STAT 211
Handout 9 (Chapter 9): Inferences Based on Two Samples
TESTS
- One-sided (One tailed) test:
- Lower tailed: H0: population characteristics claimed constant value
(Left-sided)Ha: population characteristics < claimed constant value
- Upper tailed: H0: population characteristics claimed constant value
(Right-sided)Ha: population characteristics > claimed constant value
- Two-sided (Two tailed) test: H0: population characteristics = claimed constant value
Ha: population characteristics claimed constant value
Hypothesis testing for:
- Independent Samples
- Population characteristics: Difference between two population means, 1-2.
0 is the claimed constant.
and are the population means for X's and Y's, respectively.
and are the sample means for X's and Y's, respectively.
m and n are the sample sizes for X's and Y's, respectively.
and are the population variances for X's and Y's, respectively.
and are the sample variances for X's and Y's, respectively.
Test statistics:
- when both popn. distributions are normal and , are known
- when there is large sample size (m>40 and n>40) and , are unknown
- when both popn. distributions are normal and at least one sample size is small where unknown , are assumed to be different () and degrees of freedom, is used to look up the critical values. Your textbook calls this two-sample t test.
- , when both popn. distributions are normal and at least one sample size is small where unknown , are assumed to be the same (=) and degrees of freedom, is used to look up the critical values. Your textbook calls this pooled t test.
Decision can be made in one of the two ways:
- Let z* or t* be the computed test statistic values.
if test statistics is z / if test statistics is t
Lower tailed test / P-value = P(z<z*) / P-value = P(t<t*)
Upper tailed test / P-value = P(z>z*) / P-value = P(t>t*)
Two-tailed test / P-value = 2P(z>|z*|)= 2P(z<-|z*|) / P-value = 2P(t > |t*| )= 2P(t <- |t*| )
In each case, you can reject H0 if P-value and fail to reject H0 (accept H0) if P-value >
- Rejection region for level test:
if test statistics is z / if test statistics is t
Lower tailed test / z -z / t -t;v
Upper tailed test / z z / t t;v
Two- tailed test / z -z/2 or z z/2 / t -t/2;v or t t/2;v
100(1-)% confidence Intervals with the same assumptions,
when you assume
when you assume =
Example: Two types of plastic are suitable for use by an electronics component manufacturer. The breaking strength of this plastic is very important. It is known that ==1 psi. From a random sample of size =10 and =12, we obtain =162.5 and =155. The company will not adopt plastic 1 unless its mean breaking strength exceeds that of plastic 2 by at least 10 psi. Based on the sample information, would they use plastic 1? Use the significance level 0.05 in reaching a decision.
Exercise 9-2:
Sample sizes are large while population variances are unknown. Notice that this is a two-tailed test.
Test statistics: =4.8462
(i) Reject H0 if z -z/2=-1.96 or z z/2 =1.96. z =4.8462 > 1.96 then reject H0. Conclude that true average tread lives for two competing brand tires are different.
(ii) P-value = 2P(Z>4.8462)=2(0)=0. Since the P-value =0.05, reject H0. Conclude that true average tread lives for two competing brand tires are different.
If you prefer to answer the question computing the confidence interval, 95% confidence interval would become
=(1250.67, 2949.33). You would see that zero does not fall into interval and you would reject H0.
Exercise 9-8:
Sample sizes are large while population variances are unknown. . Notice that this is a upper-tailed test.
versus
Test statistics: =28.57
(i) Reject H0 if z z =1.645 if =0.05. z =28.57 > 1.645 and reject H0. Conclude that the data provide compelling evidence that the true average strength for the 1078 grade exceeds that for the 1064 grade by more than 10.
(ii) P-value = P(Z>28.57)=0. Since the P-value =0.05, reject H0. Conclude that the data provide compelling evidence that the true average strength for the 1078 grade exceeds that for the 1064 grade by more than 10.
Exercise 9-18:
Sample sizes are small and population variances are unknown. . Notice that this is a two-tailed test.
Test statistics: =6.166
(i) Reject H0 if t -t/2;v= -2.447 or t t/2;v =2.447 where =0.05 and v=6.886. t =6.166 > 2.447 and reject H0. Conclude that true average densities for two different types of brick are different.
(ii) P-value = 2P(t>6.166)=2(0)=0. Since the P-value =0.05, reject H0. Conclude that true average densities for two different types of brick are different.
- Population characteristics: Difference between two population proportions, p1-p2.
p0 is the claimed constant.
and are the sample proportions for X's and Y's, respectively.
and are the population proportions for X's and Y's, respectively.
m and n are the large sample sizes for X's and Y's, respectively.
The estimator for p is
Test statistics:
Decision can be made in one of the two ways:
(a)Let z* or t* be the computed test statistic values.
Lower tailed test / P-value = P(z<z*)Upper tailed test / P-value = P(z>z*)
Two-tailed test / P-value = 2P(z > |z*| )=2P(z <- |z*| )
In each case, you can reject H0 if P-value and fail to reject H0 (accept H0) if P-value >
(b)Rejection region for level test:
Lower tailed test / z -zUpper tailed test / z z
Two- tailed test / z -z/2 or z z/2
100(1-)% large sample confidence Interval:
Example: Two different types of injection-molding machines are used to form plastic parts. A part is considered defective if it has excessive shrinkage or is discolored. Two random samples, each of size 300 are selected and 15 defective parts are found from machine 1 while 8 defective parts are found in the sample from machine 2. Is it reasonable to conclude that both machines produce the same fraction of defective parts, using the significance 0.05? Find also the P-value after answering the question.
Exercise 9-48(a):
Sample sizes are large enough to satisfy the assumptions and it is a two-tailed test.
=0.2875 where =0.21 and =0.4167
Test statistics: =-4.844
(i) Reject H0 if z -z/2=-1.96 or z z/2 =1.96. |z| =4.844 > 1.96 and reject H0. Conclude that it is different for two groups of residents.
(ii) P-value = 2P(Z>4.844)=2(0)=0. Since the P-value =0.05, reject H0. Conclude that it is different for two groups of residents.
- Population characteristics: Ratio of the two population variances, or standard deviations, .
X and Y's are random sample from a normal distribution.
and are the population variances for X's and Y's, respectively.
and are the sample variances for X's and Y's, respectively.
m and n are the sample sizes for X's and Y's, respectively.
Test statistics:
Decision can be made in one of the two ways:
(a)Let F* be the computed test statistic values.
Lower tailed test / P-value = P(F<F*)Upper tailed test / P-value = P(F>F*)
Two-tailed test / P-value = 2P(F > F*)
In each case, you can reject H0 if P-value and fail to reject H0 (accept H0) if P-value >
(b)Rejection region for level test:
Lower tailed test / F F1-;m-1,n-1Upper tailed test / F F;m-1,n-1
Two- tailed test / F F1-/2;m-1,n-1 or F F/2;m-1,n-1
Notice that F1-/2;m-1,n-1 = 1 / F/2;n-1,m-1
100(1-)% confidence Interval for:
Exercise 9.57:
(a) On the F-table, column for 5 and row for 8 will give the area on the right 0.05 with F0.05:5,8 = 3.69
(d) F0.95:8,5 = 1/ F0.05:5,8 = 1/3.69 =0.271
(e) The percentile is the area on the left of the value and it means the area on the right of the value is 0.01. On the F-table, look at column for 10 and row for 12 with the area on the right 0.01.
P( F F0.01:10,12 )= 0.99 then F0.01:10,12 = 4.31
(h) P(0.177 F 4.74) = P(F0.99:10,5 F F0.05:10,5) =1-(0.01+0.05)=0.94 where F0.99:10,5 = 1/F0.01:5,10 = 1/5.64=0.177
Example: A study was performed to determine whether men and women differ in their repeatability in assembling components on printed circuit boards. Two samples of 25 men and 21 women were selected and each subject assembled the units. The two sample standard deviations of assembly time were smen=0.98 min and swomen=1.02 min. Is there evidence to support the claim that men and women differ in repeatability for this assembly task? Use the significance level 0.02 and state any necessary assumptions about underlying distribution of the data.
Exercise 9.62:
when
when
=0.10, m=48, n=45
Test statistics:
Reject H0 if F F1-/2;m-1,n-1 = F0.95;47,44 =1/1.6336=0.61214 or F F/2;m-1,n-1= F0.05;47,44 =1.6412
Since the table on your book is not very detailed, you can use both degrees of freedom as 40 to look at the table.
Since 1.2219 is not larger than 1.6412 or smaller than 0.61214, fail to reject H0 . Conclude that the variances are the same.
- Dependent Samples- Paired Data
Population characteristics: Difference between two population means, D =1-2.
0 is the claimed constant.
Assumption: the difference distribution should be normal.
Test statistics: where D=X-Y and and are the corresponding sample average and the standard deviation of D. Both X and Y must have n observations.
Decision can be made in one of the two ways:
(a) Let z* or t* be the computed test statistic values.
Lower tailed test / P-value = P(t<t*)Upper tailed test / P-value = P(t>t*)
Two-tailed test / P-value = 2P(t > |t*| )=2P(t <- |t*| )
In each case, you can reject H0 if P-value and fail to reject H0 (accept H0) if P-value >
(b) Rejection region for level test:
Lower tailed test / t -t;n-1Upper tailed test / t t;n-1
Two- tailed test / t -t/2;n-1 or t t/2;n-1
100(1-)% confidence Intervals with the same assumptions,
Example: The manager of a fleet of automobiles is testing two brands of radial tires. He assigns one tire of each brand at random to the two rear wheels of eight cars and runs the cars until the tires wear out. The descriptive statistics for the data are shown below (in kilometers). Find the 99% confidence interval on the difference in mean life. Which brand would you prefer based on this calculation? Is there an alternative method to answer this question instead of computing the confidence interval?
Variable N Mean Median StDev SE Mean Minimum Maximum Q1 Q3
Brand1 8 38479 37067 5590 1976 32100 48360 34185 43525
Brand2 8 37611 36655 5244 1854 31950 47800 33491 41214
Difference 8 868 475 1290 456.1 -805 3020 N/A N/A
Brand1 / 36925 / 45300 / 36240 / 32100 / 37210 / 48360 / 38200 / 33500Brand2 / 34318 / 42280 / 35500 / 31950 / 38015 / 47800 / 37810 / 33215
Difference / 2607 / 3020 / 740 / 150 / -805 / 560 / 390 / 285
Exercise 9.44:
or
or
(a) the difference distribution should be normal.
(b) Computed differences, d=X-Y: 8151084681550 535 786 1162 517 910
which gives the sample mean, =782.2222 and the standard deviation, =236.736.
Test statistics: =3.5764
Since this is an upper tailed test, reject H0 if t t;n-1= t0.05;8 =1.86 or if the P-value=P(t>3.5764) is less than =0.05. Notice that test statistics, 3.5764 is large than 1.86. Also the 0.005 <P-value < 0.001 by looking at the table. Reject H0 and there is sufficient evidence to conclude that the true average yield in winter wheat is more than 500 higher than for spring wheat.