Chapter 10 10-1

Learning Objectives

In this chapter, you learn how to use hypothesis testing for comparing the difference between:

• The means of two independent populations

• The means of two related populations

• The proportions of two independent Chapter 10 populations

Two-Sample Tests • The variances of two independent populations by testing the ratio of the two variances

Prof. Shuguang Liu Chap 10-1 Prof. Shuguang Liu Chap 10-2

Two-Sample Tests Difference Between Two Means

DCOVA DCOVA

Two-Sample Tests Population means, Goal: Test hypothesis or form independent * a confidence interval for the samples difference between two Population Population population means, µ – µ Means, Means, Population Population 1 2 Proportions Variances Independent Related σ1 and σ2 unknown, Samples Samples assumed equal The point estimate for the difference is Examples: Group 1 vs. Same group Proportion 1 vs. Variance 1 vs. Group 2 before vs. after Proportion 2 Variance 2 X1 – X2 treatment σ1 and σ2 unknown, not assumed equal

Prof. Shuguang Liu Chap 10-3 Prof. Shuguang Liu Chap 10-4

Difference Between Two Means: Hypothesis Tests for Independent Samples Two Population Means

DCOVA DCOVA § Different data sources Two Population Means, Independent Samples Population• Unrelated means, independent• Independent * Lower-tail test: Upper-tail test: Two-tail test: samplesu Sample selected from one population has no effect on the sample selected from the other population H0: µ1 ≥ µ2 H0: µ1 ≤ µ2 H0: µ1 = µ2 H1: µ1 < µ2 H1: µ1 > µ2 H1: µ1 ≠ µ2

Use S to estimate unknown i.e., i.e., i.e., p σ and σ unknown, 1 2 σ. Use a Pooled-Variance t H : µ – µ ≤ 0 H : µ – µ = 0 assumed equal H0: µ1 – µ2 ≥ 0 0 1 2 0 1 2 test. H : µ – µ > 0 H : µ – µ ≠ 0 H1: µ1 – µ2 < 0 1 1 2 1 1 2 Use S and S to estimate σ1 and σ2 unknown, 1 2 not assumed equal unknown σ1 and σ2. Use a Separate-Variance t test. Prof. Shuguang Liu Chap 10-5 Prof. Shuguang Liu Chap 10-6 Chapter 10 10-2

Hypothesis tests for µ1 - µ2 with σ1 and

Hypothesis tests for µ1 – µ2 σ2 unknown and assumed equal

DCOVA DCOVA Two Population Means, Independent Samples Population means, Assumptions: Lower-tail test: Upper-tail test: Two-tail test: independent § Samples are randomly and H : µ – µ 0 H : µ – µ ≤ 0 H : µ – µ = 0 samples 0 1 2 ≥ 0 1 2 0 1 2 independently drawn H1: µ1 – µ2 < 0 H1: µ1 – µ2 > 0 H1: µ1 – µ2 ≠ 0 § Populations are normally σ and σ unknown, α α α/2 α/2 1 2 * distributed or both sample assumed equal sizes are at least 30

-tα tα -tα/2 tα/2 § Population variances are unknown but assumed equal Reject H0 if tSTAT < -tα Reject H0 if tSTAT > tα Reject H0 if tSTAT < -tα/2 σ1 and σ2 unknown, or tSTAT > t /2 α not assumed equal

Prof. Shuguang Liu Chap 10-7 Prof. Shuguang Liu Chap 10-8

Hypothesis tests for µ - µ with σ and 1 2 1 Confidence interval for µ1 - µ2 with σ1 σ2 unknown and assumed equal and σ2 unknown and assumed equal

(continued) DCOVA DCOVA • The pooled variance is: Population means, 2 2 Population means, 2 (n1 − 1)S1 + (n2 − 1)S2 independent S = independent samples p (n 1) (n 1) samples 1 − + 2 − The confidence interval for

• The test statistic is: µ1 – µ2 is: σ and σ unknown, 1 2 * σ1 and σ2 unknown, assumed equal * (X1 − X2 )− (µ1 − µ2 ) assumed equal tSTAT = 2 ⎛ 1 1 ⎞ X1 X2 t S ⎜ ⎟ ( − ) ± α/2 p ⎜ + ⎟ 2 ⎛ 1 1 ⎞ n n S ⎝ 1 2 ⎠ p ⎜ + ⎟ ⎝ n1 n2 ⎠ σ and σ unknown, σ and σ unknown, 1 2 1 2 Where tα/2 has d.f. = n1 + n2 – 2 not assumed equal • Where tSTAT has d.f. = (n1 + n2 – 2) not assumed equal

Prof. Shuguang Liu Chap 10-9 Prof. Shuguang Liu Chap 10-10

Pooled-Variance t Test Example: Pooled-Variance t Test Example Calculating the Test Statistic

You are a financial analyst for a brokerage firm. Is thereDCOV a A (continued) difference in dividend yield between stocks listed on the H0: µ1 - µ2 = 0 i.e. (µ1 = µ2) DCOVA NYSE & NASDAQ? You collect the following data: H1: µ1 - µ2 ≠ 0 i.e. (µ1 ≠ µ2) NYSE NASDAQ Number 21 25 The test statistic is: Sample mean 3.27 2.53 Sample std dev 1.30 1.16 (X1 − X2 )− (µ1 − µ2 ) (3.27 − 2.53)− 0 tSTAT = = = 2.040 ⎛ 1 1 ⎞ ⎛ 1 1 ⎞ S2 ⎜ + ⎟ 1.5021⎜ + ⎟ p ⎜ ⎟ 21 25 Assuming both populations are ⎝ n1 n2 ⎠ ⎝ ⎠ approximately normal with

equal variances, is 2 2 2 2 2 (n1 −1)S1 + (n2 −1)S2 (21 −1)1.30 + (25 −1)1.16 there a difference in mean S = = = 1.5021 P (n −1) + (n −1) (21-1) + (25 −1) yield (α = 0.05)? 1 2

Prof. Shuguang Liu Chap 10-11 Prof. Shuguang Liu Chap 10-12 Chapter 10 10-3

Pooled-Variance t Test Example: Excel Pooled-Variance t test Comparing NYSE & Hypothesis Test Solution NASDAQ

DCOVA H0: µ1 - µ2 = 0 i.e. (µ1 = µ2) DCOVA Reject H0 Reject H0 Pooled-Variance t Test for the Difference Between Two Means H1: µ1 - µ2 ≠ 0 i.e. (µ1 ≠ µ2) (assumes equal populaon variances) Data Hypothesized Difference 0 α = 0.05 Level of Significance 0.05 Decision: .025 .025 Populaon 1 Sample df = 21 + 25 - 2 = 44 Sample Size 21 =COUNT(DATA!$A:$A) Sample Mean 3.27 =AVERAGE(DATA!$A:$A) Reject H0 at α = 0.05 Critical Values: t = ± 2.0154 Sample Standard Deviaon 1.3 =STDEV(DATA!$A:$A) -2.0154 0 2.0154 Populaon 2 Sample t Sample Size 25 =COUNT(DATA!$B:$B) Conclusion: Sample Mean 2.53 =AVERAGE(DATA!$B:$B) Sample Standard Deviaon 1.16 =STDEV(DATA!$B:$B) Test Statistic: 2.040 There is evidence of a Intermediate Calculaons Populaon 1 Sample Degrees of Freedom 20 =B7 - 1 difference in means. Populaon 2 Sample Degrees of Freedom 24 =B11 - 1 Decision: Total Degrees of Freedom 44 =B16 + B17 3.27 − 2.53 Pooled Variance 1.502 =((B16 * B9^2) + (B17 * B13^2)) / B18 Reject H at α = 0.05 Standard Error 0.363 =SQRT(B19 * (1/B7 + 1/B11)) tSTAT = = 2.040 0 Difference in Sample Means 0.74 =B8 - B12 ⎛ 1 1 ⎞ t Test Stasc 2.040 =(B21 - B4) / B20 1.5021 ⎜ + ⎟ Conclusion: Two-Tail Test 21 25 Lower Crical Value -2.015 =-TINV(B5, B18) ⎝ ⎠ There is evidence of a Upper Crical Value 2.015 =TINV(B5, B18) p-value 0.047 =TDIST(ABS(B22),B18,2) Reject the null hypothesis =IF(B27

Prof. Shuguang Liu Chap 10-13 Prof. Shuguang Liu Chap 10-14

Minitab Pooled-Variance t test Comparing NYSE & Pooled-Variance t Test Example: NASDAQ Confidence Interval for µ1 - µ2

DCOVA DCOVA Two-Sample T-Test and CI Since we rejected H0 can we be 95% confident that µNYSE Sample N Mean StDev SE Mean 1 21 3.27 1.30 0.28 > µNASDAQ? 2 25 2.53 1.16 0.23

Difference = mu (1) - mu (2) 95% Confidence Interval for µNYSE - µNASDAQ Estimate for difference: 0.740

95% CI for difference: (0.009, 1.471)

T-Test of difference = 0 (vs not =): T-Value = 2.04 P-Value = 0.047 DF = 44 2 ⎛ 1 1 ⎞ Both use Pooled StDev = 1.2256 X1 − X2 ± t /2 Sp ⎜ + ⎟ = 0.74 ± 2.0154× 0.3628 = (0.009, 1.471) ( ) α ⎜ ⎟ ⎝ n1 n 2 ⎠ Decision:

Reject H0 at α = 0.05 Since 0 is less than the entire interval, we can be 95% Conclusion: confident that µ > µ There is evidence of a NYSE NASDAQ

difference in means. Prof. Shuguang Liu Chap 10-15 Prof. Shuguang Liu Chap 10-16

Hypothesis tests for µ1 - µ2 with σ1 and Hypothesis tests for µ1 - µ2 with σ1 and σ2 unknown, not assumed equal σ2 unknown and not assumed equal

(continued) DCOVA DCOVA The test statistic is: Population means, Assumptions: Population means,

independent independent X1 X2 µ µ § Samples are randomly and ( − )− ( 1 − 2 ) samples samples tSTAT = independently drawn S2 S2 2 1 2 ⎛ S 2 S 2 ⎞ ⎜ 1 + 2 ⎟ + ⎜ n n ⎟ ⎝ 1 2 ⎠ n1 n2 ν = 2 2 § Populations are normally ⎛ S 2 ⎞ ⎛ S 2 ⎞ σ1 and σ2 unknown, ⎜ 1 ⎟ ⎜ 2 ⎟ σ1 and σ2 unknown, ⎜ ⎟ ⎜ ⎟ n1 n2 distributed or both sample assumed equal ⎝ ⎠ + ⎝ ⎠ assumed equal t n1 − 1 hasn2 − 1 d.f. ν = sizes are at least 30 STAT

§ Population variances are unknown and cannot be σ and σ unknown, σ and σ unknown, 1 2 assumed to be equal 1 2 not assumed equal * not assumed equal *

Prof. Shuguang Liu Chap 10-17 Prof. Shuguang Liu Chap 10-18 Chapter 10 10-4

Related Populations Related Populations

The Paired Difference Test The Paired Difference Test DCOVA DCOVA th (continued) Tests Means of 2 Related Populations The i paired difference is Di , where – Paired or matched samples Related – Repeated measures (before/after) Related Di = X1i - X2i samples – Use difference between paired values: samples n The point estimate for the D paired difference ∑ i Di = X1i - X2i D = i=1 § Eliminates Variation Among Subjects population mean µD is D : n

§ Assumptions: n The sample standard (D D)2 • Both Populations Are Normally Distributed ∑ i − deviation is SD S = i=1 • Or, if not Normal, use large samples D n − 1 n is the number of pairs in the paired sample

Prof. Shuguang Liu Chap 10-19 Prof. Shuguang Liu Chap 10-20

The Paired Difference Test:

Finding tSTAT The Paired Difference Test: Possible Hypotheses

DCOVA DCOVA § The test statistic for µ is: D Paired Samples

Paired Lower-tail test: Upper-tail test: Two-tail test: samples H : µ ≥ 0 H : µ ≤ 0 H : µ = 0 D − µ D 0 D 0 D 0 D tSTAT = H1: µD < 0 H1: µD > 0 H1: µD ≠ 0 SD n α α α/2 α/2

-tα tα -tα/2 tα/2 n Where tSTAT has n - 1 d.f. Reject H0 if tSTAT < -tα Reject H0 if tSTAT > tα Reject H0 if tSTAT < -tα/2 or tSTAT > tα/2 Where tSTAT has n - 1 d.f.

Prof. Shuguang Liu Chap 10-21 Prof. Shuguang Liu Chap 10-22

The Paired Difference Confidence Interval Paired Difference Test: DCOVA Example DCOVA The confidence interval for µ is § Assume you send your salespeople to a “customer D service” training workshop. Has the training made a difference in the number of complaints? You collect the Paired following data: samples

SD D ± tα / 2 Number of Complaints: (2) - (1) Σ Di Salesperson Before (1) After (2) Difference, Di D = n n C.B. 6 4 - 2 = -4.2 T.F. 20 6 -14 n (D D)2 M.H. 3 2 - 1 ∑ i − 2 where i=1 R.K. 0 0 0 (Di − D) SD = ∑ M.O. 4 0 - 4 SD = n −1 n −1 -21 = 5.67

Prof. Shuguang Liu Chap 10-23 Prof. Shuguang Liu Chap 10-24 Chapter 10 10-5

Paired Difference Test: Paired t Test In Excel DCOVA DCOVA § Has the training made a Solutiondifference in the number of complaints (at the 0.01 level)? Paired t Test Since -2.776 < -1.66 < 2.776 Data Hypothesized Mean Diff. 0 we do not reject the null hypothesis. Reject Reject Level of Significance 0.05 H0: µD = 0 Or Intermediate Calculaons H1: µD ≠ 0 Sample Size 5 =COUNT(I2:I6) Since p-value = 0.173 > 0.05 we α/2 α/2 Dbar -4.2 =AVERAGE(I2:I6) do not reject the null hypothesis. - 4.604 4.604 Degrees of Freedom 4 =B8 - 1 α = .01 D = - 4.2 SD 5.67 =STDEV(I2:I6) - 1.66 Standard Error 2.54 =B11/SQRT(B8) Thus we conclude that there is t = ± 4.604 t-Test Stasc -1.66 =(B9 - B4)/B12 0.005 Decision: Do not reject H0 Insufficient evidence to conclude d.f. = n - 1 = 4 Two-Tail Test there is a difference in the average (tstat is not in the reject region) Lower Crical Value -2.776 =-TINV(B5,B10) number of complaints. Test Statistic: Upper Crical Value 2.776 =TINV(B5,B10) Conclusion: There is p-value 0.173 =TDIST(ABS(B13),B10,2) D − µ − 4.2 − 0 insufficient evidence there Do not reject the null =IF(B18

Paired t Test In Minitab Yields The Same Conclusions Two Population Proportions

DCOVA Paired T-Test and CI: After, Before DCOVA Goal: test a hypothesis or form a Paired T for After - Before confidence interval for the difference

N Mean StDev SE Mean between two population proportions, After 5 2.40 2.61 1.17 π – π Before 5 6.60 7.80 3.49 1 2 Difference 5 -4.20 5.67 2.54 Population

95% CI for mean difference: (-11.25, 2.85) proportions T-Test of mean difference = 0 (vs not = 0): T-Value = -1.66 P-Value = 0.173

The point estimate for the difference is p1 − p2

Prof. Shuguang Liu Chap 10-27 Prof. Shuguang Liu Chap 10-28

Two Population Proportions Two Population Proportions

DCOVA (continued) In the null hypothesis we assume the The test statistic for DCOVA null hypothesis is true, so we assume π Population 1 Population π1 – π2 is a Z statistic: proportions = π2 and pool the two sample estimates proportions The pooled estimate for the overall proportion is: ( p1 − p2 )− ( π1 − π2 ) Z STAT = X + X ⎛ 1 1 ⎞ p = 1 2 p(1− p)⎜ + ⎟ ⎝ n1 n2 ⎠ n1 + n2

where X1 and X2 are the number of items of X1 + X 2 X1 X 2 interest in samples 1 and 2 where p = , p1 = , p2 = n1 + n2 n1 n2

Prof. Shuguang Liu Chap 10-29 Prof. Shuguang Liu Chap 10-30 Chapter 10 10-6

Hypothesis Tests for Hypothesis Tests for Two Population Proportions Two Population Proportions

DCOVA (continued) Population proportions Population proportions DCOVA Lower-tail test: Upper-tail test: Two-tail test: Lower-tail test: Upper-tail test: Two-tail test: H : π – π ≥ 0 H : π – π ≤ 0 H : π – π = 0 0 1 2 0 1 2 0 1 2 H1: π1 – π2 < 0 H1: π1 – π2 > 0 H1: π1 – π2 ≠ 0 H0: π1 ≥ π2 H0: π1 ≤ π2 H0: π1 = π2 H1: π1 < π2 H1: π1 > π2 H1: π1 ≠ π2 /2 /2 i.e., i.e., i.e., α α α α

H : π – π ≤ 0 H : π – π = 0 H0: π1 – π2 ≥ 0 0 1 2 0 1 2 H : π – π > 0 H : π – π ≠ 0 -z z -z z H1: π1 – π2 < 0 1 1 2 1 1 2 α α α/2 α/2

Reject H0 if ZSTAT < -Zα Reject H0 if ZSTAT > Zα Reject H0 if ZSTAT < -Zα/2 or ZSTAT > Zα/2

Prof. Shuguang Liu Chap 10-31 Prof. Shuguang Liu Chap 10-32

Hypothesis Test Example: Two population Proportions Hypothesis Test Example: Two population Proportions Is there a significant difference between the proportion of (continued) DCOVA § The hypothesis test is: men and the proportion of women who will vote Yes on DCOVA Proposition A? H0: π1 – π2 = 0 (the two proportions are equal) H1: π1 – π2 ≠ 0 (there is a significant difference between proportions)

§ In a random sample, 36 of 72 men and 35 of 50 women n The sample proportions are: indicated they would vote Yes n Men: p1 = 36/72 = 0.50

n Women: p2 = 35/50 = 0.70 § Test at the .05 level of significance § The pooled estimate for the overall proportion is: X + X 36 + 35 71 p = 1 2 = = = 0.582 n1 + n2 72 + 50 122

Prof. Shuguang Liu Chap 10-33 Prof. Shuguang Liu Chap 10-34

Hypothesis Test Example: Two population Proportions Two Proportion Test In Excel DCOVA (continued) DCOVA Z Test for Differences in Two Proporons Reject H0 Reject H0 Data Since -2.20 < -1.96 The test statistic for π1 – π2 is: Hypothesized Difference 0 .025 .025 Level of Significance 0.05 Group 1 Or p p π π Number of items of interest 36 ( 1 − 2 )− ( 1 − 2 ) zSTAT = Sample Size 72 ⎛ 1 1 ⎞ Group 2 Since p-value = 0.028 < 0.05 -1.96 1.96 Number of items of interest 35 p(1− p)⎜ + ⎟ ⎜ n n ⎟ Sample Size 50 ⎝ 1 2 ⎠ -2.20 We reject the null hypothesis Intermediate Calculaons ( .50 − .70)− ( 0) Group 1 Proporon 0.5 =B7/B8 = = − 2.20 Group 2 Proporon 0.7 =B10/B11 1 1 Decision: Reject H0 Difference in Two Proporons -0.2 =B14 - B15 ⎛ ⎞ Decision: Reject H0 .582(1− .582 )⎜ + ⎟ Average Proporon 0.582 =(B7 + B10)/(B8 + B11) 72 50 ⎝ ⎠ Conclusion: There is Z Test Stasc -2.20 =(B16-B4)/SQRT((B17*(1-B17))*(1/B8+1/B11)) evidence of a difference in Two-Tail Test Conclusion: There is Critical Values = ±1.96 Lower Crical Value -1.96 =NORMSINV(B5/2) evidence of a difference in proportions who will vote Upper Crical Value 1.96 =NORMSINV(1 - B5/2) proportions who will vote For α = .05 p-value 0.028 =2*(1 - NORMSDIST(ABS(B18))) yes between men and Reject the null hypothesis =IF(B23 < B5,"Reject the null hypothesis", yes between men and "Do not reject the null hypothesis") women. women. Prof. Shuguang Liu Chap 10-35 Prof. Shuguang Liu Chap 10-36 Chapter 10 10-7

Two Proportion Test In Minitab Shows The Same Confidence Interval for Conclusions Two Population Proportions

DCOVA DCOVA

Population Test and CI for Two Proportions The confidence interval for proportions Sample X N Sample p π1 – π2 is: 1 36 72 0.500000 2 35 50 0.700000

Difference = p (1) - p (2) Estimate for difference: -0.2 95% CI for difference: (-0.371676, -0.0283244) p1(1− p1 ) p2 (1− p2 ) Test for difference = 0 (vs not = 0): Z = -2.28 P-Value = 0.022 (p1 − p2 ) ± Zα/2 + n1 n2

Prof. Shuguang Liu Chap 10-37 Prof. Shuguang Liu Chap 10-38

Testing for the Ratio Of Two Population Variances The F Distribution

DCOVA DCOVA § The F critical value is found from the F table Hypotheses FSTAT § There are two degrees of freedom required: numerator Tests for Two 2 2 H0: σ1 = σ2 and denominator Population * 2 2 2 H1: σ1 ≠ σ2 S Variances 1 § The larger sample variance is always the numerator 2 2 2 H0: σ1 ≤ σ2 S2 H : σ 2 > σ 2 1 1 2 § When 2 F test statistic S 1 df = n – 1 ; df = n – Where: FSTAT = 2 1 1 2 2 2 S S 1 = Variance of sample 1 (the larger sample variance) 2 1

n1 = sample size of sample 1 § In the F table, 2 S = Variance of sample 2 (the smaller sample variance) 2 • numerator degrees of freedom determine the column

n2 = sample size of sample 2 • denominator degrees of freedom determine the row n1 –1 = numerator degrees of freedom

n2 – 1 = denominator degrees of freedom

Prof. Shuguang Liu Chap 10-39 Prof. Shuguang Liu Chap 10-40

Finding the Rejection Region F Test: An Example

DCOVA DCOVA You are a financial analyst for a brokerage firm. You want 2 2 2 2 H0: σ1 = σ2 H0: σ1 ≤ σ2 to compare dividend yields between stocks listed on the H : σ 2 ≠ σ 2 H : σ 2 > σ 2 1 1 2 1 1 2 NYSE & NASDAQ. You collect the following data: NYSE NASDAQ Number 21 25 α/2 Mean 3.27 2.53 F α Std dev 1.30 1.16 0 0 Do not Reject H 0 Do not Reject H0 F reject H0 reject H Fα/2 0 Fα Is there a difference in the variances between the NYSE & NASDAQ at the α = 0.05 level? Reject H0 if FSTAT > Fα/2 Reject H0 if FSTAT > Fα

Prof. Shuguang Liu Chap 10-41 Prof. Shuguang Liu Chap 10-42 Chapter 10 10-8

F Test: Example Solution F Test: Example Solution DCOVA DCOVA § Form the hypothesis test: § The test statistic is: (continued) (there is no difference between variances) H : σ 2 = σ 2 H : σ 2 σ 2 0 1 2 0 1 = 2 (there is a difference between variances) 2 2 2 2 H1: σ1 ≠ σ2 H : σ 2 ≠ σ 2 S 1.30 1 1 2 F = 1 = =1.256 STAT S 2 1.162 n 2 Find the F critical value for α = 0.05: α/2 = .025

0 n F Numerator d.f. = n1 – 1 = 21 –1 = 20 Do not Reject H0 reject H 0 F =2.33 n FSTAT = 1.256 is not in the rejection 0.025 region, so we do not reject H n Denominator d.f. = n2 – 1 = 25 –1 = 24 0

n Conclusion: There is insufficient evidence of n Fα/2 = F.025, 20, 24 = 2.33 a difference in variances at α = .05

Prof. Shuguang Liu Chap 10-43 Prof. Shuguang Liu Chap 10-44

Two Variance F Test In Minitab Yields The Same Two Variance F Test In Excel Conclusion

DCOVA DCOVA Test and CI for Two Variances

F Test for Differences in Two Variables Null hypothesis Sigma(1) / Sigma(2) = 1 Conclusion: Alternative hypothesis Sigma(1) / Sigma(2) not = 1 Significance level Alpha = 0.05 Data There is insufficient evidence Level of Significance 0.05 Statistics Larger-Variance Sample of a difference in variances Sample N StDev Variance Sample Size 21 1 21 1.300 1.690 Sample Variance 1.6900 =1.3^2 at α = .05 because: 2 25 1.160 1.346 Smaller-Variance Sample Sample Size 25 Ratio of standard deviations = 1.121 Sample Variance 1.3456 =1.16^2 Ratio of variances = 1.256 F statistic =1.256 < 2.327 F Intermediate Calculaons α/2 F Test Stasc 1.256 1.255945 or 95% Confidence Intervals Populaon 1 Sample Degrees of Freedom 20 =B6 - 1 Populaon 2 Sample Degrees of Freedom 24 =B9 - 1 CI for p-value = 0.589 > 0.05 = α. Distribution CI for StDev Variance Two-Tail Test of Data Ratio Ratio Upper Crical Value 2.327 =FINV(B4/2,B14,B15) Normal (0.735, 1.739) (0.540, 3.024) p-value 0.589 =2*FDIST(B13,B14,B15) Do not reject the null hypothesis =IF(B19

Prof. Shuguang Liu Chap 10-45 Prof. Shuguang Liu Chap 10-46

Chapter Summary Chapter Summary

§ Compared two independent samples § Compared two population proportions(continued) • Performed pooled-variance t test for the • Formed confidence intervals for the difference difference in two means between two population proportions • Performed separate-variance t test for Performed Z-test for two population difference in two means • proportions • Formed confidence intervals for the difference between two means § Performed F test for the ratio of two § Compared two related samples (paired population variances samples) • Performed paired t test for the mean difference • Formed confidence intervals for the mean difference Prof. Shuguang Liu Chap 10-47 Prof. Shuguang Liu Chap 10-48