two population means hypothesis testing and confidence intervals with known standard deviations
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Two Population MeansTwo Population Means
Hypothesis Testing and Hypothesis Testing and Confidence IntervalsConfidence Intervals
With Known Standard DeviationsWith Known Standard Deviations
SITUATION: 2 PopulationsSITUATION: 2 Populations
Population 1
Mean = 1
St’d Dev. = 1
Population 2
Mean = 2
St’d Dev. = 2
Salaries in Chicago Salaries in St. Louis
Women’s Test Scores Men’s Test Scores
Lakers Attendance Clippers Attendance
Anaheim Sales Irvine Sales
KEY ASSUMPTIONSKEY ASSUMPTIONSSampling is done from two populations.
– Population 1 has mean µ1 and variance σ12.
– Population 2 has mean µ2 and variance σ22.
– A sample of size n1 will be taken from population 1.
– A sample of size n2 will be taken from population 2.
– Sampling is random and both samples are drawn independently.
– Either the sample sizes will be large or the populations are assumed to be normally distribution.
1
21
1
111 n
σ variance,
n
σ deviation standard ,μ mean :X variableRandom
2
22
2
222 n
σ variance,
n
σ deviation standard ,μ mean :X variableRandom
The ProblemThe Problem1 and 2 are unknown
1 and 2 may or may not be known
(In this module we assume they are known.)
OBJECTIVESOBJECTIVES• Test whether 1 > 2 (by a certain amount)
– or whether 1 2
• Determine a confidence interval for the difference in the means: 1 - 2
21 XX
Key Concepts About the Key Concepts About the Random Variable .Random Variable .
• is the difference in two sample means.1. Its meanmean is the difference of the two individual means:
2. If the variables are independent (which we assumed), the variancevariance (not the standard deviation) of the random variable of the differences = the sum (not the difference) of the two variances:
3. Thus its standard deviationstandard deviation is:
4. Its distributiondistribution is:• Normal if σ1 and σ2 are known • t if σ1 and σ2 are unknown
21 XX
2
22
1
21
n
σ
n
σ
21 μμ
2
22
1
21
n
σ
n
σ
Random Variable XDistribution normal normal normal
Mean 1 2
Standard Deviation
21 XX X
n
σ
2
22
1
21
n
σ
n
σ
21 XX ,X X, Variable Random
Hypothesis Test Statistics forHypothesis Test Statistics forDifference in Means, Known Difference in Means, Known σσ’s’s
• We will be performing hypothesis tests with null hypotheses, H0, of the form:
• From the general form of a test statistic, the required test statistic will be:
2
22
1
21
21
nσ
nσ
v)xx(
ErrorStandard
Value) zed(Hypothesi - imate)(Point Estz
HH00: : µµ11 - µ - µ22 = v = v
Confidence Intervals for Confidence Intervals for µµ11 - µ - µ22
Known Known σσ’s’s• Recall the general form of a confidence interval is:
Thus when the σ’s are known this becomes:
(Point Estimate) ± zα/2(Appropriate Standard Error)
2
22
1
21
α/221 n
σ
n
σzxx
EXAMPLEEXAMPLEHypothesis Test: Hypothesis Test: 11, , 2 2 KnownKnown
Test whether starting salaries for secretaries in Chicago are at least $5 more per week than those in St. Louis.
GIVEN:Salaries assumed to be normal
Standard Deviations known: Chicago $10; St. Louis $15
Sample ResultsSampled 100 secretaries in Chicago; 75 secretaries in St. Louis
Sample averages: Chicago - $550, St. Louis -$540
Hypothesis TestHypothesis Test
H0: 1 - 2 = 5
HA: 1 - 2 > 5
Use = .05
Reject H0 (Accept HA) if z > z.05 = 1.645
Calculating zCalculating zRemember
Deviation Standard eAppropriat
Value) zed(HypothesiEstimate)(Point z
2
22
1
21
21
nσ
nσ
5)x-x(z
2.5
75
15
100
10
5540)-(550z
22
ConclusionConclusion
• Since 2.5 > 1.645– It can be concluded that the average starting salary for
secretaries in Chicago is at least $5 per week greater than the average starting salary in St. Louis.
• The p-value:– The area above z= 2.5 on the normal curve = 1 - .9938
= .0062– Since .0062 is low (compared to α), it can be concluded
that the average starting salary for secretaries in Chicago is at least $5 per week greater than the average starting salary in St. Louis.
EXAMPLEEXAMPLEConfidence Interval: Confidence Interval: 11, , 2 2 KnownKnown
• Construct a 95% confidence for the difference in average between weekly starting salaries for secretaries in Chicago and St. Louis.
2
22
1
21
.02521 n
σ
n
σz )x-x(
75
15
100
101.96 540)-(550
22
$10 ± $3.92
$6.08 ↔ $13.92
Excel ApproachExcel Approach
• Suppose, as shown on the next slide the data for Chicago is given in column A (A2:A101) and the data for St. Louis is given in column B (B2:B76).
• The analysis can be done using an entry from the Data Analysis Menu:
z-test: Two Sample for Meansz-test: Two Sample for Means
Select
z-Test: Two Sample For Means
Go to Data
Then Data Analysis
Select
z-Test: Two Sample For Means
For 1-tail tests, input columns so that the test is
a “>” test.Enter
Hypothesized Difference
Enter Variances
Not
Standard Deviations
Check
Labels
Enter
Beginning Cell
For Output
=(E4-F4)-NORMSINV(0.975)*SQRT(E5/E6+F5/F6)
Highlight formula in cell E15—press F4.
Drag to cell E16 and change “-” to “+”.
Estimating Sample SizesEstimating Sample Sizes• Usual Assumptions:
– Same sample size from each pop.: n1 = n2 = n
– Standard deviations, 1 2 known
• Calculate n from the “±” part of the confidence interval for known 1and 2
En
σσ1.96
n
σ
n
σ1.96
22
21
22
21
ExampleExample• How many workers would have to be
surveyed in Chicago and St. Louis to estimate the true average difference in starting weekly salary to within $3?
surveyd! be would total workers278
cityeach in 139138.72(11.78)n
78.113
3251.96 n
3n
15101.96
n
σ
n
σ1.96
2
2222
21
ReviewReview• Mean and standard deviation for X1 -X2
• Assumptions for tests and confidence intervals
• z-tests for differences in means when 1 and 2 are known: – By Formula– By Excel data analysis tool
• Confidence intervals for differences in means when 1 and 2 are known:
– By Formula– By Excel data analysis tool
• Estimating Sample Sizes
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