week 9 october 27-31

Week 9 October 27-31

Four Mini-Lectures QMM 510Fall 2014

10-2

Two-Sample Hypothesis Tests

Chapter Contents10.1 Two-Sample Tests

10.2 Comparing Two Means: Independent Samples

10.3 Confidence Interval for the Difference of Two Means, 1 2

10.4 Comparing Two Means: Paired Samples

10.5 Comparing Two Proportions

10.6 Confidence Interval for the Difference of Two Proportions, 1 2

10.7 Comparing Two Variances

Chapter 10

So many topics, so little time …

10-3

• A two-sample test compares two sample estimates with each other.

• A one-sample test compares a sample estimate to a nonsample benchmark.

What Is a Two-Sample Test

Basis of Two-Sample Tests• Two-sample tests are especially useful because they possess

a built-in point of comparison.

• The logic of two-sample tests is based on the fact that two samples drawn from the same population may yield different estimates of a parameter due to chance.

Chapter 10

Two-Sample Tests

10-4

If the two sample statistics differ by more than the amount attributable to chance, then we conclude that the samples came from populations with different parameter values.

What Is a Two-Sample Test

Chapter 10

Two-Sample Tests

10-5

• The hypotheses for comparing two independent population means µ1 and µ2 are:

Format of Hypotheses

Chapter 10

Comparing Two Means: ML 9.1Independent Samples

10-6

• When the population variances 12 and 2

2 are known, use the normal distribution for the test (assuming a normal population).

• The test statistic is:

Case 1: Known Variances

Chapter 10

Comparing Two Means: Independent Samples

10-7

• If the variances are unknown, they must be estimated and the Student’s t distribution used to test the means.

• Assuming the population variances are equal, s12 and s2

2 can be used to estimate a common pooled variance sp

2.

Case 2: Unknown Variances, Assumed Equal

Chapter 10


10-8

• If the population variances cannot be assumed equal, the distribution of the random variable is uncertain (Behrens-Fisher problem)..

• The Welch-Satterthwaite test addresses this difficulty by estimating each variance separately and then adjusting the degrees of freedom.

A quick rule for degrees of freedom is to use min(n1 – 1, n2 – 1). You will get smaller d.f. but avoid the tedious formula above.

Chapter 10


Case 3: Unknown Variances, Assumed Unqual

1 2x x

10-9

• If the population variances 12 and 2

2 are known, then use the normal distribution. Of course, we rarely know 1

2 and 22 .

• If population variances are unknown and estimated using s12 and s2

2, then use the Student’s t distribution (Case 2 or Case 3)

• If you are testing for zero difference of means (H0: µ1−µ2 = 0) the formulas are simplified to:

Test Statistic

Chapter 10


10-10

• If the sample sizes are equal, the Case 2 and Case 3 test statistics will be identical, although the degrees of freedom may differ and therefore the p-values may differ.

• If the variances are similar, the two tests will usually agree.• If no information about the population variances is available, then the

best choice is Case 3.• The fewer assumptions, the better.

Which Assumption Is Best?

Chapter 10

Must Sample Sizes Be Equal?• Unequal sample sizes are common and the formulas still apply.


10-11

Large Samples• If both samples are large (n1 30 and n2 30) and the population is

not badly skewed, it is reasonable to assume normality for the difference in sample means and use Appendix C.

Chapter 10

• Assuming normality makes the test easier. However, it is not conservative to replace t with z.

• Excel does the calculations, so we should use t whenever population variances are unknown (i.e., almost always).


10-12

Three Caveats:

• In small samples, the mean may not be a reliable indicator of central tendency and the t-test will lack power.

• In large samples, a small difference in means could be “significant” but may lack practical importance.

Chapter 10


• Are the populations severely skewed? Are there outliers? Check using histograms and/or dot plots of each sample. t tests are OK if moderately skewed, while outliers are more serious.

10-13

Example: Order Size

Chapter 10

Summary statistics in 8 spreadsheet cells and use MegaStat:

Hypothesis Test: Independent Groups (t-test, pooled variance)Friday Saturday22.32 25.56 mean4.35 6.16 std. dev.

13 18 n

29 df-3.24000 difference (Friday - Saturday)30.07397 pooled variance5.48397 pooled std. dev.1.99604 standard error of difference

0 hypothesized difference

-1.623 t.1154 p-value (two-tailed)

Hypothesis Test: Independent Groups (t-test, unequal variance)Friday Saturday22.32 25.56 mean4.35 6.16 std. dev.

13 18 n

28 df-3.24000 difference (Friday - Saturday)1.88777 standard error of difference

0 hypothesized difference

-1.716 t.0972 p-value (two-tailed)

Friday Saturday22.32 25.56

4.35 6.1613 18

Assuming either Case 2 or Case 3, we would not reject H0 at α = .05 (because the p-value exceeds .05)

H0: μ1 = μ2

H0: μ1 ≠ μ2

Are the means equal? Test the hypotheses:


10-14

Paired Data• Data occur in matched pairs when the same item is observed

twice but under different circumstances.• For example, blood pressure is taken before and after a

treatment is given.• Paired data are typically displayed in columns.

Chapter 10

Comparing Two Means: ML 9.2Paired Samples

10-15

Paired t Test• Paired data typically come from a before/after experiment.• In the paired t test, the difference between x1 and x2 is

measured as d = x1 – x2

• The mean and standard deviation for the differences d are:

• The test statistic becomes just a one-sample t-test.

Chapter 10

Comparing Two Means: Paired Samples

10-16

• Step 1: State the hypotheses. For example:H0: µd = 0H1: µd ≠ 0

• Step 2: Specify the decision rule. Choose (the level of significance) and determine the critical values from Appendix D or with use of Excel.

• Step 3: Calculate the test statistic t.

• Step 4: Make the decision. Reject H0 if the test statistic falls in the rejection region(s) as defined by the critical values.

Steps in Testing Paired Data

Chapter 10


10-17

A two-tailed test for a zero difference is equivalent to asking whether the confidence interval for the true mean difference µd includes zero.

Analogy to Confidence Interval

Chapter 10


10-18

Example: Exam Scores

Chapter 10


Hypothesis Test: Paired Observations0.000 hypothesized value

84.000 mean Post-Test81.833 mean Pre-Test2.167 mean difference (Post-Test - Pre-Test)5.345 std. dev.2.182 std. error

6 n5 df

0.993 t.1832 p-value (one-tailed, upper)

-3.442 confidence interval 95.% lower7.776 confidence interval 95.% upper5.609 margin of error

confidence interval includes zero

=T.DIST.RT(0.9930,5)calc2.1667

/ (5.3448) / 6d

dts n

Name Post-Test Pre-Test DiffCecil 85 79 6David 97 87 10Edward 81 78 3Fred 77 82 -5Gary 96 96 0Henry 68 69 -1

2.16675.3448

H 0: μ d = 0 (no change in mean) t calc 0.9930

H 1: μ d > 0 (improved mean score) t .05 2.015p -value 0.1832

Mean differenceSt dev of differences

Right-tailed test to see if mean scores improved

Do not reject H 0 because t calc does not exceed t .05 (p > .05).

Using MegaStat:

10-19

To test for equality of two population proportions, 1, 2, use the following hypotheses:

Testing for Zero Difference: 1 2 = 0

Chapter 10

Comparing Two Proportions ML 9.3

10-20

The sample proportion p1 is a point estimate of 1 and p2 is a point estimate of 2:

Chapter 10

Sample Proportions

Comparing Two Proportions


10-21

If H0 is true, there is no difference between 1 and 2, so the samples are pooled (or averaged) in order to estimate the common population proportion.

Pooled Proportion

Chapter 10



10-22

• If the samples are large, p1 – p2 may be assumed normally distributed.• The test statistic is the difference of the sample proportions divided

by the standard error of the difference.• The standard error is calculated by using the pooled proportion.• The test statistic for the hypothesis 1 2 = 0 is:

Test Statistic

Chapter 10 Testing for Zero Difference: 1 2 = 0


10-23

Example: Hurricanes

Chapter 10


1 21 2

1 2

19 45=.4130 .642946 70

x xp pn n

1 2

1 2

19 45 64 .551746 70 116

x xpn n

2.435

Hypothesis test for two independent proportionsp1 p2 p c

0.413 0.6429 0.5517 p (as decimal)19/46 45/70 64/116 p (as fraction)

19. 45. 64. X46 70 116 n

-0.2298 difference0. hypothesized difference

0.0944 std. error-2.435 z.01491 p-value (two-tailed)

-0.468 confidence interval 99.% lower0.0084 confidence interval 99.% upper0.2382 margin of error

… or using MegaStat:

=2*NORM.S.DIST(-2.435,1)

10-24

• We have assumed a normal distribution for the statistic p1 – p2.

• This assumption can be checked.

• For a test of two proportions, the criterion for normality is n 10 and n(1 − ) 10 for each sample, using each sample proportion in place of .

• If either sample proportion is not normal, their difference cannot safely be assumed normal.

• The sample size rule of thumb is equivalent to requiring that each sample contains at least 10 “successes” and at least 10 “failures.”

Chapter 10

Checking for Normality



10-25

Chapter 10 Testing for Nonzero Difference


10-26

We may need to test whether two population variances are equal.

Format of Hypotheses

Chapter 10

Comparing Two Variances ML 9.4

10-27

• The test statistic is the ratio of the sample variances:

The F Test

• If the variances are equal, this ratio should be near unity: F = 1.

Chapter 10

Comparing Two Variances

10-28

• If the test statistic is far below 1 or above 1, we would reject the hypothesis of equal population variances.

• The numerator s12 has degrees of freedom df1 = n1 – 1 and the denominator

s22 has degrees of freedom df2 = n2 – 1.

• The F distribution is skewed with mean > 1 and mode < 1.

The F Test

Chapter 10


Example: 5% right-tailed area for F11,8

10-29

• For a two-tailed test, critical values for the F test are denoted FL (left tail) and FR (right tail).

• A right-tail critical value FR may be found from Appendix F using df1 and df2 degrees of freedom.

FR = Fdf1, df2

• A left-tail critical value FL may be found by reversing the numerator and denominator degrees of freedom, finding the critical value from Appendix F and taking its reciprocal:

FL = 1/Fdf2, df1

F Test: Critical Values

Chapter 10


Excel function is:=F.INV.RT(α, df1, df2)

Excel function is:=F.INV(α, df1, df2)

10-30

• Step 1: State the hypotheses:H0: 1

2 = 22

H1: 12 ≠ 2

2

• Step 2: Specify the decision rule.Degrees of freedom are:

Numerator: df1 = n1 – 1 Denominator: df2 = n2 – 1

Choose α and find the left-tail and right-tail critical values from Appendix F or from Excel.

Two-Tailed F-Test:

Chapter 10


• Step 3: Calculate the test statistic.

• Step 4: Make the decision. Reject H0 if the test statistic falls in the rejection regions as defined by the critical values.

10-31

One -Tailed F-Test

• Step 1: State the hypotheses. For example: H0: 1

2 22

H1: 12 < 2

2

• Step 2: State the decision rule. Degrees of freedom are:Numerator: df1 = n1 – 1 Denominator: df2 = n2 – 1

Choose α and find the critical value from Appendix F or Excel.

Chapter 10


• Step 3: Calculate the test statistic.

• Step 4: Make the decision. Reject H0 if the test statistic falls in the rejection region as defined by the critical value.

Example: 5% left-tailed area for F11,8

=F.INV(0.05,11,8)

10-32

EXCEL’s F Test

Chapter 10


Note: Excel uses a left-tailed test if s1

2 < s22

So, if you want a two-tailed test, you must double Excel’s one-tailed p-value.

Conversely, Excel uses a right-tailed test if s1

2 > s22

10-33

Assumptions of the F Test

• The F test assumes that the populations being sampled are normal. It is sensitive to nonnormality of the sampled populations.

• MINITAB reports both the F test and a robust alternative called Levene’s test along with its p-values.

Chapter 10


week 9 october 27-31

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