1 Objective Compare of two matched-paired means using two samples from each population. Hypothesis Tests and Confidence Intervals of two dependent means.

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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

Section 9.4Inferences About Two Means

(Matched Pairs)

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Definition

Two samples are dependent if 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.

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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.

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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.

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Notation

d Individual difference between two matched paired values

μd Population mean for the difference of the two values.

n Number of paired values in sample

d Mean value of the differences in sample

sd Standard deviation of differences in sample

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(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) orThe differences have a normal distribution

Requirements

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

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Tests for Two Dependent Means

Goal: Compare the mean of the differences

H0 : μd =

0

H1 : μd ≠ 0

Two tailed Left tailed Right tailed

H0 : μd =

0

H1 : μd < 0

H0 : μd =

0

H1 : μd > 0

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t = d – µdsdn

degrees of freedom: df = n – 1

Note: d

= 0 according to H0

Finding the Test Statistic

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Test Statistic

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

Degrees of freedom df = n – 1

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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

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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 mean change in weight from September to April is 0 kg (i.e. on average, there is no change)

Claim: μd = 0 using α = 0.05

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H0 : µd = 0

H1 : µd ≠ 0t = 0.186

tα/2 = 2.78

t-dist.df = 4

Test Statistic

Critical Value

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.

-tα/2 = -2.78

Example 1

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

d Data: -1 -1 4 -2 1

Sample Stats

n = 5 d = 0.2 sd = 2.387

Use StatCrunch: Stat – Summary Stats – Columns

Two-TailedH0 = Claim

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H0 : µd = 0

H1 : µd ≠ 0Two-TailedH0 = Claim

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.

Example 1 d Data: -1 -1 4 -2 1Sample Stats

n = 5 d = 0.2 sd = 2.387

Use StatCrunch: Stat – Summary Stats – Columns

Null: proportion=

Alternative

Sample mean:

Sample std. dev.:

Sample size:

● Hypothesis Test0.2

2.387

5

0

≠

P-value = 0.8605

Stat → T statistics→ One sample → With summary

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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 )

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Example 2 Find the 95% Confidence Interval Estimate of μd from the data in Example 1

Sample Stats

n = 5 d = 0.2 sd = 2.387

CI = (-2.8, 3.2)

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

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Example 2 Find the 95% Confidence Interval Estimate of μd from the data in Example 1

Sample Stats

n = 5 d = 0.2 sd = 2.387

Level:Sample mean:

Sample std. dev.:

Sample size:

● Confidence Interval0.2

2.387

5

0.95

Stat → T statistics→ One sample → With summary

CI = (-2.8, 3.2)

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18

Objective

Compare of two population variances using two samples from each population.

Hypothesis Tests and Confidence Intervals of two variances use the F-distribution

Section 9.5Comparing Variation in Two Samples

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(1) The two populations are independent

(2) The two samples are random samples

(3) The two populations are normally distributed(Very strict!)

Requirements

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

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Important

The first sample must have a larger sample standard deviation s1 than the second sample. i.e. we must have

s1 ≥ s2

If this is not so, i.e. if s1 < s2 , then we will need to switch the indices 1 and 2

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Notationσ1 First population standard deviation

s1 First sample standard deviation

n1 First sample size

σ2 Second population standard deviation

s2 Second sample standard deviation

n2 Second sample size

Note: Use index 1 on sample/population with the larger sample standard deviation (s)

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Tests for Two Proportions

H0 : σ1 = σ2

H1 : σ1 ≠ σ2

Two tailed Right tailed

Note: We do not consider σ1 < σ2

(since we used indexes 1 and 2 such that s1 is larger)

Note: We only test the relation between σ1 and σ2

(not the actual numerical values)

The goal is to compare the two population variances (or standard deviations)

H0 : σ1 = σ2

H1 : σ1 > σ2

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The F-Distribution

Similar to the χ2-dist.

•Not symmetric

•Non-negative values (F ≥ 0)

•Depends on two degrees of freedom

df1 = n1 – 1 (Numerator df )

df2 = n2 – 1 (Denominator df )

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The F-Distribution

On StatCrunch: Stat – Calculators – F

df1 = n1 – 1

df2 = n2 – 1

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s1F = s2

2

2

Test Statistic for Hypothesis Tests with Two Variances

Where s12 is the first (larger) of the two sample variances

Because of this, we will always have F ≥ 1

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If the two populations have equal variances,

then F = s12/s2

2 will be close to 1

(Since s12 and s2

2 will be close in value)

If the two populations have different variances,

then F = s12/s2

2 will be greater than 1

(Since s12 will be larger than s2

2)

Use of the F Distribution

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Conclusions from the F-Distribution

Values of F close to 1 are evidence in favor

of the claim that the two variances are equal.

Large values of F, are evidence against this

claim (i.e. it suggest there is some difference

between the two)

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Steps for Performing a Hypothesis Test on Two Independent Means

• Write what we know

• Index the variables such that s1 ≥ s2 (important!)

• State H0 and H1

• Draw a diagram

• Find the Test Statistic

• Find the two degrees of freedom

• Find the Critical Value(s)

• State the Initial Conclusion and Final Conclusion

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Example 1Below are sample weights (in g) of quarters made before 1964 and weights of quarters made after 1964.

When designing coin vending machines, we must consider the standard deviations of pre-1964 quarters and post-1964 quarters.

Use a 0.05 significance level to test the claim that the weights of pre-1964 quarters and the weights of post-1964 quarters are from populations with the same standard deviation.

Claim: σ1 = σ2 using α = 0.05

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H0 : σ1 = σ2

H1 : σ1 ≠ σ2

Test Statistic

Critical Value

Initial Conclusion: Since F is in the critical region, reject H0

Final Conclusion: We reject the claim that the weights of the pre-1964 and post-1964 quarters have the same standard

deviation

Example 1

Using StatCrunch: Stat – Calculators – F

Fα/2 = F0.025 = 1.891

Two-TailedH0 = Claim

n1 = 40 n2 = 40 α = 0.05 s1 = 0.08700 s2 = 0.06194 (Note: s1≥s2)

2

2

Degrees of Freedomdf1 = n1 – 1 = 39 df2 = n2 – 1 = 39

Fα/2 = 1.891

F = 1.973

F is in the critical region

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H0 : σ1 = σ2

H1 : σ1 ≠ σ2

Initial Conclusion: Since P-value is less than α (0.05), reject H0

Final Conclusion: We reject the claim that the weights of the pre-1964 and post-1964 quarters have the same standard

deviation

Example 1

Two-TailedH0 = Claim

n1 = 40 n2 = 40 α = 0.05 s1 = 0.08700 s2 = 0.06194 (Note: s1≥s2)

Null: variance ratio=

Alternative

Sample 1: Variance:

Size:

Sample 2: Variance:

Size:

● Hypothesis Test0.007569

P-value = 0.0368

Stat → Variance → Two sample → With summary

s12 = 0.007569 s2

2 = 0.003837

0.003837

40

40

1

≠