject H0
.0250
-1.96 -1.15 0
z-stat
1.96
We cannot conclude that support from female voters is
different from support from male voters.
Matched Samples
Example: Express Deliveries
A Chicago-based firm has documents that must be quickly
distributed to district offices throughout the U.S. The firm must
decide between two delivery services, UPX (United Parcel
Express) and INTEX (International Express), to transport its
documents.
In testing the delivery times of the two services, the firm sent
two reports to a random sample of its district offices with one
report carried by UPX and the other report carried by INTEX. Do
the data on the next slide indicate a difference in mean delivery
times for the two services? Use a 5% level of significance.
Matched Samples
District Office
Seattle
Los Angeles
Boston
Cleveland
New York
Houston
Atlanta
St. Louis
Milwaukee
Denver
Delivery Time (Hours)
UPX INTEX Difference (d)
32
30
19
16
15
18
14
10
7
16
25
24
15
15
13
15
15
8
9
11
7
6
4
1
2
3
-1
2
-2
5
Sum = 27
d  2.7
Matched Samples
di
di  d
di  d
(di  d )2
7
6
4
1
2
3
-1
2
-2
5
7– 2.7
6 – 2.7
4 – 2.7
1 – 2.7
2 – 2.7
3 – 2.7
-1 – 2.7
2 – 2.7
-2 – 2.7
5 – 2.7
4.3
3.3
1.3
-1.7
-0.7
0.3
-3.7
-0.7
-4.7
2.3
18.49
10.89
1.69
2.89
0.49
0.09
13.69
0.49
22.09
5.29
sd 
 (d
i
d)
n 1
2
Sum =
sd2 =
sd =
76.1
8.5
2.9
Matched Samples
1. Develop the hypotheses with d = 1 – 2
H0: d = 0
Ha: d  
No
difference exists
exists
A difference
2. Determine the critical values with  = .050. Since this is a
two-tailed test
(the column of the t-table)
 /2 = .025
df = 10 – 1 = 9 (the row of the t-table)
 t.025 = 2.262
3. Compute the value of the test statistic.
d2.7
 0 0
t -stat  2.94
s2.9
n10
d
t.025 = 2.262
Matched Samples
4. Reject or do not reject the null hypothesis
Reject H0
Do Not Reject
H0: d = 0
.0250
2.262
t.025
Reject H0
.0250
0
At the 5% level of significance, the sample
evidence indicates there is a difference in
mean delivery times for the two services.
2.262 2.94
t.025 t-stat