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Section 9.4 Inferences About Two Means (Matched Pairs). Objective Compare of two matched-paired means using two samples from each population. Hypothesis Tests and Confidence Intervals of two dependent means use the t -distribution. Definition.

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section 9 4 inferences about two means matched pairs
Section 9.4Inferences About Two Means(Matched Pairs)

Objective

Compare of two matched-paired means using two samples from each population.

Hypothesis Tests and Confidence Intervals of two dependent means use the t-distribution

definition
Definition

Two samples are dependentif there is some relationship between the two samples so that each value in one sample is paired with a corresponding value in the other sample.

Two samples can be treated as the matched pairs of values.

examples
Examples
  • Blood pressure of patients before they are given medicine and after they take it.
  • Predicted temperature (by Weather Forecast) and the actual temperature.
  • Heights of selected people in the morning and their heights by night time.
  • Test scores of selected students in Calculus-I and their scores in Calculus-II.
slide4

Example 1

First sample: weights of 5 students in April

Second sample: their weights in September

These weights make 5 matched pairs

Third line: differences between April weights and September weights (net change in weight for each student, separately)

In our calculations we only use differences (d), not the values in the two samples.

notation
dIndividual difference between two matched paired values

μdPopulation mean for the difference of the two values.

nNumber of paired values in sample

dMean value of the differences in sample

sdStandard deviation of differences in sample

Notation
requirements
(1) The sample data are dependent (i.e. they make matched pairs)

(2) Either or both the following holds:

The number of matched pairs is large (n>30)

or The differences have a normal distribution

Requirements

All requirements must be satisfied to make a Hypothesis Test or to find a Confidence Interval

tests for two dependent means
Tests for Two Dependent Means

Goal: Compare the mean of the differences

H0:μd=0

H1:μd≠0

H0:μd=0

H1:μd<0

H0:μd=0

H1:μd>0

Two tailed

Left tailed

Right tailed

finding the test statistic

d –µd

t=

sd

n

degrees of freedom: df = n – 1

Finding the Test Statistic

Note: md= 0according to H0

test statistic
Test Statistic

Degrees of freedomdf= n – 1

Note: Hypothesis Tests are done in same way as in Ch.8-5

steps for performing a hypothesis test on two independent means
Steps for Performing a Hypothesis Test on Two Independent Means
  • Write what we know
  • State H0 and H1
  • Draw a diagram
  • Calculate the Sample Stats
  • Find the Test Statistic
  • Find the Critical Value(s)
  • State the Initial Conclusion and Final Conclusion

Note: Same process as in Chapter 8

slide11

Example 1

Assume the differences in weight form a normal distribution.

Use a 0.05 significance level to test the claim that for the population of students, the meanchangein weight from September to April is 0 kg (i.e. on average, there is no change)

Claim: μd = 0 using α = 0.05

slide12

d Data: -1 -1 4 -2 1

Example 1

H0:µd=0

H1:µd≠0

t-dist.

df = 4

Two-Tailed

H0 =Claim

t = 0.186

-tα/2 = -2.78

tα/2 = 2.78

Sample Stats

n = 5 d = 0.2sd = 2.387

Use StatCrunch: Stat – Summary Stats – Columns

Test Statistic

Critical Value

tα/2 = t0.025 = 2.78 (Using StatCrunch, df = 4)

Initial Conclusion:Since t is not in the critical region, accept H0

Final Conclusion: We accept the claim that mean change in weight from September to April is 0 kg.

slide13

d Data: -1 -1 4 -2 1

Example 1

Sample Stats

H0:µd=0

H1:µd≠0

n = 5 d = 0.2sd = 2.387

Use StatCrunch: Stat – Summary Stats – Columns

Two-Tailed

H0 =Claim

Stat → T statistics→ One sample → With summary

● Hypothesis Test

Sample mean:

Sample std. dev.:

Sample size:

0.2

2.387

5

Null: proportion=

Alternative

0

P-value = 0.8605

Initial Conclusion:Since P-value is greater than α (0.05), accept H0

Final Conclusion: We accept the claim that mean change in weight from September to April is 0 kg.

confidence interval estimate
Confidence Interval Estimate

We can observe how the two proportions relate by looking at the Confidence Interval Estimate of μ1–μ2

CI = (d – E,d + E)

slide15

Find the 95% Confidence Interval Estimate of μd from the data in Example 1

Example 2

Sample Stats

n = 5 d = 0.2 sd = 2.387

tα/2 = t0.025 = 2.78 (Using StatCrunch, df = 4)

CI =(-2.8, 3.2)

slide16

Find the 95% Confidence Interval Estimate of μd from the data in Example 1

Example 2

Sample Stats

n = 5 d = 0.2 sd = 2.387

Stat → T statistics→ One sample → With summary

● Confidence Interval

Sample mean:

Sample std. dev.:

Sample size:

0.2

2.387

5

Level:

0.95

CI =(-2.8, 3.2)