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Sampling Distribution of 1 2x x
1 2-
2 21 2
1 2n n
2x1xIf and are normally distributed and samples 1 and 2 are independent, their difference is
Sampling Distribution of 1 2x x
1 2-
2 21 2
1 2
s s
n n
2x1xIf and are normally distributed and samples 1 and 2 are independent, their difference is
22 21 2
1 22 22 2
1 2
1 1 2 2
1 11 1
s sn n
dfs s
n n n n
The normal distribution is not used if s1 and s2 are unknown. The t-distribution is used instead with
Sampling Distribution of
n1p1 > 5 n1(1 - p1) > 5
n2p2 > 5 n2(1 - p2) > 5
1 2p p
21 1 1(1 )p p 2
2 2 2(1 )p p
1 2-p p
2 21 2
1 2n n
If and are normally distributed and samples 1 and 2 are independent, their difference is
1p 2p
The sample proportions are pooled (averaged) when testing the equality of proportions.
Sampling Distribution of 1 2p p
21 (1 )p p 2
2 (1 )p p
1 2-p p
2 21 2
1 2n n
If and are normally distributed and samples 1 and 2 are independent, their difference is
1p 2p
The sample proportions are pooled (averaged) when testing whether the proportions are equal or not.
1 21 2
1 2
n nnp p
np
In a test of driving distance using a mechanicaldriving device, a sample of Par golf balls wascompared with a sample of golf balls made by Rap,Ltd., a competitor. The sample statistics appear on thenext slide.
Par, Inc. is a manufacturer of golf equipment and has developed a new golf ball that has been designed to provide “extra distance.”
Example 1
Interval estimate of 1 2
Sample SizeSample Mean
Sample #1Par, Inc.
Sample #2Rap, Ltd.
120 balls 80 balls275 yards 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
Example 1
Interval estimate of 1 2
/
2 21 2
1 21 2
2x xn n
z
17 + 5.14 yards
11.86 yards to 22.14 yards
.
2 2
0250
(15) (20)275 258
120 80z
1.9617 6.875
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.
Example 1
Interval estimate of 1 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.
Example 2
nM Cars J Cars
24 cars 28 cars29.8 mpg 27.3 mpg2.56 mpg 1.81 mpg
Point estimate of 1 - 2
= 29.8 – 27.3= 2.5 MPG
= x1 – x2
xs
Interval estimate of 1 2
22 2
2 22 2
2.56 1.81
2.5
( ) ( )24 28
1 ( ) 1 ( )24 1 24 28 1 2
6 1.8
81
The degrees of freedom are:
df
With 1-a = .9000
t.0500 = 1.684
a/2 =
df =
a = .1000
40.585
2
2 2
.2731 .1170
.2731 .1170
23 27
(the row of t-table)
.0500(the column of t-table)
40
40
-t.0500 = -1.684
Example 2
Interval estimate of 1 2
2.5 + 1.051
2 22.529.8
6( ) ( ) 1.684
227.
4 28
1.813
2 2
/ 21
1 21
1 2
( ) tn
s s
nx x
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
Example 2
Interval estimate of 1 2
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 surveyconducted immediately after the new campaign showed 120of 250 households “aware” of the client’s product. Develop a 95% confidence interval estimate of the difference betweenthe proportion of households that are aware of the client’s product.
2p60150
.40
Example 3
1p120250
.48 1 2p p .08
Interval estimate of 1 2p p
.48(1 .48) .40(1 .40).48 .40 1.96
250 150
z.0250 = 1.96
.08 + .10
1 1 2 21 2 / 2
1 2
(1 ) (1 )a
p p p pp p z
n n
1-a = .9500 a/2 =a = .0500 .0250
Example 3
Interval estimate of 1 2p p
We are 95% confident that the “change in awareness”due to the campaign is between -0.02 and 0.18…
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 2
1 2 2 20
1 2 0
Hypothesis Tests About m1 - m2
Ha: 1 – 2 > 0
1. Develop the hypotheses.
H0: 1 – 2 < 0
Example 1 (continued)
2. Determine the critical value
a = .0500
3. Compute the value of the test statistic.
-statz 1 2 0
2 21 2
1 2
( )
n n
Dx x
6.49
2 2
( )
(15) (20)120
275 258 0
80
a / 1 = .0500 z.0500 = 1.645
Hypothesis Tests About m1 - m2
Do Not Reject H0 Reject H0
z 1.645 6.49
z-stat0
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.
.050
4. Reject or do not reject the null hypothesis
Hypothesis Tests About m1 - m2
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
Hypothesis Tests About m1 - m2
nM Cars J Cars
24 cars 28 cars29.8 mpg 27.3 mpg2.56 mpg 1.81 mpg
xs
1. Develop the hypotheses. H0: 1 - 2 < 0Ha: 1 - 2 > 0
Example 2 (continued)
a /1 = .050 (column) t.050 = 1.684
df = 40 (row)
Hypothesis Tests About m1 - m2
3. Compute the value of the test statistic.
1 2
22
2
1 2
0
1
( )-stat
x
s
x Dt
n ns
2 22.56
( )
( ) ( )24
029.8 27
1.828
.3
1
4.003
2. Determine the critical value.
Do Not Reject H0 Reject H0
t 1.684
t050
4.003 t-stat
0
At 5% significance, fuel economy of M cars is greater than
the mean fuel economy of
J cars.
.050
4. Reject or do not reject the null hypothesis
Hypothesis Tests About m1 - m2
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?
1
120250
.48p 2
60150
.40p
Hypothesis Tests About p1 - p2
Example 3 (continued)
1. Develop the hypotheses. H0: p1 - p2 < 0
Ha: p1 - p2 > 0
2. Determine the critical value. a = .0500 z.05 = 1.645
3. Compute the value of the test statistic.
1.57..080510
.48 .( )-s
0t
.05104
at0
z
z
Do Not Reject H0 Reject H0
0
We cannot conclude that the proportion of households aware of the client’s product increased after the new campaign.
1.57 z-stat
1.645z.050
.050
4. Reject or do not reject the null hypothesis
Hypothesis Tests About p1 - p2
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.
1
310620
.5p 2
362680
.532p
1 1 22
1 2
p pn nn n
0.5 0.532620 680620 680
310 362620 680
.5169p
Hypothesis Tests About p1 - p2
Example 4
1. Develop the hypotheses.
Ha: p1 - p2 ≠ 0
H0: p1 - p2 = 0
2. Determine the critical value.
3. Compute the value of the test statistic.
a = .0500 a / 2 = .0250 ± z.0250 = ± 1.96
1.15
.0
.0322775
.5 .5( )-
0st
.032
5at
277z
Hypothesis Tests About p1 - p2
1 2p ps .5169(1 .5169) .5169(1 .5169)
620 680
.02775
-1.96 1.96
1 – a = .9700
.0250.0250
0
4. Reject or do not reject the null hypothesis
-1.15z-stat
Reject H0 Do Not Reject H0 Reject H0
We cannot conclude that support from female voters is different from support from male voters.
Hypothesis Tests About p1 - p2
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
3230191615181410 716
25241515131515 8 911
UPX INTEX Difference (d)District Office
SeattleLos AngelesBostonClevelandNew YorkHoustonAtlantaSt. LouisMilwaukeeDenver
Delivery Time (Hours)
7
6
4
1 2 3 -1 2 -2
5 Sum = 27
2.7d
Matched Samples
7 6 4 1 2 3 -1 2 -2 5
7– 2.76 – 2.74 – 2.71 – 2.72 – 2.73 – 2.7
-1 – 2.72 – 2.7
-2 – 2.75 – 2.7
4.33.31.3
-1.7-0.70.3
-3.7-0.7-4.72.3
18.4910.891.692.890.490.09
13.690.49
22.095.29
id id d 2( )id did d
Sum = 76.1
sd = 8.5
sd = 2.9
22( )
1di d
sd
n
Matched Samples
H0: d = 0
with d = m1 – m21. Develop the hypotheses
A difference existsNo difference exists
2. Determine the critical values with a = .050.
a /2 = .025 (the column of the t-table)
df = 10 – 1 = 9
Since this is a
two-tailed test
- t.025 = -2.262 t.025 = 2.262
(the row of the t-table)
Ha: d
02.72.9 10
-statt 2.94
0
d
ds n
3. Compute the value of the test statistic.
Matched Samples
-2.262 2.262 -t.025 t.025
2.94 t-stat
0
At the 5% level of significance, the sample evidence indicates there is a difference in mean delivery times for the two services.
Reject H0 Do Not Reject Reject H0
.0250.0250
H0: d = 0
4. Reject or do not reject the null hypothesis
Matched Samples