people.highline.edu · i • ~ . • two cities, bradford and kane are separated only by the...
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• Two cities, Bradford and Kane are separated only by the Conewango River • The local paper recently reported that the mean household income in Bradford is $38,010 from a sample
of 40 households. The population standard deviation (past data) is $6,000. r • The same article reported the mean income in Kane is $35,006 from a sample of 35 households. The
population standard deviation (past data) is $7,000. • From the sample data, create a 95% Confidence Interval to estimate population difference.
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• Two cities, Bradford and Kane are separated only by the Conewango River • The local paper recently reported that the mean household income in Bradford is $38,010 from a sample
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population standard deviation (past data)j s $7,000. M or~ > • At the .05 significance level can we conclude the.mean income in Bradford is..more?
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• Samples are Independent, Distributions for samples are normal
• At the 95% Confidence Level, construct a Confidence Interval for the Population Difference between Imported Auto MPG and Domestic Auto MPG
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• A sample of 20 imported compact cars revealed a mean of 36.52 MPG with a standard deviation of 3.43 MPG
• Assume: 1} Samples are random and independent, 2) Distributions for samples are normal.
• At the .05 significance level can the EPA conclude that the MPG is higher on the imported cars?
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Step 1: List Null and Alternative Hypotheses 1C F"lt'p
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Step 3: Sample, Calculate, Draw Picture, Calculate Test Statistic
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A B c D E
Ho : ~1 = ~2 = ~3 = ~4
Ha = Not all population means are the same (at least two are different)
Decision Rule: p-value < alpha, Reject HO & Accept Ha or Test Statistic F >Critical F,
Reject HO & Accept Ha, otherwise we fail to Reject HO r-~------~---------r----------r---------~
Treatment 1 Treatment 2 Treatment 3 Treatment 4 I
Method 1 Method 2 Method 3 Method 4
12.56 11.87 15.57 12.94
15.92 12.94 10.64 12.01
10.42 11.23 14.05 10.21
14.41 15.97 13.08 10.18
12 10.32 17.14 16.71 11.21
13 Sample Size = n = 5 5 5 5
14 Sample Size= Xbar = 12.726 13.83 14.01 11.31
15 Sample Variance = s2 = 6.042.88 6.73235 5.48675 1.41095
16
17 Grand Over All Count= nT = 20
18 Grand Over All Sample Mean = XbarG = 12.969 Check:!~... __ 1_2_.9_6.....~91 19 #Treatments = k = 4
20 Calculations for Total Variance for All Observations=
Check: 101.87338 101.87338 21 Sum of Squares Total =SST= 101.87338 ~----------~--------~
22 dfT = 19
23 Calculations for Variance Between Treatments=
24 Sum of Squares of Treatments = SSTR =
25 dfTR =
Estimate of Variance Between Treatments
26 = MSTR = SSTR/dfTR
27 Calculations for Variance Within Treatments=
28 Sum of Squares Error= SSE =
Estimate of Variance Within Treatments=
23.18166
3
7.72722
78.69172
16
30 MSE = SSE/dfE 4.9182325
31 F Test Statistic= MSTR/MSE = 1.571137599
32 Alpha = 0.05
33 p-value = 0.235295845
34 F Critical 3.238871517
35
Check: ~...1 __ 7_8._6_91_7.....~21
Because the F Calculated Test Statistic is not past our F Critical, We Fail To Reject HO.
37 It seems reasonable to assume that the pop means are the same.
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