MTH3003 PJJ SEM II 2014/2015 F2F II 12/4/2015
MTH3003 PJJ SEM II 2014/2015 F2F II 12/4/2015. ASSIGNMENT :25% Assignment 1 (10%) Assignment 2 (15%) Mid exam :30% Part A (Objective) Part B (Subjective)
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MTH3003 PJJSEM II 2014/2015
F2F II 12/4/2015
ASSIGNMENT :25%
Assignment 1 (10%)
Assignment 2 (15%) Mid exam :30%
Part A (Objective)
Part B (Subjective) Final Exam: 40%
Part A (Objective)Part B (Subjective - Short)Part C (Subjective – Long)
Key ConceptsI. Types of EstimatorsI. Types of Estimators
1. Point estimator: a single number is calculated to estimate the population parameter.2. Interval estimatorInterval estimator: two numbers are calculated to form an interval that contains the parameter.
II. Properties of Good EstimatorsII. Properties of Good Estimators1. Unbiased: the average value of the estimator equals the parameter to be estimated.2. Minimum variance: of all the unbiased estimators, the best estimator has a sampling distribution with the smallest standard error.3. The margin of error measures the maximum distance between the estimator and the true value of the parameter.
The Margin of Error
estimator theoferror std2 z estimator theoferror std2 z
Margin of error: Margin of error: The maximum error of estimation, is the maximum likely difference
observed between sample mean x and true population mean µ, calculated as :
1.645 1.96 2.33 2.575
Key ConceptsIII. Large-Sample Point EstimatorsIII. Large-Sample Point Estimators
To estimate one of four population parameters when the sample sizes are large, use the following point estimators with the appropriate margins of error.
Example 1
• A homeowner randomly samples 64 homes similar to her own and finds that the average selling price is $252,000 with a standard deviation of $15,000. Estimate the average selling price for all similar homes in the city.
Point estimator of : 252, 000
15, 000Margin of error : 1.96 1.96 3675
64
μ x
s
n
Point estimator of : 252, 000
15, 000Margin of error : 1.96 1.96 3675
64
μ x
s
n
A quality control technician wants to estimate the proportion of soda cans that are underfilled. He randomly samples 200 cans of soda and finds 10 underfilled cans.
03.200
)95)(.05(.96.1
ˆˆ96.1
05.200/10ˆ
200
n
qp
x/npp
pn
:error of M argin
: ofestimator Point
cans dunderfille of proportion
03.200
)95)(.05(.96.1
ˆˆ96.1
05.200/10ˆ
200
n
qp
x/npp
pn
:error of M argin
: ofestimator Point
cans dunderfille of proportion
Example 2
Key ConceptsIV. Large-Sample Interval EstimatorsIV. Large-Sample Interval Estimators
To estimate one of four population parameters when the sample sizes are large, use the following interval estimators.
Example 3
A random sample of n = 50 males showed a mean average daily intake of dairy products equal to 756 grams with a standard deviation of 35 grams. Find a 95% confidence interval for the population average .
n
szx 205.0
50
3596.17 56 7 0.97 56
grams. 65.70 746.30or 7
1.96
Of a random sample of n = 150 college students, 104 of the students said that they had played on a soccer team during their K-12 years. Estimate the proportion of college students who played soccer in their youth with a 98% confidence interval.
n
qpzp
ˆˆˆ 202.0
150
)31(.69.33.2
104
150
0 9.. 69 .60or .7 8. p
Example 4
2.33
Example 5
• Compare the average daily intake of dairy products of men and women using a 95% confidence interval.
78.126
.78.6 18.78-or 21
Avg Daily Intakes Men Women
Sample size 50 50
Sample mean 756 762
Sample Std Dev 35 30
2
22
1
21
205.021 )(n
s
n
szxx
2 235 30(756 762) 1.96
50 50
1.96
Example 6
• Compare the proportion of male and female college students who said that they had played on a soccer team during their K-12 years using a 99% confidence interval.
2
22
1
11201.021
ˆˆˆˆ)ˆˆ(
n
qp
n
qpzpp
70
)44(.56.
80
)19(.81.575.2)
70
39
80
65( 19.062.0
45.0 0.07or 21 pp
Youth Soccer Male Female
Sample size 80 70
Played soccer 65 39
2.575
CHAPTER 9LARGE SAMPLE TESTS OF HYPOTHESES
PART ITesting the single mean & single proportion
PART IITesting the difference between two means
&difference between two proportions
Key ConceptsI.I. Parts of a Statistical TestParts of a Statistical Test
1. Null hypothesis: a contradiction of the alternative hypothesis
2. Alternative hypothesis: the hypothesis the researcher wants to support.
3. Test statistic and its p-value: sample evidence calculated from sample data.
4. Rejection region—critical values and significance levels: values that separate rejection and nonrejection of the null hypothesis
5. Conclusion: Reject or do not reject the null hypothesis, stating the practical significance of your conclusion.
Key ConceptsII.II. Errors and Statistical SignificanceErrors and Statistical Significance
1. The significance level is the probability if rejecting H 0 when it is in fact true.
2. The p-value is the probability of observing a test statistic as extreme as or more than the one observed; also, the
smallest value of for which H 0 can be rejected.
3. When the p-value is less than the significance level ,the null hypothesis is rejected. This happens when the
test statistic exceeds the critical value.
4. In a Type II error, is the probability of accepting H 0 when it is in fact false. The power of the test is (1 ), the probability of rejecting H 0 when it is false.
Key ConceptsIII.III. Large-Sample Test Large-Sample Test
Statistics Using the Statistics Using the zz DistributionDistributionTo test one of the four population parameters when the sample sizes are large, use the following test statistics:
statistic-hypothesized value
standard error of statisticz
statistic-hypothesized value
standard error of statisticz
Example 1 (testing the single mean)
The daily yield for a chemical plant has averaged 880 tons for several years. The quality control manager wants to know if this average has changed. She randomly selects 50 days and records an average yield of 871 tons with a standard deviation of 21 tons.
880:H
880:H
a
0
880:H
880:H
a
0
03.350/21
880871
/
:statisticTest
0
ns
xz
03.350/21
880871
/
:statisticTest
0
ns
xz
Using critical value approach:
What is the critical value of z that cuts off exactly /2 = .01/2 = .005 in the tail of the z distribution?
Rejection Region: Reject H0 if z > 2.58 or z < -2.58. If the test statistic falls in the rejection region, its p-value will be less than a = .01.
Rejection Region: Reject H0 if z > 2.58 or z < -2.58. If the test statistic falls in the rejection region, its p-value will be less than a = .01.
For our example, z = -3.03 falls in the rejection region and H0 is rejected at the 1% significance level.
0024.)0012(.2)03.3(2
)03.3()03.3( :value-
zP
zPzPp0024.)0012(.2)03.3(2
)03.3()03.3( :value-
zP
zPzPp
This is an unlikely occurrence, which happens about 2 times in 1000, assuming = 880!
Using p-value approach:
Example 2 (testing the single proportion)
Regardless of age, about 20% of Americanadults participate in fitness activities at least
twice a week. A random sample of 100 adults over 40 years old found 15 who exercised at least twice a week. Is this evidence of a decline in participation after age 40? Use = .05.
2.:H
2.:H
a
0
p
p2.:H
2.:H
a
0
p
p
25.1
100)8(.2.
2.15.ˆ
:statisticTest
00
0
nqp
ppz 25.1
100)8(.2.
2.15.ˆ
:statisticTest
00
0
nqp
ppz
Using critical value approach:What is the critical value of z that cuts off exactly = .05 in the left-tail of the z distribution?
Rejection Region: Reject H0 if z < -1.645. If the test statistic falls in the rejection region, its p-value will be less than = .05.
Rejection Region: Reject H0 if z < -1.645. If the test statistic falls in the rejection region, its p-value will be less than = .05.
For our example, z = -1.25 does not fall in the rejection region and H0 is not rejected. There is not enough evidence to indicate that p is less than .2 for people over 40.
Example 3 (testing difference between two means)
• Is there a difference in the average daily intakes of dairy products for men versus women? Use a = .05.
Avg Daily Intakes
Men Women
Sample size 50 50
Sample mean 756 762
Sample Std Dev 35 30
(same) 0:H 210
2
22
1
21
21 0
:statisticTest
ns
ns
xxz
)(different 0:H 21a
92.
5030
5035
076275622
Using p-value approach:
3576.)1788(.2
)92.()92.( :value-
zPzPp
3576.)1788(.2
)92.()92.( :value-
zPzPp
Since the p-value is greater than = .05, H0 is not rejected. There is insufficient evidence to indicate that men and women have different average daily intakes.
Example 4 (testing difference between two proportions)
Compare the proportion of male and female college students who said that they had played on a soccer team during their K-12 years using a test of hypothesis.
Youth Soccer Male Female
Sample size 80 70
Played soccer 65 39
(same) 0:H 210 pp )(different 0:H 21a pp
81.80/65ˆ Calculate 1 p 56.70/39ˆ 2 p
69.150
104ˆ
21
21
nn
xxp
Using p-value approach:Youth Soccer Male Female
Sample size 80 70
Played soccer 65 39
21
21
11ˆˆ
0ˆˆ
:statisticTest
nnqp
ppz 30.3
701
801
)31(.69.
56.81.
001.)0005(.2)30.3()30.3( :value- zPzPp 001.)0005(.2)30.3()30.3( :value- zPzPp
Since the p-value is less than = .01, H0 is rejected. The results are highly significant. There is evidence to indicate that the rates of participation are different for boys and girls.
CHAPTER 10INFERENCE FROM SMALL SAMPLE
PART ITesting the single mean
& difference between two means
PART IITesting the single variance
&ratio of two variances
Key ConceptsI. Experimental Designs for Small SamplesI. Experimental Designs for Small Samples
1. Single random sample: The sampled population must be normal.2. Two independent random samples: Both sampled populations must be normal.
a. Populations have a common variance 2.
b. Populations have different variances 3. Paired-difference or matched-pairs design: The samples are not independent.
Key ConceptsII. Statistical Tests of SignificanceII. Statistical Tests of Significance
1. Based on the t, F, and 2 distributions
2. Use the same procedure as in Chapter 93. Rejection region—critical values and significance levels: based on the t, F, and
2 distributions with the appropriate degrees of freedom4. Tests of population parameters: a single mean, the difference between two means, a single variance, and the ratio of two variances
III. Small Sample Test StatisticsIII. Small Sample Test StatisticsTo test one of the population parameters when the sample sizes are small, use the following test statistics:
Key Concepts
Testing the single mean
The basic procedures are the same as those used for large samples. For a test of hypothesis:
ns
xt
/
statisticTest
0
ns
xt
/
statisticTest
0
One-tailed (lower-tail)
One-tailed (upper-tail)
Two-tailed
Using p-values or a rejection region based on t distribution with df = n-1
01
00
:
:
H
H
01
00
:
:
H
H
01
00
:
:
H
H
For a 100(1)% confidence interval for the population mean
.1on with distributi- ta of tailin the
/2 area off cuts that of value theis where 2/
2/
ndf
ttn
stx
.1on with distributi- ta of tailin the
/2 area off cuts that of value theis where 2/
2/
ndf
ttn
stx
Confidence Interval
A sprinkler (sprayer) system is designed so that the average time for the sprinklers to activate after being turned on is no more than 15 seconds. A test of 5 systems gave the following times:
17, 31, 12, 17, 13, 25 Is the system working as specified? Test using = .05.
specified) as ng(not worki 15:H
specified) as (working 15:H
a
0
specified) as ng(not worki 15:H
specified) as (working 15:H
a
0
Example 1
SolutionData:Data: 17, 31, 12, 17, 13, 25First, calculate the sample mean and standard deviation, using your calculator or the formulas in Chapter 2.
387.75
6115
2477
1
)(
167.196
115
222
nnx
xs
n
xx i
387.75
6115
2477
1
)(
167.196
115
222
nnx
xs
n
xx i
Data:Data: 17, 31, 12, 17, 13, 25Calculate the test statistic and find the rejection region for =.05.
5161 38.16/387.7
15167.19
/
:freedom of Degrees :statisticTest
0
ndfns
xt
5161 38.16/387.7
15167.19
/
:freedom of Degrees :statisticTest
0
ndfns
xt
Rejection Region: Reject H0 if t > 2.015. If the test statistic falls in the rejection region, its p-value will be less than = .05.
Rejection Region: Reject H0 if t > 2.015. If the test statistic falls in the rejection region, its p-value will be less than = .05.
Data:Data: 17, 31, 12, 17, 13, 25Compare the observed test statistic to the rejection region, and draw conclusions.
.015.2 if HReject
:RegionRejection
38.1 :statisticTest
0
t
t
.015.2 if HReject
:RegionRejection
38.1 :statisticTest
0
t
t
Conclusion: For our example, t = 1.38 does not fall in the rejection region and H0 is not rejected. There is insufficient evidence to indicate that the average activation time is greater than 15.
15:H
15:H
a
0
15:H
15:H
a
0
Testing the difference between two means (Independent Samples)
• As in Chapter 9, independent random samples of size n1 and n2 are drawn from population 1 and population 2 with means 1 dan 2,and variances and .
• Since the sample sizes are small, the two populations must be normal
21 2
2
The basic procedures are the same as those used for large samples. For a test of hypothesis:
One-tailed (lower-tail)
One-tailed (upper-tail)
Two-tailed
0211
0210
:
:
H
H
0211
0210
:
:
H
H
0211
0210
:
:
H
H
11 2
2
2221
2
121
2
2221
21
nnsnns
nsnsdf
Interval Estimate of 1 - 2:Small-Sample Case (n1 < 30 and/or n2 < 30)
• Interval Estimate with 2 Unknown
21;221 xxdf stxx
21
2 1121 nn
ss xx
2
11
21
222
21122
nn
snsnss p
22
21 2
221
2
22
1
21
21 n
s
n
ss xx
• Instead of estimating each population variance separately, we estimate the common variance with
2
)1()1(
21
222
2112
nn
snsns
2
)1()1(
21
222
2112
nn
snsns
21
2
021
11
nns
xxt
21
2
021
11
nns
xxt
has a t distribution with n1+n2-2 degrees of freedom.
•And the resulting test statistic,
Test Statistics ( )22
21
• How to check the reasonable equality of variance assumption?
Test Statistics (cont’d)
Rule of Thumb
3smaller
larger 2
2
s
s Assume that the variance are equal
3smaller
larger 2
2
s
s Do Not Assume that the variance are equal
• If the population variances cannot be assumed equal, the test statistic is
• It has an approximate t distribution with degrees of freedom of
2
22
1
21
21
ns
ns
xxt
2
22
1
21
21
ns
ns
xxt
1)/(
1)/(
2
22
22
1
21
21
2
2
22
1
21
nns
nns
ns
ns
df
1)/(
1)/(
2
22
22
1
21
21
2
2
22
1
21
nns
nns
ns
ns
df
Test Statistics ( )22
21
Confidence Interval ( ) •You can also create a 100(1-)%
confidence interval for 1-2.
2
)1()1( with
21
222
2112
nn
snsns
2
)1()1( with
21
222
2112
nn
snsns
21
22/21
11)(
nnstxx
21
22/21
11)(
nnstxx
Remember the three assumptions:
1.Original populations normal
2.Samples random and independent
3.Equal population variances.
Remember the three assumptions:
1.Original populations normal
2.Samples random and independent
3.Equal population variances.
22
21
Example 2
Two training procedures are compared by measuring the time that it takes trainees to assemble a device. A different group of trainees are taught using each method. Is there a difference in the two methods? Use = 0.01.
Time to Assemble Method 1 Method 2
Sample size 10 12
Sample mean 35 31
Sample Std Dev 4.9 4.5
99.1
121
101
942.21
3135
t
942.2120
)5.4(11)9.4(9
2
)1()1(
:Calculate
22
21
222
2112
nn
snsns
Solution
0:H
0:H
210
210
Hypothesis
Equality of Variances Checking
2
2
2
2
5.4
9.4
smaller
larger
s
s3186.1
25.20
01.24
21
2
21
11
0
nns
xxt
Test Statistics
Using critical value approach:
What is the critical value of t that cuts off exactly /2 = .01/2 = .005 in the tail of the t distribution?
Critical value:
Rejection Region: Reject H0 if t > 2.845 or t < -2.845. If the test statistic falls in the rejection region, its p-value will be less than = .01
Rejection Region: Reject H0 if t > 2.845 or t < -2.845. If the test statistic falls in the rejection region, its p-value will be less than = .01
For our example, t = 1.99 falls in the rejection region and H0 is rejected at the 1% significance level.
845.220;201.0;2 tt df
The Paired-Difference Test (dependent samples)
•Sometimes the assumption of independent samples is intentionally violated, resulting in a matched-pairsmatched-pairs or paired-difference testpaired-difference test.
•By designing the experiment in this way, we can eliminate unwanted variability in the experiment by analyzing only the differences,
ddii = = xx11ii – – xx22ii
•to see if there is a difference in the two population means, 1-2.
Example 3
• One Type A and one Type B tire are randomly assigned to each of the rear wheels of five cars. Compare the average tire wear for types A and B using a test of hypothesis.
Car 1 2 3 4 5
Type A 10.6 9.8 12.3 9.7 8.8
Type B 10.2 9.4 11.8 9.1 8.3
0:H
0:H
21a
210
0:H
0:H
21a
210
• But the samples are not
independent. The pairs of responses are linked because measurements are taken on the same car.
The Paired-DifferenceTest
.1on with distributi- ta
on basedregion rejection aor value- theUse
. s,difference theofdeviation standard andmean
theare and pairs, ofnumber where
/
statistic test theusing
:H test we:H test To
0
0d00210
ndf
p
d
sdn
ns
dt
i
d
d
.1on with distributi- ta
on basedregion rejection aor value- theUse
. s,difference theofdeviation standard andmean
theare and pairs, ofnumber where
/
statistic test theusing
:H test we:H test To
0
0d00210
ndf
p
d
sdn
ns
dt
i
d
d
Car 1 2 3 4 5
Type A 10.6 9.8 12.3 9.7 8.8
Type B 10.2 9.4 11.8 9.1 8.3
Difference .4 .4 .5 .6 .5
0:H
0:H
a
0
d
d
0:H
0:H
a
0
d
d
.0837
1
.48 Calculate
22
nn
dd
s
n
dd
ii
d
i
.0837
1
.48 Calculate
22
nn
dd
s
n
dd
ii
d
i
8.125/0837.
048.
/
0
:statisticTest
ns
dt
d
8.125/0837.
048.
/
0
:statisticTest
ns
dt
d
Solution
CarCar 1 2 3 4 5
Type A 10.6 9.8 12.3 9.7 8.8
Type B 10.2 9.4 11.8 9.1 8.3
Difference 0.4 0.4 0.5 0.6 0.5
Rejection region: Reject H0 if |t| > 2.776.
Conclusion:
Since t table= 12.8, H0 is rejected. There is a difference in the average tire wear for the two types of tires.
Solution
Confidence interval (dependent samples)
You can construct a 100(1-)% confidence interval for a paired experiment using
n
std d
2/n
std d
2/
Testing a single variance
.1on with distributi square-chi a
on basedregion rejection awith )1(
statistic test theuse we
tailedor two one :H versus:H test To
20
22
120
20
ndf
sn
.1on with distributi square-chi a
on basedregion rejection awith )1(
statistic test theuse we
tailedor two one :H versus:H test To
20
22
120
20
ndf
sn
2)2/1(
22
22/
2 )1()1(
:interval Confidence
snsn2
)2/1(
22
22/
2 )1()1(
:interval Confidence
snsn
A cement manufacturer claims that his cement has a compressive strength with a standard deviation of 10 kg/cm2 or less. A sample of n = 10 measurements produced a mean and standard deviation of 312 and 13.96, respectively. Do these data produce sufficient evidence to reject the manufacturer’s claim? Use = .05A test of hypothesis:
H0: 2 = 100 (claim is correct)
H1: 2 > 100 (claim is wrong)
A test of hypothesis:
H0: 2 = 100 (claim is correct)
H1: 2 > 100 (claim is wrong)
uses the test statistic:uses the test statistic:
55.17100
)96.13(9
10
)1( 2
2
22
sn
Example 4
Rejection region: Reject H0 if 16.91905.
Conclusion:
Since = 17.55, H0 is rejected. The standard deviation of the cement strengths is more than 10.
.1 and 1
on with distributi an on basedregion rejection awith
. variancessample two theoflarger theis where
statistic test theuse We
tailedor two one :H
:H
2211
212
2
21
1
22
210
ndfndf
F
ss
sF
.1 and 1
on with distributi an on basedregion rejection awith
. variancessample two theoflarger theis where
statistic test theuse We
tailedor two one :H
:H
2211
212
2
21
1
22
210
ndfndf
F
ss
sF
12
21
,22
21
22
21
,22
21 1
:interval Confidence
dfdfdfdf
Fs
s
Fs
s
12
21
,22
21
22
21
,22
21 1
:interval Confidence
dfdfdfdf
Fs
s
Fs
s
Testing the ratio of two variances
An experimenter has performed a lab experiment using two groups of rats. He wants to test H0: 1 = 2, but first he wants to make sure that the population variances are equal. Standard (2) Experimental (1)
Sample size 10 11
Sample mean 13.64 12.42
Sample Std Dev 2.3 5.8
22
211
22
210 :H versus:H
:y testPreliminar
22
211
22
210 :H versus:H
:y testPreliminar
Example 5
Standard (2) Experimental (1)
Sample size 10 11
Sample Std Dev 2.3 5.8
22
211
22
210
:H
:H
22
211
22
210
:H
:H
36.63.2
8.5
:statisticTest
2
2
22
21 s
sF 36.6
3.2
8.5
:statisticTest
2
2
22
21 s
sF
We designate the sample with the larger standard deviation as sample 1, to force the test statistic into the upper tail of the F distribution.
We designate the sample with the larger standard deviation as sample 1, to force the test statistic into the upper tail of the F distribution.
Solution
22
211
22
210
:H
:H
22
211
22
210
:H
:H
36.63.2
8.5
:statisticTest
2
2
22
21 s
sF 36.6
3.2
8.5
:statisticTest
2
2
22
21 s
sF
The rejection region is two-tailed, with = .05, but we only need to find the upper critical value, which has = .025 to its right.
From Table 6, with df1=10 and df2 = 9, we reject H0 if F > 3.96.
CONCLUSION: Reject H0. There is sufficient evidence to indicate that the variances are unequal. Do not rely on the assumption of equal variances for your t test!
The rejection region is two-tailed, with = .05, but we only need to find the upper critical value, which has = .025 to its right.
From Table 6, with df1=10 and df2 = 9, we reject H0 if F > 3.96.
CONCLUSION: Reject H0. There is sufficient evidence to indicate that the variances are unequal. Do not rely on the assumption of equal variances for your t test!
Solution
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