1. Inferences on Two Samples
Chapter 11
Inferences on Two Samples

2. Inference about Two Means: Dependent Samples
Section
11.1
Inference about Two Means: Dependent Samples

3. Objectives Distinguish between independent and dependent sampling
Test hypotheses regarding matched-pairs data
Construct and interpret confidence intervals about the population mean difference of matched-pairs data

4. Objective 1
Distinguish between Independent and Dependent Sampling

5. A sampling method is independent when the individuals selected for one sample do not dictate which individuals are to be in a second sample. A sampling method is dependent when the individuals selected to be in one sample are used to determine the individuals to be in the second sample.
Dependent samples are often referred to as matched-pairs samples.

6. Parallel Example 1: Distinguish between Independent and
Parallel Example 1: Distinguish between Independent and Dependent Sampling
For each of the following, determine whether the sampling method is independent or dependent.
A researcher wants to know whether the price of a one night stay at a Holiday Inn Express is less than the price of a one night stay at a Red Roof Inn. She randomly selects 8 towns where the location of the hotels is close to each other and determines the price of a one night stay.
A researcher wants to know whether the “state” quarters (introduced in 1999) have a mean weight that is different from “traditional” quarters. He randomly selects 18 “state” quarters and 16 “traditional” quarters and compares their weights.

7. Solution
The sampling method is dependent since the 8 Holiday Inn Express hotels can be matched with one of the 8 Red Roof Inn hotels by town.
The sampling method is independent since the “state” quarters which were sampled had no bearing on which “traditional” quarters were sampled.

8. Objective 2
Test Hypotheses Regarding Matched-Pairs Data

9. “In Other Words”
Statistical inference methods on matched-pairs data use the same methods as inference on a single population mean with  unknown, except that the differences are analyzed.

10. Testing Hypotheses Regarding the Difference
of Two Means Using a Matched-Pairs Design
To test hypotheses regarding the mean difference of matched-pairs data, the following must be satisfied:
the sample is obtained using simple random sampling
the sample data are matched pairs,
the differences are normally distributed with no outliers or the sample size, n, is large (n ≥ 30).

11. Step 1: Determine the null and alternative. hypotheses
Step 1: Determine the null and alternative hypotheses. The hypotheses can be structured in one of three ways, where d is the population mean difference of the matched-pairs data.

12. Step 2: Select a level of significance, , based
on the seriousness of making a Type I error.

13. Step 3: Compute the test statistic
which approximately follows Student’s t-distribution with n-1 degrees of freedom. The values of and sd are the mean and standard deviation of the differenced data.

14. Classical Approach
Step 4: Use Table VI to determine the critical
value using n-1 degrees of freedom.

15. Classical Approach
Two-Tailed
(critical value)

16. Classical Approach
Left-Tailed
(critical value)

17. Classical Approach
Right-Tailed
(critical value)

18. Classical Approach
Step 5: Compare the critical value with the test statistic:

19. P-Value Approach
Step 4: Use Table VI to determine the P-value
using n-1 degrees of freedom.

20. P-Value Approach
Two-Tailed

21. P-Value Approach
Left-Tailed

22. P-Value Approach
Right-Tailed

23. P-Value Approach
Step 5: If P-value < , reject the null hypothesis.

24. Step 6: State the conclusion.

25. These procedures are robust, which means that minor departures from normality will not adversely affect the results. However, if the data have outliers, the procedure should not be used.

26. Parallel Example 2: Testing a Claim Regarding Matched-Pairs Data
The following data represent the cost of a one night stay in Hampton Inn Hotels and La Quinta Inn Hotels for a random sample of 10 cities. Test the claim that Hampton Inn Hotels are priced differently than La Quinta Hotels at the =0.05 level of significance.

27. City
Hampton Inn
La Quinta
Dallas
129
105
Tampa Bay
149 96
St. Louis 49
Seattle
189
San Diego
109
119
Chicago
160 89
New Orleans 72
Phoenix 59
Atlanta 90
Orlando 69

28. Solution
This is a matched-pairs design since the hotel prices come from the same ten cities. To test the hypothesis, we first compute the differences and then verify that the differences come from a population that is approximately normally distributed with no outliers because the sample size is small.
The differences (Hampton - La Quinta) are:
with = 51.4 and sd =

29. Solution
No violation of normality assumption.

30. Solution
No outliers.

31. Solution Step 1: We want to determine if the prices differ:
H0: d = 0 versus H1: d  0
Step 2: The level of significance is =0.05.
Step 3: The test statistic is

32. Solution: Classical Approach
Step 4: This is a two-tailed test so the critical values at the =0.05 level of significance with n-1=10-1=9 degrees of freedom are -t0.025= and t0.025=2.262.

33. Solution: Classical Approach
Step 5: Since the test statistic, t0=5.27 is greater than the critical value t.025=2.262, we reject the null hypothesis.

34. Solution: P-Value Approach
Step 4: Because this is a two-tailed test, the P-value is two times the area under the t-distribution with n-1=10-1=9 degrees of freedom to the right of the test statistic t0= That is, P-value = 2P(t > 5.27) ≈ 2( )= (using technology). Approximately 5 samples in 10,000 will yield results as extreme as we obtained if the null hypothesis is true.

35. Solution: P-Value Approach
Step 5: Since the P-value is less than the level of significance =0.05, we reject the null hypothesis.

36. Solution
Step 6: There is sufficient evidence to conclude that Hampton Inn hotels and La Quinta hotels are priced differently at the =0.05 level of significance.

37. Objective 3
Construct and Interpret Confidence Intervals for the Population Mean Difference of Matched-Pairs Data

38. Confidence Interval for Matched-Pairs Data
A (1-)100% confidence interval for d is given by
Lower bound:
Upper bound:
The critical value t/2 is determined using n-1 degrees of freedom.

39. Confidence Interval for Matched-Pairs Data
Note: The interval is exact when the population is normally distributed and approximately correct for nonnormal populations, provided that n is large.

40. Parallel Example 4: Constructing a Confidence Interval for
Parallel Example 4: Constructing a Confidence Interval for Matched-Pairs Data
Construct a 90% confidence interval for the mean difference in price of Hampton Inn versus La Quinta hotel rooms.

41. Solution
We have already verified that the differenced data come from a population that is approximately normal with no outliers.
Recall = 51.4 and sd =
From Table VI with  = 0.10 and 9 degrees of freedom, we find t/2 =

42. Solution Thus, Lower bound = Upper bound =
We are 90% confident that the mean difference in hotel room price for Ramada Inn versus La Quinta Inn is between $33.53 and $69.27.

43. Inference about Two Means: Independent Samples
Section
11.2
Inference about Two Means: Independent Samples

44. Objectives
Test hypotheses regarding the difference of two independent means
Construct and interpret confidence intervals regarding the difference of two independent means

45. Sampling Distribution of the Difference of Two
Means: Independent Samples with Population
Standard Deviations Unknown (Welch’s t)
Suppose that a simple random sample of size n1 is taken from a population with unknown mean 1 and unknown standard deviation 1. In addition, a simple random sample of size n2 is taken from a population with unknown mean 2 and unknown standard deviation 2. If the two populations are normally distributed or the sample sizes are sufficiently large (n1 ≥ 30, n2 ≥ 30) , then
approximately follows Student’s t-distribution with the smaller of n1-1 or n2-1 degrees of freedom where is the sample mean and si is the sample standard deviation from population i.

46. Objective 1
Test Hypotheses Regarding the Difference of Two Independent Means

47. Testing Hypotheses Regarding the Difference of Two Means
To test hypotheses regarding two population means, 1 and 2, with unknown population standard deviations, we can use the following steps, provided that:
the samples are obtained using simple random sampling;
the samples are independent;
the populations from which the samples are drawn are normally distributed or the sample sizes are large (n1 ≥ 30, n2 ≥ 30).

48. Step 1: Determine the null and alternative. hypotheses
Step 1: Determine the null and alternative hypotheses. The hypotheses are structured in one of three ways:

49. Step 2: Select a level of significance, , based
on the seriousness of making a Type I error.

50. Step 3: Compute the test statistic
which approximately follows Student’s t- distribution.

51. Classical Approach
Step 4: Use Table VI to determine the critical
value using the smaller of n1 -1 or n2 -1
degrees of freedom.

52. Classical Approach
Two-Tailed
(critical value)

53. Classical Approach
Left-Tailed
(critical value)

54. Classical Approach
Right-Tailed
(critical value)

55. Classical Approach
Step 5: Compare the critical value with the test statistic:

56. P-Value Approach
Step 4: Use Table VI to determine the P-value
using the smaller of n1 -1 or n2 -1 degrees of freedom.

57. P-Value Approach
Two-Tailed

58. P-Value Approach
Left-Tailed

59. P-Value Approach
Right-Tailed

60. P-Value Approach
Step 5: If P-value < , reject the null hypothesis.

61. Step 6: State the conclusion.

62. These procedures are robust, which means that minor departures from normality will not adversely affect the results. However, if the data have outliers, the procedure should not be used.

63. Parallel Example 1: Testing Hypotheses Regarding Two Means
A researcher wanted to know whether “state” quarters had a weight that is more than “traditional” quarters. He randomly selected 18 “state” quarters and 16 “traditional” quarters, weighed each of them and obtained the following data.

65. Test the claim that “state” quarters have a mean weight that is more than “traditional” quarters at the =0.05 level of significance.
NOTE: A normal probability plot of “state” quarters indicates the population could be normal. A normal probability plot of “traditional” quarters indicates the population could be normal

66. No outliers.

67. Solution
Step 1: We want to determine whether state quarters weigh more than traditional quarters:
H0: 1 = 2 versus H1: 1 > 2
Step 2: The level of significance is =0.05.
Step 3: The test statistic is

68. Solution: Classical Approach
Step 4: This is a right-tailed test with =0.05. Since n1-1=17 and n2-1=15, we will use 15 degrees of freedom. The corresponding critical value is t0.05=1.753.

69. Solution: Classical Approach
Step 5: Since the test statistic, t0=2.53 is greater than the critical value t.05=1.753, we reject the null hypothesis.

70. Solution: P-Value Approach
Step 4: Because this is a right-tailed test, the P-value is the area under the t-distribution to the right of the test statistic t0= That is, P-value = P(t > 2.53) ≈ 0.01.

71. Solution: P-Value Approach
Step 5: Since the P-value is less than the level of significance =0.05, we reject the null hypothesis.

72. Solution
Step 6: There is sufficient evidence at the =0.05 level to conclude that the state quarters weigh more than the traditional quarters.

73. NOTE: The degrees of freedom used to determine the critical value in the last example are conservative. Results that are more accurate can be obtained by using the following degrees of freedom:

74. Objective 3
Construct and Interpret Confidence Intervals Regarding the Difference of Two Independent Means

75. Constructing a (1-) 100% Confidence Interval for the Difference of Two Means
A simple random sample of size n1 is taken from a population with unknown mean 1 and unknown standard deviation 1. Also, a simple random sample of size n2 is taken from a population with unknown mean 2 and unknown standard deviation 2. If the two populations are normally distributed or the sample sizes are sufficiently large (n1≥30 and n2≥30), a (1-)100% confidence interval about 1 - 2 is given by
Lower bound:
Upper bound:

76. Parallel Example 3: Constructing a Confidence Interval for
Parallel Example 3: Constructing a Confidence Interval for the Difference of Two Means
Construct a 95% confidence interval about the difference between the population mean weight of a “state” quarter versus the population mean weight of a “traditional” quarter.

77. Solution
We have already verified that the populations are approximately normal and that there are no outliers.
Recall = 5.702, s1 = , = and s2 =
From Table VI with  = 0.05 and 15 degrees of freedom, we find t/2 =

78. Solution
Thus,
Lower bound =
Upper bound =

79. Solution
We are 95% confident that the mean weight of the “state” quarters is between and ounces more than the mean weight of the “traditional” quarters. Since the confidence interval does not contain 0, we conclude that the “state” quarters weigh more than the “traditional” quarters.

80. When the population variances are assumed to be equal, the pooled t-statistic can be used to test for a difference in means for two independent samples. The pooled t-statistic is computed by finding a weighted average of the sample variances and using this average in the computation of the test statistic.
The advantage to this test statistic is that it exactly follows Student’s t-distribution with n1+n2-2 degrees of freedom.
The disadvantage to this test statistic is that it requires that the population variances be equal.

81. Inference about Two Population Proportions
Section
11.3
Inference about Two Population Proportions

82. Objectives Test hypotheses regarding two population proportions
Construct and interpret confidence intervals for the difference between two population proportions
Use McNemar’s Test to compare two proportions from matched-pairs data
Determine the sample size necessary for estimating the difference between two population proportions within a specified margin of error.

83. Sampling Distribution of the Difference between Two Proportions
Suppose that a simple random sample of size n1 is taken from a population where x1 of the individuals have a specified characteristic, and a simple random sample of size n2 is independently taken from a different population where x2 of the individuals have a specified characteristic. The sampling distribution of , where and , is approximately normal, with mean and standard deviation
provided that and

84. Sampling Distribution of the Difference between Two Proportions
The standardized version of is then written as
which has an approximate standard normal distribution.

85. Objective 1
Test Hypotheses Regarding Two Population Proportions

86. The best point estimate of p is called the pooled estimate of p, denoted , where
Test statistic for Comparing Two Population Proportions

87. Hypothesis Test Regarding the Difference
between Two Population Proportions
To test hypotheses regarding two population proportions, p1 and p2, we can use the steps that follow, provided that:
the samples are independently obtained using simple random sampling,
and and
3. n1 ≤ 0.05N1 and n2 ≤ 0.05N2 (the sample size is no more than 5% of the population size); this requirement ensures the independence necessary for a binomial experiment.

88. Step 1: Determine the null and alternative hypotheses
Step 1: Determine the null and alternative hypotheses. The hypotheses can be structured in one of three ways:

89. Step 2: Select a level of significance, , based
on the seriousness of making a Type I error.

90. Step 3: Compute the test statistic
where

91. Classical Approach
Step 4: Use Table V to determine the critical
value.

92. Classical Approach
Two-Tailed
(critical value)

93. Classical Approach
Left-Tailed
(critical value)

94. Classical Approach
Right-Tailed
(critical value)

95. Classical Approach
Step 5: Compare the critical value with the test statistic:

96. P-Value Approach
Step 4: Use Table V to estimate the P-value..

97. P-Value Approach
Two-Tailed

98. P-Value Approach
Left-Tailed

99. P-Value Approach
Right-Tailed

100. P-Value Approach
Step 5: If P-value < , reject the null hypothesis.

101. Step 6: State the conclusion.

102. Parallel Example 1: Testing Hypotheses Regarding Two
Parallel Example 1: Testing Hypotheses Regarding Two Population Proportions
An economist believes that the percentage of urban households with Internet access is greater than the percentage of rural households with Internet access. He obtains a random sample of 800 urban households and finds that 338 of them have Internet access. He obtains a random sample of 750 rural households and finds that 292 of them have Internet access. Test the economist’s claim at the =0.05 level of significance.

103. Solution We must first verify that the requirements are satisfied:
The samples are simple random samples that were obtained independently.
x1=338, n1=800, x2=292 and n2=750, so
3. The sample sizes are less than 5% of the population size.

104. H0: p1 - p2=0 versus H1: p1 - p2 > 0
Solution
Step 1: We want to determine whether the percentage of urban households with Internet access is greater than the percentage of rural households with Internet access. So,
H0: p1 = p2 versus H1: p1 > p2
or, equivalently,
H0: p1 - p2=0 versus H1: p1 - p2 > 0
Step 2: The level of significance is  = 0.05.

105. Solution
Step 3: The pooled estimate of is:
The test statistic is:

106. Solution: Classical Approach
Step 4: This is a right-tailed test with =0.05. The critical value is z0.05=1.645.

107. Solution: Classical Approach
Step 5: Since the test statistic, z0=1.33 is less than the critical value z.05=1.645, we fail to reject the null hypothesis.

108. Solution: P-Value Approach
Step 4: Because this is a right-tailed test, the P-value is the area under the normal to the right of the test statistic z0= That is, P-value = P(Z > 1.33) ≈ 0.09.

109. Solution: P-Value Approach
Step 5: Since the P-value is greater than the level of significance =0.05, we fail to reject the null hypothesis.

110. Solution
Step 6: There is insufficient evidence at the =0.05 level to conclude that the percentage of urban households with Internet access is greater than the percentage of rural households with Internet access.

111. Objective 2
Construct and Interpret Confidence Intervals for the Difference between Two Population Proportions

112. Constructing a (1-) 100% Confidence Interval for the Difference between Two Population Proportions
To construct a (1-)100% confidence interval for the difference between two population proportions, the following requirements must be satisfied:
the samples are obtained independently using simple random sampling,
, and
3. n1 ≤ 0.05N1 and n2 ≤ 0.05N2 (the sample size is no more than 5% of the population size); this requirement ensures the independence necessary for a binomial experiment.

113. Constructing a (1-) 100% Confidence Interval for the Difference between Two Population Proportions
Provided that these requirements are met, a (1-)100% confidence interval for p1-p2 is given by
Lower bound:
Upper bound:

114. Parallel Example 3: Constructing a Confidence Interval for
Parallel Example 3: Constructing a Confidence Interval for the Difference between Two Population Proportions
An economist obtains a random sample of 800 urban households and finds that 338 of them have Internet access. He obtains a random sample of 750 rural households and finds that 292 of them have Internet access. Find a 99% confidence interval for the difference between the proportion of urban households that have Internet access and the proportion of rural households that have Internet access.

115. Solution
We have already verified the requirements for constructing a confidence interval for the difference between two population proportions in the previous example.
Recall

116. Solution
Thus,
Lower bound =
Upper bound =

117. Solution
We are 99% confident that the difference between the proportion of urban households that have Internet access and the proportion of rural households that have Internet access is between and Since the confidence interval contains 0, we are unable to conclude that the proportion of urban households with Internet access is greater than the proportion of rural households with Internet access.

118. Objective 3
Use McNemar’s Test to Compare Two Proportions from Matched-Pairs Data

119. McNemar’s Test is a test that can be used to compare two proportions with matched-pairs data (i.e., dependent samples)

120. Testing a Hypothesis Regarding the Difference of Two Population Proportions: Dependent Samples
To test hypotheses regarding two population proportions
p1 and p2, where the samples are dependent, arrange the
data in a contingency table as follows:
Treatment A
Treatment B
Success
Failure
f11
f12
f21
f22

121. Testing a Hypothesis Regarding the Difference of Two Population Proportions: Dependent Samples
We can use the steps that follow provided that:
the samples are dependent and are obtained
randomly and
the total number of observations where the
outcomes differ must be greater than or equal to 10.
That is, f12 + f21 ≥ 10.

122. Step 1: Determine the null and alternative
hypotheses.
H0: the proportions between the two
populations are equal (p1 = p2)
H1: the proportions between the two
populations differ (p1 ≠ p2)

123. Step 2: Select a level of significance, , based
on the seriousness of making a Type I error.

124. Step 3: Compute the test statistic

125. Classical Approach
Step 4: Use Table V to determine the critical
value. This is a two tailed test. However, z0 is always positive, so we only need to find the right critical value z/2.

126. Classical Approach

127. Classical Approach
Step 5: If z0 > z/2, reject the null hypothesis.

128. P-Value Approach
Step 4: Use Table V to determine the P-value..
Because z0 is always positive, we find the area to the right of z0 and then double this area (since this is a two-tailed test).

129. P-Value Approach

130. P-Value Approach
Step 5: If P-value < , reject the null hypothesis.

131. Step 6: State the conclusion.

132. Parallel Example 4: Analyzing the Difference of Two
Parallel Example 4: Analyzing the Difference of Two Proportions from Matched-Pairs Data
A recent General Social Survey asked the following two questions of a random sample of 1483 adult Americans under the hypothetical scenario that the government suspected that a terrorist act was about to happen:
Do you believe the authorities should have the right to tap people’s telephone conversations?
Do you believe the authorities should have the right to detain people for as long as they want without putting them on trial?

133. Tap Phone Agree Disagree
Parallel Example 4: Analyzing the Difference of Two Proportions from Matched-Pairs Data
The results of the survey are shown below:
Do the proportions who agree with each scenario differ significantly? Use the =0.05 level of significance.
Detain
Tap
Phone
Agree
Disagree
572
237
224
450

134. Solution
The sample proportion of individuals who believe that the authorities should be able to tap phones is The sample proportion of individuals who believe that the authorities should have the right to detain people is
. We want to determine whether the difference in sample proportions is due to sampling error or to the fact that the population proportions differ.

135. Solution
The samples are dependent and were obtained randomly. The total number of individuals who agree with one scenario, but disagree with the other is =461, which is greater than 10. We can proceed with McNemar’s Test.
Step 1: The hypotheses are as follows
H0: the proportions between the two populations are equal (pT = pD)
H1: the proportions between the two populations differ (pT ≠ pD)
Step 2: The level of significance is  = 0.05.

136. Solution
Step 3: The test statistic is:

137. Solution: Classical Approach
Step 4: The critical value with an =0.05 level of significance is z0.025=1.96.

138. Solution: Classical Approach
Step 5: Since the test statistic, z0=0.56 is less than the critical value z.025=1.96, we fail to reject the null hypothesis.

139. Solution: P-Value Approach
Step 4: The P-value is two times the area under the normal curve to the right of the test statistic z0= That is, P-value = 2*P(Z > 0.56) ≈

140. Solution: P-Value Approach
Step 5: Since the P-value is greater than the level of significance =0.05, we fail to reject the null hypothesis.

141. Solution
Step 6: There is insufficient evidence at the =0.05 level to conclude that there is a difference in the proportion of adult Americans who believe it is okay to phone tap versus detaining people for as long as they want without putting them on trial in the event that the government believed a terrorist plot was about to happen.

142. Objective 4
Determine the Sample Size Necessary for Estimating the Difference between Two Population Proportions within a Specified Margin of Error

143. Sample Size for Estimating p1-p2
The sample size required to obtain a (1-)100%
confidence interval with a margin of error, E, is given by
rounded up to the next integer, if prior estimates of p1
and p2, , are available. If prior estimates of
p1 and p2 are unavailable, the sample size is
rounded up to the next integer.

144. Parallel Example 5: Determining Sample Size
A doctor wants to estimate the difference in the proportion of year old mothers that received prenatal care and the proportion of year old mothers that received prenatal care. What sample size should be obtained if she wished the estimate to be within 2 percentage points with 95% confidence assuming:
she uses the results of the National Vital Statistics Report results in which 98% of the year old mothers received prenatal care and 99.2% of year old mothers received prenatal care.
she does not use any prior estimates.

145. Solution We have E=0.02 and z/2=z0.025=1.96. Letting ,
The doctor must sample 265 randomly selected year old mothers and 265 randomly selected year old mothers.

146. Solution b) Without prior estimates of p1 and p2, the sample size is
The doctor must sample 4802 randomly selected year old mothers and 4802 randomly selected year old mothers. Note that having prior estimates of p1 and p2 reduces the number of mothers that need to be surveyed.

147. Inference about Two Population Standard Deviations
Section
11.4
Inference about Two Population Standard Deviations

148. Objectives Find critical values of the F-distribution
Test hypotheses regarding two population standard deviations

149. Objective 1
Find Critical Values of the F-distribution

150. Requirements for Testing Claims Regarding Two Population Standard Deviations
1. The samples are independent simple random samples.

151. Requirements for Testing Claims Regarding Two Population Standard Deviations
1. The samples are independent simple random samples.
2. The populations from which the samples are drawn are normally distributed.

152. CAUTION! If the populations from which the samples are drawn
are not normal, do not use the inferential procedures
discussed in this section.

153. Notation Used When Comparing Two Population Standard Deviations
: Variance for population 1
: Variance for population 2
: Sample variance for population 1
: Sample variance for population 2
n1 : Sample size for population 1
n2 : Sample size for population 2

154. Fisher's F-distribution
If and and are sample variances from
independent simple random samples of size n1 and
n2, respectively, drawn from normal populations,
then
follows the F-distribution with n1-1 degrees of
freedom in the numerator and n2-1 degrees of
freedom in the denominator.

155. Characteristics of the F-distribution
1. It is not symmetric. The F-distribution is skewed right.

156. Characteristics of the F-distribution
1. It is not symmetric. The F-distribution is skewed right.
2. The shape of the F-distribution depends upon the degrees of freedom in the numerator and denominator. This is similar to the 2 distribution and Student’s t-distribution, whose shapes depend upon their degrees of freedom.

157. Characteristics of the F-distribution
1. It is not symmetric. The F-distribution is skewed right.
2. The shape of the F-distribution depends upon the degrees of freedom in the numerator and denominator. This is similar to the distribution and Student’s t-distribution, whose shape depends upon their degrees of freedom.
3. The total area under the curve is 1.

158. Characteristics of the F-distribution
1. It is not symmetric. The F-distribution is skewed right.
2. The shape of the F-distribution depends upon the degrees of freedom in the numerator and denominator. This is similar to the distribution and Student’s t-distribution, whose shape depends upon their degrees of freedom.
3. The total area under the curve is 1.
4. The values of F are always greater than or equal to zero.

160. is the critical F with n1 – 1 degrees of freedom in the numerator and n2 – 1 degrees of freedom in the denominator and an area of  to the right of the critical F.

161. To find the critical F with an area of α to the left, use the following:

162. Find the critical F-value:
Parallel Example 1: Finding Critical Values for the F-distribution
Find the critical F-value:
for a right-tailed test with  =0.1, degrees of freedom in the numerator = 8 and degrees of freedom in the denominator = 4.
for a two-tailed test with  =0.05, degrees of freedom in the numerator = 20 and degrees of freedom in the denominator = 15.

163. Solution
a) F0.1,8,4 = 3.95
b) F.025,20,15 = 2.76; = 0.39

164. NOTE:
If the number of degrees of freedom is not found in the table, we follow the practice of choosing the degrees of freedom closest to that desired. If the degrees of freedom is exactly between two values, find the mean of the values.

165. Objective 2
Test Hypotheses Regarding Two Population Standard Deviations

166. Test Hypotheses Regarding Two Population Standard Deviations
To test hypotheses regarding two population standard deviations, 1 and 2, we can use the following steps, provided that
the samples are obtained using simple random sampling,
the sample data are independent, and
the populations from which the samples are drawn are normally distributed.

167. Step 1: Determine the null and alternative hypotheses
Step 1: Determine the null and alternative hypotheses. The hypotheses can be structured in one of three ways:
Two-Tailed
Left-Tailed
Right-Tailed
H0: 1 = 2
H1: 1 ≠ 2
H1: 1 < 2
H1: 1 > 2
Note: 1 is the population standard deviation for population 1
and 2 is the population standard deviation for population 2.

168. Step 2: Select a level of significance, , based
on the seriousness of making a Type I error.

169. Step 3: Compute the test statistic
which follows Fisher’s F-distribution with n1-1 degrees of freedom in the numerator and n2-1 degrees of freedom in the denominator.

170. Classical Approach
Step 4: Use Table VIII to determine the critical
value(s) using n1-1 degrees of freedom in the numerator and n2-1 degrees of freedom in the denominator.

171. Classical Approach
Two-Tailed
(critical value)

172. Classical Approach
Left-Tailed
(critical value)

173. Classical Approach
Right-Tailed
(critical value)

174. Step 5: Compare the critical value with the test statistic:
Classical Approach
Step 5: Compare the critical value with the test statistic:
Two-Tailed
Left-Tailed
Right-Tailed If
or ,
reject the null hypothesis.
If ,
If ,

175. P-Value Approach
Step 4: Use technology to determine the
P-value.

176. P-Value Approach
Step 5: If P-value < , reject the null hypothesis.

177. Step 6: State the conclusion.

178. CAUTION! The procedures just presented are not robust,
minor departures from normality will adversely affect the results of the test. Therefore, the test should be used only when the requirement of normality has been verified.

179. Parallel Example 2: Testing Hypotheses Regarding Two
Parallel Example 2: Testing Hypotheses Regarding Two Population Standard Deviations
A researcher wanted to know whether “state” quarters had a standard deviation weight that is less than “traditional” quarters. He randomly selected 18 “state” quarters and 16 “traditional” quarters, weighed each of them and obtained the data on the next slide. A normal probability plot indicates that the sample data could come from a population that is normal. Test the researcher’s claim at the  = 0.05 level of significance.

181. Solution
Step 1: The researcher wants to know if “state” quarters have a standard deviation weight that is less than “traditional” quarters. Thus
H0: 1= 2 versus H1: 1< 2
This is a left-tailed test.
Step 2: The level of significance is  = 0.05.

182. Solution
Step 3: The standard deviation of “state” quarters was found to be and the standard deviation of “traditional” quarters was found to be The test statistic is then

183. Solution: Classical Approach
Step 4: Since this is a left-tailed test, we determine the critical value at the 1- = = 0.95 level of significance with n1-1=18-1=17 degrees of freedom in the numerator and n2-1=16-1=15 degrees of freedom in the denominator. Thus,
Note: we used the table value F0.05,15,15 for the above calculation since this is the closest to the required degrees of freedom available from Table VIII.

184. Solution: Classical Approach
Step 5: Since the test statistic F0= 0.52 is greater than the critical value F0.95,17,15=0.42, we fail to reject the null hypothesis.

185. Solution: P-Value Approach
Step 4: Using technology, we find that the P-value is If the statement in the null hypothesis were true, we would expect to get the results obtained about 10 out of 100 times. This is not very unusual.

186. Solution: P-Value Approach
Step 5: Since the P-value is greater than the level of significance,  = 0.05, we fail to reject the null hypothesis.

187. Solution
Step 6: There is not enough evidence to conclude that the standard deviation of weight is less for “state” quarters than it is for “traditional” quarters at the  = 0.05 level of significance.

188. Putting It Together: Which Method Do I Use?
Section
11.5
Putting It Together: Which Method Do I Use?

189. Objectives
Determine the appropriate hypothesis test to perform

190. Objective 1
Determine the Appropriate Hypothesis Test to Perform

191. What parameter is addressed in the hypothesis?
Proportion, p
 or 2
Mean, 

192. Proportion, p
Is the sampling Dependent or Independent?

193. Proportion, p Dependent samples:
Provided the samples are obtained randomly and the
total number of observations where the outcomes
differ is at least 10, use the normal distribution with

194. Proportion, p Independent samples: Provided for each sample and the
sample size is no more than 5% of the population
size, use the normal distribution with
where

195.  or 2 Provided the data are normally distributed, use the
F-distribution with

196. Mean, 
Is the sampling Dependent or Independent?

197. Mean,  Dependent samples:
Provided each sample size is greater than 30 or the
differences come from a population that is
normally distributed, use Student’s t-distribution
with n-1 degrees of freedom with

198. Mean,  Independent samples:
Provided each sample size is greater than 30 or each
population is normally distributed, use Student’s
t-distribution