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STATISTICAL INFERENCE PART VIII. HYPOTHESIS TESTING - APPLICATIONS – TWO POPULATION TESTS. PARAMETERS: 1 ,. PARAMETERS: 2 ,. Sample size: n 2. Sample size: n 1. Statistics:. Statistics:. INFERENCE ABOUT THE DIFFERENCE BETWEEN TWO SAMPLES. INDEPENDENT SAMPLES. POPULATION 1.

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## STATISTICAL INFERENCE PART VIII

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**STATISTICAL INFERENCEPART VIII**HYPOTHESIS TESTING - APPLICATIONS – TWO POPULATION TESTS**PARAMETERS:**1, PARAMETERS: 2, Sample size: n2 Sample size: n1 Statistics: Statistics: INFERENCE ABOUT THE DIFFERENCE BETWEEN TWO SAMPLES • INDEPENDENT SAMPLES POPULATION 1 POPULATION 2**SAMPLING DISTRIBUTION OF**• Consider random samples of n1 and n2 from two normal populations. Then, • For non-normal distributions, we can use Central Limit Theorem for n130 and n230.**CONFIDENCE INTERVAL FOR m1- m21 AND 2 ARE KNOWN FOR**NORMAL DISTRIBUTION OR LARGE SAMPLE • A 100(1-C.I. for is given by: • If and are unknown and unequal, we can replace them with s1 and s2. We are still using Z table because this is large sample or normal distribution situation.**EXAMPLE**• Is there any significant difference between mean family incomes of two groups? • Set up a 95% CI for 2 - 1. • 95% CI for 2 - 1:**INTERPRETATION**• With 95% confidence, mean family income in the second group may exceed that in the first group by between $363 and $2397.**Test Statistic for m1- m2when s1 and s2 are known**• Test statistic: • If and are unknown and unequal, we can replace them with s1 and s2.**EXAMPLE**• Two different procedures are used to produce battery packs for laptop computers. A major electronics firm tested the packs produced by each method to determine the number of hours they would last before final failure. • The electronics firm wants to know if there is a difference in the mean time before failure of the two battery packs.=0.10**SOLUTION**• STEP 1: H0: 1 = 2 H0: 1 - 2 = 0 HA: 1 2 HA : 1 - 2 0 • STEP 2: Test statistic: • STEP 3: Decision Rule = Reject H0 if z<-z/2=-1.645 or z>z/2=1.645. • STEP 4: Not reject H0. There is not sufficient evidence to conclude that there is a difference in the mean life of the 2 types of battery packs.**1 AND 2 ARE UNKNOWN if 1 = 2**• A 100(1-C.I. for is given by: where**Test Statistic for m1- m2when s1 = s2 and unknown**• Test Statistic: where**EXAMPLE**• The statistics obtained from random sampling are given as • It is thought that 1 < 2. Test the appropriate hypothesis assuming normality with = 0.01.**SOLUTION**• n1< 30 and n2< 30 t-test • Because s1 and s2 are not much different from each other, use equal-variance t-test (More formally, we can test Ho: σ²1=σ²2). H0: 1 = 2 HA: 1 < 2 (1 - 2<0)**Decision Rule: Reject H0 if t < -t0.01,8+9-2=-2.602**• Conclusion: Since t = -19.13 < -t0.01,8+9-2=-2.602, reject H0 at = 0.01.**Test Statistic for m1- m2when s1 s2 and unknown**• Test Statistic: with the degree of freedom**EXAMPLE**• Does consuming high fiber cereals entail weight loss? 30 people were randomly selected and asked what they eat for breakfast and lunch. They were divided into those consuming and those not consuming high fiber cereals. The statistics are obtained as n1=10; n2=20**SOLUTION**• Because s1 and s2 are too different from each other and the population variances are not assumed equal, we can use a t statistic with degrees of freedom**H0: 1 - 2 = 0**HA: 1 - 2 < 0 • DECISION RULE: Reject H0 if t < -t,df = -t0.05, 25 = -1.708. • CONCLUSION: Since t =-2.31< -t0.05, 25=-1.708, reject H0 at = 0.05.**MINITAB OUTPUT**• Two Sample T-Test and Confidence Interval Twosample T for Consmers vs Non-cmrs N Mean StDev SE Mean Consmers 10 595.8 35.7 11 Non-cmrs 20 661 116 26 • 95% C.I. for mu Consmers - mu Non-cmrs:( -123, -7) T-Test mu Consmers = mu Non-cmrs (vs <): T= -2.31 P=0.015 DF= 25**Inference about the Difference of Two Means: Matched Pairs**Experiment • Data are generated from matched pairs; not independent samples. • Let Xi and Yi denote the measurements for the i-th subject. Thus, (Xi, Yi) is a matched pair observations. • Denote Di = Yi-Xior Xi-Yi. • If there are n subjects studied, we have D1, D2,…, Dn. Then,**CONFIDENCE INTERVAL FOR D= 1 - 2**• A 100(1-C.I. for D=is given by: • For n 30, we can use z instead of t.**HYPOTHESIS TESTS FOR D= 1 - 2**• The test statistic for testing hypothesis about Dis given by with degree of freedom n-1.**EXAMPLE**• Sample data on attitudes before and after viewing an informational film. i Xi Di=Yi-Xi Yi**90% CI for D= 1- 2:**• With 90% confidence, the mean attitude measurement after viewing the film exceeds the mean attitude measurement before viewing by between 0.36 and 14.92 units. t0.05, 9**EXAMPLE**• How can we design an experiment to show which of two types of tires is better? Install one type of tire on one wheel and the other on the other (front) wheels. The average tire (lifetime) distance (in 1000’s of miles) is: with a sample difference s.d. of • There are a total of n=20 observations**SOLUTION**H0: mD=0 HA:mD>0 • Test Statistics: Reject H0 if t>t.05,19=1.729, Conclusion: Reject H0 at =0.05**Inference About the Difference of Two Population Proportions**Independent populations Population 1 Population 2 PARAMETERS: p1 PARAMETERS: p2 Sample size: n2 Sample size: n1 Statistics: Statistics:**SAMPLING DISTRIBUTION OF**• A point estimator of p1-p2 is • The sampling distribution of is if nipi 5 and niqi 5, i=1,2.**STATISTICAL TESTS**• Two-tailed test Ho:p1=p2 HA:p1p2 Reject H0 if z < -z/2 or z > z/2. • One-tailed tests Ho:p1=p2 HA:p1 > p2 Reject H0 if z > z Ho:p1=p2 HA:p1 < p2 Reject H0 if z < -z**EXAMPLE**• A manufacturer claims that compared with his closest competitor, fewer of his employees are union members. Of 318 of his employees, 117 are unionists. From a sample of 255 of the competitor’s labor force, 109 are union members. Perform a test at = 0.05.**SOLUTION**H0: p1 = p2 HA: p1 < p2 and , so pooled sample proportion is Test Statistic:**Decision Rule: Reject H0 if z < -z0.05=-1.96.**• Conclusion: Because z = -1.4518 > -z0.05=-1.96, not reject H0 at =0.05. Manufacturer is wrong. There is no significant difference between the proportions of union members in these two companies.**Example**• In a study, doctors discovered that aspirin seems to help prevent heart attacks. Half of 22,000 male participants took aspirin and the other half took a placebo. After 3 years, 104 of the aspirin and 189 of the placebo group had heart attacks. Test whether this result is significant. • p1: proportion of all men who regularly take aspirin and suffer from heart attack. • p2: proportion of all men who do not take aspirin and suffer from heart attack**Test of Hypothesis for p1-p2**H0: p1-p2 = 0 HA: p1-p2 < 0 • Test Statistic: Conclusion: Reject H0 since p-value=P(z<-5.02) 0**Confidence Interval for p1-p2**A 100(1-C.I. for pp is given by:**Inference About Comparing Two Population Variances**Independent populations Population 1 Population 2 PARAMETERS: PARAMETERS: Sample size: n2 Sample size: n1 Statistics: Statistics:**SAMPLING DISTRIBUTION OF**• For independent r.s. from normal populations • (1-α)100% CI for**STATISTICAL TESTS**• Two-tailed test Ho: (or ) HA: Reject H0 if F < F/2,n1-1,n2-1 or F > F 1- /2,n1-1,n2-1. • One-tailed tests Ho: HA: Reject H0 if F > F 1- , n1-1,n2-1 Ho: HA: Reject H0 if F < F,n1-1,n2-1**Example**• A scientist would like to know whether group discussion tends to affect the closeness of judgments by property appraisers. Out of 16 appraisers, 6 randomly selected appraisers made decisions after a group discussion, and rest of them made their decisions without a group discussion. As a measure of closeness of judgments, scientist used variance. • Hypothesis: Groups discussion will reduce the variability of appraisals.**Example, cont.**Ho: versus H1: Reject Ho. Group discussion reduces the variability in appraisals.

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