Sampling Distributions:(6.4)
1) Sampling Distributions of sample Mean 𝑥̅
2) Sampling Distribution of the sample Proportion 𝑝̂ :
3) Sampling Distribution of the two sample means 𝑥̅1 − 𝑥̅2 ‫المحاضره االضافيه‬
4) Sampling Distribution of the two sample Proportions 𝑝̂1 − 𝑝̂2 ‫المحاضره‬
‫االضافيه‬
(ask about probability of sample statistics x̅ , p̂ , x̅1 − x̅2 , p̂1 − p̂2 )and
give information about population parameters
Steps to answer:
 Compute means (𝜇𝑥̅ , 𝜇𝑥̅1−𝑥̅2 , 𝜇𝑝̂ , 𝜇𝑝̂1−𝑝̂2 )
 Compute variance (𝜎 2 𝑥̅ , 𝜎 2 𝑥̅1−𝑥̅2 , 𝜎 2 𝑝̂ , 𝜎 2 𝑝̂1−𝑝̂2 )
 Compute standard deviation or "sd" Standard error s.e "(𝜎𝑥̅ , 𝜎𝑥̅1−𝑥̅2 ,
𝜎𝑝̂ , 𝜎𝑝̂1−𝑝̂2 )

Use 𝑧
=
𝑣𝑎𝑙𝑢𝑒−𝑚𝑒𝑎𝑛
𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑒𝑟𝑟𝑜𝑟
,(Standard deviation = Standard error)
Symbol
sample
mean
variance
Standard deviation
proportion
𝑥̅
𝑠2
s
𝑝̂
population
µ
𝝈𝟐
σ
p
Estimate population parameters
 point estimate
 interval estimate
point estimate:
Point estimate for
Point estimate for
Point estimate for
Point estimate for
Point estimate for
(µ ) is 𝑥̅
(σ ) is S
̂
(p ) is 𝒑
(µ𝟏 -𝝁𝟐 ) is 𝑥̅1 − 𝑥̅2
(𝒑𝟏 -𝒑𝟐 ) is 𝑝̂1 − 𝑝̂2
Interval estimate(ch6)
1)
2)
3)
4)
5)
interval for population mean µ
interval for two population means µ𝟏 -𝝁𝟐 (not related)
interval for population proportion p
interval for two population proportions 𝒑𝟏 -𝒑𝟐
interval for two population means µ𝟏 -𝝁𝟐 (related or paired)(in
chapter 7)
(ask about population parameters µ , µ𝟏 -𝝁𝟐 , p , 𝒑𝟏 -𝒑𝟐 and give
information about sample statistics )
The general formula:
point estimate ± (table value(z or t)) × √
𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒
𝑛
**************************************************
𝑠𝑝2 = 𝑒𝑠𝑡𝑖𝑚𝑎𝑡𝑒 𝑝𝑜𝑜𝑙𝑒𝑑 𝑐𝑜𝑚𝑚𝑜𝑛 𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒
𝑠𝑝 = 𝑒𝑠𝑡𝑖𝑚𝑎𝑡𝑒 𝑝𝑜𝑜𝑙𝑒𝑑 𝑐𝑜𝑚𝑚𝑜𝑛 𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛
𝑧1.𝛼 , 𝑡1.𝛼 = reliability coefficient , 𝑡𝑎𝑏𝑙𝑒 𝑣𝑎𝑙𝑢𝑒
2
2
𝑝̅= pooled estimate proportion
Test Hypotheses (ch7)
1) test for population mean µ
2) test for two population means µ𝟏 -𝝁𝟐 (not related)
note: degree of freedom for T is (𝑑𝑓 = 𝑛1 + 𝑛2 − 2)
(when use T-test)
3) test for population proportion p
4) test for two population proportions 𝒑𝟏 -𝒑𝟐
5) test for two population means µ𝟏 -𝝁𝟐 (related or paired)
note: degree of freedom for T is (𝑑𝑓 = 𝑛 − 1)
ask about population parameters µ , µ𝟏 -𝝁𝟐 , p , 𝒑𝟏 -𝒑𝟐 and give
information about sample statistics
Steps
1)
2)
3)
4)
5)
6)
data
assumptions
hypotheses
test statistic
decision
conclusion
‫الجداول من التعديل الذي أرسلناه لكم‬
Note :
(1) example 7.3.1
p_value= 0.102 (‫)خطأ في الملزمه‬
:‫الصحيح‬
p_value= 0.012
(2) Example 7.3.2
P_value=-1.33( ‫)خطأ في الملزمه‬
‫الصحيح‬
P_value=0.716
Paired Sample
1) Confidence interval
2) test hypotheses
1)Confidence interval(Paired or related population)
Use
̅±𝑡 𝛼
𝐷
1− ,𝑛−1
2
𝑠𝐷
√𝑛
(df=n-1)
And
𝑛
̅ = ∑𝑖=1 𝐷𝑖
𝐷
𝑛
𝑠𝐷2 =
𝑠𝐷 = √𝑠𝑑2
( mean of difference)
̅ 2
∑𝑛
𝑖=1(𝐷𝑖 −𝐷 )
𝑛−1
(variance of difference)
(𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒)
2)Test hypotheses(Paired or related population)
1) data
2)Assumption: normal + paired
3)Hypotheses:
we have three cases
Case I : H0: μ 1 = μ2 → μ 1 - μ2 = 0 → 𝜇𝑑 = 0
HA: μ 1 ≠ μ 2 → μ 1 - μ 2 ≠ 0 → 𝜇𝑑 ≠ 0
e.g. we want to test that the mean for first
population is different from second population mean.
Case II : H0: μ 1 = μ2 → μ 1 - μ2 = 0 → 𝜇𝑑 = 0
HA: μ 1 > μ 2 → μ 1 - μ 2 > 0 → 𝜇𝑑 > 0
e.g. we want to test that the mean for first
population is greater than second population mean.
Case III : H0: μ 1 = μ2 → μ 1 - μ2 = 0 → 𝜇𝑑 = 0
HA: μ 1 < μ 2
→ μ 1 - μ 2 < 0 → 𝜇𝑑 < 0
e.g. we want to test that the mean for first population is
less than second population
4)𝑇𝑒𝑠𝑡:
𝑇=
̅
𝐷
𝑠𝑑
√𝑛
5)Decision
Reject 𝐻0 if :
Case1:
𝑇𝑐 < −𝑇1−𝛼,𝑛−1 𝑜𝑟
2
Case2:
𝑇𝑐 > 𝑇1−𝛼,𝑛−1
Case3:
𝑇𝑐 < −𝑇1−𝛼,𝑛−1
𝑇𝑐 > 𝑇1−𝛼,𝑛−1
2
Example
Page 146
Solution:
3)Hypotheses:
H0: μ 1 = μ2 → μ 1 - μ2 = 0 → 𝜇𝑑 = 0
HA: μ 1 ≠ μ 2 → μ 1 - μ 2 ≠ 0 → 𝜇𝑑 ≠ 0
̅ , 𝑠𝑑 ‫)من المثال السابق لحساب فترة الثقة‬
4)test statistic ( 𝐷
𝑇=
̅
𝐷
𝑠𝑑
√𝑛
=
5.45
7.09
√10
= 2.43
5) Decision
Reject 𝐻0 if :
𝑇𝑐 < −𝑇1−𝛼,𝑛−1
2
𝑜𝑟
𝑇𝑐 > 𝑇1−𝛼,𝑛−1
2
2.43 < −2.262
0r 2.43 > 2.262 ‫تحقق‬
So Reject 𝐻0 and accept 𝐻1
We can conclude the diet program is effective at 𝛼 = 0.05