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Probability and Statistics Estimation
Chapter 7 Section 4 Estimating u1 - u2 and p1 - p2
Essential Question: How are differences of confidence intervals calculated? Student Objectives: The student will distinguish between dependent and independent
samples. The student will determine the confidence interval using the formula
for the difference in population means when the standard deviations are known (large samples).
The student will determine the confidence interval using the calculator for the difference in population means when the standard deviations are known (large samples).
The student will determine the confidence interval using the formula for the difference in population means when the standard deviations are unknown (small samples).
The student will determine the confidence interval using the calculator for the difference in population means when the standard deviations are known (small samples).
The student will determine the confidence interval using the formula for the difference in population proportions.
The student will determine the confidence interval using the calculator for the difference in population proportions.
The student will interpret the meaning and implications of all- positive, all-negative, or mixed signed intervals.
Terms: c
Degrees of freedom
Dependent
independent
Margin of error
n
Pooled
p̂
r
s
Satterthwaite’s formula
Graphing Calculator Skills: Calculating the confidence intervals using the stat package from the graphing calculator Enter a list of data Enter the stats of the sample data Performing a 2-SampZInt (option 9) on the calculator Performing a 2-SampTInt (option 0) on the calculator Performing a 2-ProbZInt (option B) on the calculator Calculating and storing the statistics for a 1-Var Stats calculation
q̂
!
tc
x
zc
Requirements:1. Let ! 1 and ! 2 be the population standard deviations from population 1 and population 2.2. Obtain two independent random samples, one from population 1 and the other from population 2.3. Let x1 and x2 be the sample means from population 1 and population 2. 4. Let n1 and n2 be the sample sizes from population 1 and population 2.5. The population distributions of population 1 and population 2 must be normal or their samples sizes must each be greater than or equal to 30 (n1 " 30 and n2 " 30).
x1 ! x2( )! zc " 12
n1
+ " 22
n2
< µ1 ! µ2 < x1 ! x2( ) + zc " 12
n1
+ " 22
n2
E = zc" 1
2
n1
+ " 22
n2
c = Confidence level (0 < c <1)zc= Critical value from the Student's t Distribution Chart
Confidence Interval for µ1 ! µ2 when using Small Samples or both " 1 and " 2 are unknown
Requirements:1. Obtain two independent random samples, one from population 1 and the other from population 2.2. Let n1 and n2 be the sample sizes from population 1 and population 2.3. Let x1 and x2 be the sample means from population 1 and population 2. 4. Let s1 and s2 be the sample standard deviations from population 1 and population 2.5. The population distributions of population 1 and population 2 are unknown and their samples sizes are each less than 30 (n1 < 30 and n2 < 30).6. The degrees of freedom is the SMALLER of n1 !1 or n2 !1.
Key Concepts:
Confidence Interval for µ1 ! µ2 when both " 1 and " 2 are known
x1 ! x2( )! tc s12
n1
+ s22
n2
< µ1 ! µ2 < x1 ! x2( ) + tc s12
n1
+ s22
n2
E = tcs1
2
n1
+ s22
n2
c = Confidence level (0 < c <1)tc= Critical value from the Student's t Distribution Chart where the degrees of freedom is the smallest of n1 !1 or n2 !1
Satterthwaite's Formula for the Degrees of Freedom This method calculates the t-value based on partial degrees of freedom. The formula to calculate the degrees of freedom is shown below (also on page 393).
df =
s12
n1+ s2
2
n2
!"#
$%&
2
s12
n1( )2 n1 '1( )+ s2
2
n2( )2 n2 '1( )
Examine problem statement forµ1 ! µ2
! 1 and ! 2
are known
Use the normal distribution with the margin of error
E = zc! 1
2
n1
+ ! 22
n2
! 1 and ! 2 are unknown
! 1 and ! 2 are unknown, but
there is reason to believe ! 1 =! 2
Use the Student's t distribution with
the margin of error
E = tcs1
2
n1
+ s22
n2
df = smaller of n1 !1 or n2 !1
Use the Student's t distribution with
the margin of error
E = tcn1 !1( )s1
2 + n2 !1( )s22
n1 + n2 ! 21n1
+ 1n2
df = n1 + n2 ! 2
Requirements:1. Consider two independent binomial experiments E1 and E2. 2. Let n1 and n2 be the number of trials in E1 and E2.3. Let r1 be the number of success out of n1 trials in E1 and r2 be the number of success out of n2 trials in E2. 4. Let p1 be the probability of success in E1 and p2 be the probability of success in E2.
5. Let p̂1 =r1n1
and p̂2 =r2n2
.
6. Let q̂1 = 1! p̂1 and q̂2 = 1! p̂2.7. The number of trials must be sufficiently large so that all four of the following inequalities are true: np̂1 > 5; nq̂1 > 5; np̂2 > 5; and nq̂2 > 5
Confidence Interval for p1 ! p2
p̂1 ! p̂2( )! zc p̂1q̂1
n1
+ p̂2q̂2
n2
< p1 ! p2 < p̂1 ! p̂2( ) + zc p̂1q̂1
n1
+ p̂2q̂2
n2
E = zcp̂1q̂1
n1
+ p̂2q̂2
n2
c = Confidence level (0 < c <1)zc= Critical value from the Student's t Distribution Chart
Suppose we construct a c% confidence interval for µ1 ! µ2 or p1 ! p2. Then three cases arise:
CASE 1:The c% confidence interval contains only negative values. In this case we conclude that µ1 ! µ2 < 0 or p1 ! p2 < 0, and we are therefore c% confident that µ1 < µ2 or p1 < p2.
CASE 2:The c% confidence interval contains only positive values. In this case we conclude that µ1 ! µ2 > 0 or p1 ! p2 > 0, and we are therefore c% confident that µ1 > µ2 or p1 > p2.
CASE 3:The c% confidence interval contains both positive and negative values. In this case, at the c% confidence level conclude that either µ1 or µ2 (or p1 or p2 ) is larger. However, if we reduce the confidence levelc to a smaller value, then the confidence interval will, in general, be shorter. Another approach is to increase the sample sizes of n1 and n2. This would also tend to make the confidence interval shorter. A shorter confidence interval might put you back into case 1 or case 2.
Sample Questions: 1. The heights (measured in inches) of 35 randomly selected women and 40 randomly
selected men were independently obtained by the student body of a local college to estimate the difference in their mean heights. The women had a mean height of 63.8 inches with a standard deviation of 1.94 inches. The men had a mean height of 70.1 inches with a standard deviation of 2.28 inches. Determine a 90% confidence interval for the difference between mean heights. Write a sentence to explain your results.
a. Perform the calculations using only your calculator.
Diet A 5 14 7 9 11 7 13 14 12 8
Diet B 5 21 16 23 4 16 13 19 9 21
b. Perform the calculations using the formula. 2. Twenty laboratory mice were randomly divided into two groups of ten. Each group was
fed a prescribed diet. At the end of three weeks, the weight gained by each animal was recorded in milligrams. Determine a 99% confidence interval on the difference between the mean weight gains.
a. Perform the calculations using only your calculator.
b. Perform the calculations using the formula. 3. In a recent pre-prom survey, in a random sample of 40 brown-haired individuals, 22
indicated that they used some sort of hair coloring. In a separate random sample of 40 blonde individuals, 28 indicated that they had used some sort of hair coloring. Determine a 95% confidence interval to determine the difference in population proportions of these groups that use hair coloring products.
a. Perform the calculations using only your calculator.
b. Perform the calculations using the formula. Homework: Pages 384 - 394 Exercises # 1 - 31, odd (skip #29) Calculator only: 15, 23, and 25 Both Formula and Calculator: #31 Exercises # 2 - 26, even (skip #28 and 30) Calculator only: 14, 16, and 26
STATTESTS2 ! SampZInt
2 ! SampZInt!7.102, ! 5.498( )x1: 63.8x2: 70.1n1: 35n2: 40
Upper: ! 5.498411964Lower: ! 7.101588036
2 ! SampZIntInpt: Stats"1: 1.94" 2: 2.28x1: 63.8n1: 35x2: 70.1n2: 40C-Level: 90Calculate
63.8 ! 70.1( )!1.6449 1.942
35+ 2.28
2
40< µ1 ! µ2 < 63.8 ! 70.1( ) +1.6449 1.942
35+ 2.28
2
40
!6.3!1.6449 3.763635
+ 5.198440
< µ1 ! µ2 < !6.3+1.6449 3.763635
+ 5.198440
!6.3!1.6449 0.1075 + 0.1300 < µ1 ! µ2 < !6.3+1.6449 0.1075 + 0.1300
!6.3!1.6449 0.2375 < µ1 ! µ2 < !6.3+1.6449 0.2375!6.3!1.6449 0.4873( ) < µ1 ! µ2 < !6.3+1.6449 0.4873( )
!6.3! 0.8016 < µ1 ! µ2 < !6.3+ 0.8016!7.1016 < µ1 ! µ2 < !5.4984
Answers to the Sample Questions: 1. The heights (measured in inches) of 35 randomly selected women and 40 randomly
selected men were independently obtained by the student body of a local college to estimate the difference in their mean heights. The women had a mean height of 63.8 inches with a standard deviation of 1.94 inches. The men had a mean height of 70.1 inches with a standard deviation of 2.28 inches. Determine a 90% confidence interval for the difference between mean heights. Write a sentence to explain your results.
a. Perform the calculations using only your calculator.
We can say with a 90% confidence level that the difference in population means heights of the women and men at the local college is ! 7.1016 inches and ! 5.4984 inches.
b. Perform the calculations using the formula.
We can say with 90% confidence level that the difference in population means heights of the women and men at the local college is ! 7.1016 inches and ! 5.4984 inches.
Diet A 5 14 7 9 11 7 13 14 12 8
Diet B 5 21 16 23 4 16 13 19 9 21
Diet A x = 10 s = 3.2318 n = 10Diet B x = 14.7 s = 6.7831 n = 10
STATTESTS2 ! SampTInt
2 ! SampTInt!11.87, 2.4679( )df = 12.28857x1: 10x2: 14.7Sx1: 3.2318Sx2: 6.7831n1: 10n2: 10
Upper: 2.467852654Lower: !11.86785265
2 ! SampTIntInpt: DataList1: L1
List2: L3
Freq1: 1Freq2: 1C-Level: 99Pooled: NoCalculate
10 !14.7( )! 3.2498 3.23182
10+ 6.7831
2
10< µ1 ! µ2 < 10 !14.7( ) + 3.2498 3.23182
10+ 6.7831
2
10
!4.7 ! 3.2498 10.444410
+ 46.011110
< µ1 ! µ2 < !4.7 + 3.2498 10.444410
+ 40.011110
!4.7 ! 3.2498 1.0444 + 4.6011 < µ1 ! µ2 < !4.7 + 3.2498 1.0444 + 4.6011
!4.7 ! 3.2498 5.6456 < µ1 ! µ2 < !4.7 + 3.2498 5.6456!4.7 ! 3.2498 2.3760( ) < µ1 ! µ2 < !4.7 + 3.2498 2.3760( )
!4.7 ! 7.7216 < µ1 ! µ2 < !4.7 + 7.7216!12.4216 < µ1 ! µ2 < 3.0216
df = 10 !1 t0.99, df=9 = 3.2498df = 9
2. Twenty laboratory mice were randomly divided into two groups of ten. Each group was fed a prescribed diet. At the end of three weeks, the weight gained by each animal was recorded in milligrams. Determine a 99% confidence interval on the difference between the mean weight gains.
a. Perform the calculations using only your calculator.
We can say with a 99% confidence level that the difference in population means for diet A and diet B is between!11.8679 milligrams and 2.4679 milligrams.
b. Perform the calculations using the formula.
We can say with a 99% confidence level that the difference in population means for diet A and diet B is between !12.4216 milligrams and 3.0216 milligrams.
p̂1 =2240
q̂1 =1840
p̂2 =2840
q̂2 =1240
p̂1 = 0.55 q̂1 = 0.45 p̂2 = 0.70 q̂2 = 0.30
2 ! PropZIntx1: 22n1: 40x2: 28n2: 40C-Level: 95Calculate
STATTESTS2 ! PropZInt
2 ! PropZInt!0.3596, 0.0596( )p̂1= 0.55p̂2 = 0.70n1: 40n2: 40
Upper: 0.0596107949Lower: ! 0.3596107949
0.55 ! 0.70( )!1.9600 0.55( ) 0.45( )40
+0.7( ) 0.3( )40
< p1 ! p2 < 0.55 ! 0.70( ) +1.9600 0.55( ) 0.45( )40
+0.7( ) 0.3( )40
!0.15 !1.9600 0.247540
+ 0.2140
< p1 ! p2 < !0.15 +1.9600 0.247540
+ 0.2140
!0.15 !1.9600 0.006188 + 0.00525 < p1 ! p2 < !0.15 +1.9600 0.006188 + 0.00525
!0.15 !1.9600 0.01144 < p1 ! p2 < !0.15 +1.9600 0.01144!0.15 !1.9600 0.1069( ) < p1 ! p2 < !0.15 +1.9600 0.1069( )
!0.15 ! 0.2096 < p1 ! p2 < !0.15 + 0.2096!0.3596 < p1 ! p2 < 0.0596
3. In a recent pre-prom survey, in a random sample of 40 brown-haired individuals, 22 indicated that they used some sort of hair coloring. In a separate random sample of 40 blonde individuals, 28 indicated that they had used some sort of hair coloring. Determine a 95% confidence interval to determine the difference in population proportions of these groups that use hair coloring products.
a. Perform the calculations using only your calculator.
We can say with a 95% confidence level that the difference in population proportions of brown-haired individuals and blonde individuals using some sort of hair coloring is between ! 0.3596 and 0.0596.
b. Perform the calculations using the formula.
We can say with a 95% confidence level that the difference in population proportions of brown-haired individuals and blonde individuals using some sort of hair coloring is between !0.3596 and 0.0596.
