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# 09 test of hypothesis small sample.ppt

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### 09 test of hypothesis small sample.ppt

1. 1. Quantitative Methods Varsha Varde
2. 2. Varsha Varde 2 • Contents: • 1. Introduction • 2. Student’s t distribution • 3. Small-sample inferences about a population mean • 4. Small-sample inferences about the difference between two means: Independent Samples • 5. Small-sample inferences about the difference between two means: Paired Samples • 6. Inferences about a population variance • 7. Comparing two population variances
3. 3. Introduction • When the sample size is small we only deal with normal populations. • For non-normal (e.g. binomial) populations different techniques are necessary Varsha Varde 3
4. 4. Summary of Test Statistics to be Used in aSummary of Test Statistics to be Used in a Hypothesis Test about a Population MeanHypothesis Test about a Population Mean nn >> 30 ?30 ? σσ known ?known ? Popul.Popul. approx.approx. normalnormal ?? σσ known ?known ? UseUse ss toto estimateestimate σσ UseUse ss toto EstimateEstimate σσ IncreaseIncrease nn toto >> 3030/ x z n µ σ − = / x z s n µ− = / x z n µ σ − = / x t s n µ− = YesYes YesYes YesYes YesYes NoNo NoNo NoNo NoNo
5. 5. Student’s t Distribution • For small samples (n < 30) from normal populations, we have • z =x¯ - µ/σ/√n • If σ is unknown, we use s instead; but we no more have a Z distribution • Assumptions. • 1. Sampled population is normal • 2. Small random sample (n < 30) • 3. σ is unknown • t =x¯ - µ/s/√n Varsha Varde 5
6. 6. Properties of the t Distribution: • (i) It has n - 1 degrees of freedom (df) • (ii) Like the normal distribution it has a symmetric mound-shaped probability distribution • (iii) More variable (flat) than the normal distribution • (iv) The distribution depends on the degrees of freedom. Moreover, as n becomes larger, t converges to Z. • (v) Critical values (tail probabilities) are obtained from the t table • Examples. • (i) Find t0.05,5 = 2.015 • (ii) Find t0.005,8 = 3.355 • (iii) Find t0.025,26 = 2.056 Varsha Varde 6
7. 7. Small-Sample Inferences About a Population Mean • Parameter of interest: µ • Sample data: n, x¯, s • Other information: µ0= target value ;α • Point estimator: x¯ • Estimator mean: µx¯ = µ • Estimated standard errorσx¯ = s/√n • Confidence Interval for µ: • x ± tα/2 ,n-1 (s/√n ) Varsha Varde 7
8. 8. Student’s t test • Test: H0 : µ = µ0 • Ha : 1) µ > µ0; 2) µ < µ0; 3) µ = µ0. • Critical value: either tα,n-1 or tα/2 ,n-1 • T.S. : t = x¯-µ0/s/√n • RR:1) Reject H0 if t > tα,n-1 • 2) Reject H0 if t < -tα,n-1 • 3) Reject H0 if t > tα/2 ,n-1 or t < -tα/2,n-1 • Decision: 1) if observed value is in RR: “Reject H0” • 2) if observed value is not in RR: “Do not reject H0” • Conclusion: At 100α% significance level there is (in)sufficient statistical evidence to “favor Ha” . • Assumptions.1. Small sample (n < 30) • 2. Sample is randomly selected • 3. Normal population • 4. Unknown variance Varsha Varde 8
9. 9. Example • Example: It is claimed that weight loss in a new diet program is at least 20 pounds during the first month. Formulate &Test the appropriate hypothesis • Sample data: n = 25, x¯ =19.3, s2 = 25, µ0 = 20, α=0.05 • Critical value: t0.05,24 = -1.711 • H0 : µ ≥ 20 (µ is greater than or equal to 20 ) • Ha : µ < 20, • T.S.:t =(x¯ - µ0 )/s/√n=(19.3 – 20)/5/√25 =-0.7 • RR: Reject H0 if t <-1.711 • Decision: Reject H0 • Conclusion: At 5% significance level there is insufficient statistical evidence to conclude that weight loss in a new diet program exceeds 20 pounds per first month. Varsha Varde 9
10. 10. Two Means: Independent Samples • Parameter of interest: µ1 - µ2 • Sample data:Sample 1: n1, x1, s1 ;Sample 2: n2, x2, s2 • Other information: D0= target value ; α • Point estimator: X¯1 – X¯2 • Estimator mean: µ X¯1 – X¯2 = µ1 - µ2 • Assumptions. • 1. Normal populations 2. Small samples ( n1 < 30; n2 < 30) • 3. Samples are randomly selected4. Samples are independent • 5. Variances are equal with common variance • σ2 = σ2 1 = σ2 2 • Pooled estimator for σ. • s = √[(n1 - 1)s2 1 + (n2 - 1)s2 2 ]/(n1 + n2 – 2) • Estimator standard error: • σX¯1 – X¯2 = σ √ (1/n1+1/n2) • Confidence Interval: • (x¯1 - x¯2) ± (tα/2 ,n1+n2-2)(s √1/n1+1/n2) Varsha Varde 10
11. 11. Two Means: Independent Samples • Test:H0 : µ1 - µ2 = D0 • Ha : 1)µ1 - µ2 > D0 or 2) µ1 - µ2 < D0 or 3) µ1 - µ2 = D0 • T.S. :t =(x¯1 - x¯2) - D0/s √1/n1+ 1/n2 • RR: 1) Reject H0 if t > tα,n1+n2-2 2) Reject H0 if t < - tα,n1+n2-2 3) Reject H0 if t > tα/2,n1+n2-2 or t < - tα/2,n1+n2-2
12. 12. Example.(Comparison of two weight loss programs) • Refer to the weight loss example. Test the hypothesis that weight loss in a new diet program is different from that of an old program. We are told that that the observed value of Test Statistics is 2.2 and we know that • Sample 1 : n1 = 7 ; Sample 2 : n2 = 8 ; α= 0.05 • Solution. • H0 : µ1 - µ2 = 0 • Ha : µ1 - µ2 ≠0 • T.S. :t =(x1 - x2) – 0/s √1/n1+ 1/n2= 2.2 • Critical value: t.025,13 = 2.16 • RR: Reject H0 if t > 2.160 or t < -2.160 • Decision: Reject H0 • Conclusion: At 5% significance level there is sufficient statistical evidence to conclude that weight loss in the two diet programs are differentVarsha Varde 12
13. 13. Small-Sample Inferences About the Difference Between Two Means: Paired Samples • Parameter of interest: µ1 - µ2 = µd • Sample of paired differences data: • Sample : n = number of pairs, d¯ = sample mean, sd • Other information: D0= target value, α • Point estimator: d¯ • Estimator mean: µd¯ = µd • Assumptions. • 1. Normal populations • 2. Small samples ( n1 < 30; n2 < 30) • 3. Samples are randomly selected • 4. Samples are paired (not independent) Varsha Varde 13
14. 14. • Sample standard deviation of the sample of n paired differences n • sd = √ Σ (di - ¯d)2 / n-1 i=1 • Estimator standard error: σd = sd/√n • Confidence Interval. • ¯d ± tα/2,n-1sd/√n Varsha Varde 14
15. 15. Test. • H0 : µ1 - µ2 = D0 (equivalently, µd = D0) • Ha : 1) µ1 - µ2 = µd > D0; • 2) µ1 - µ2 = µd < D0; • 3) µ1 - µ2 = µd = D0, • T.S. :t =¯d - D0/sd/√n • RR: • 1) Reject H0 if t > tα, n-1 • 2) Reject H0 if t < -tα,n-1 • 3) Reject H0 if t > tα/2,n-1 or t < -tα/2,n-1 Varsha Varde 15
16. 16. Example. • A manufacturer wishes to compare wearing qualities of two different types of tires, A and B. For the comparison a tire of type A and one of type B are randomly assigned and mounted on the rear wheels of each of five automobiles. The automobiles are then operated for a specified number of miles, and the amount of wear is recorded for each tire. These measurements are tabulated below. • Automobile Tire A Tire B • 1 10.6 10.2 • 2 9.8 9.4 • 3 12.3 11.8 • 4 9.7 9.1 • 5 8.8 8.3 • Varsha Varde 16
17. 17. • x¯1 = 10.24 x¯2 = 9.76 • Using the previous section test we would have t = 0.57 resulting in an insignificant test which is inconsistent with the data. • Automobile Tire A Tire B d=A-B • 1 10.6 10.2 0.4 • 2 9.8 9.4 0.4 • 3 12.3 11.8 0.5 • 4 9.7 9.1 0.6 • 5 8.8 8.3 0.5 • x¯1 = 10.24 x¯2 = 9.76 d¯ =0 .48 • Q1: Provide a summary of the data in the above table. • Sample summary: n = 5, d ¯= .48, sd = .0837 Varsha Varde 17
18. 18. • Q2: Do the data provide sufficient evidence to indicate a difference in average wear for the two tire types. • Test. (parameter µd = µ1 - µ2) • H0 : µd = 0 • Ha : µd ≠ 0 • T.S. :t =d ¯- D0/sd/√n • =.48 – 0/.0837/√5 = 12.8 • RR: Reject H0 if t > 2.776 or t < -2.776 ( t.025,4 = 2.776) • Decision: Reject H0 • Conclusion: At 5% significance level there is sufficient statistical evidence to to conclude that the average amount of wear for type A tire is different from that for type B tire. • Exercise. Construct a 99% confidence interval for the 18
19. 19. Inferences About a Population Variance • Chi-square distribution. When a random sample of size n is drawn from a normal population with mean µ and standard deviation σ, the sampling distribution of S2 depends on n. The standardized distribution of S2 is called the chi-square distribution and is given by • χ2 =(n - 1)s2 /σ2 • Degrees of freedom (df): ν = n - 1 • Graph: Non-symmetrical and depends on df • Critical values: using χ2 tables Varsha Varde 19
20. 20. Inferences About a Population Variance • Test.H0 : σ2 = σ2 0 • Ha : σ2 ≠ σ2 0 (two-tailed test). • Ha : σ2 <σ2 0 (One –tailed test ) • T.S. :X2 =(n - 1)s2 / σ2 0 • RR: (two-tailed test). • Reject H0 if X2 > X2 α/2 or X2 < X2 1-α/2 where • X2 is based on (n - 1) degrees of freedom. • RR: (One-tailed test). • Reject H0 if X2 < X2 1-α where • X2 is based on (n - 1) degrees of freedom • Assumptions. • 1. Normal population • 2. Random sample Varsha Varde 20
21. 21. Example • Future Technologies ltd. manufactures high resolution telescopes .The management wants its products to have a variation of less than 2 sd in resolution while focusing on objects which are beyond 500 light years. When they tested their newly manufactured telescope for 30 times to focus on an object 500 light years away they found sample sd to be 1.46. State the Hypothesis and test it at 1% level of significance. Can the management accept to sell this product? Varsha Varde 21
22. 22. Solution • Given ,n=30 and S2 =(1.46)2 ,σ2 0 =4We set up hypothesis • H0: σ2 =4 • Ha: σ2< 4 • Significance Level α =1%=.01 • This is a one tailed test • T.S = X2 =(n - 1)s2 / σ2 0= (30-1)(1.46)2/4=15.45 • Referring to X2 tables we find at 29 d.f the value of X2 that leaves an area of 1-.01 =.99 in the upper tail and .01 in the lower tail is 14.256.Since calculated value 15.45 is greater than the cut off level 14.256 we accept null hypothesis and conclude that sd=2. There fore management will not allow sale of its telescope. Varsha Varde 22
23. 23. Comparing Two Population Variances • F-distribution. When independent samples are drawn from two normal populations with equal variances then S2 1/S2 2 possesses a sampling distribution that is known as an • F distribution. That is • F = S2 1/S2 2 • Degrees of freedom (df): ν1 = n1 - 1; ν2 = n2 - 1 • Graph: Non-symmetrical and depends on df • Critical values: using F tables • Test. • H0 : σ2 1 = σ2 2 • Ha : σ2 1≠ σ2 2(two-tailed test). • T.S. :F = S2 1/S2 2 • Where S2 1is the larger sample variance. • Note: F = larger sample variance/smaller sample variance • RR: Reject H0 if F > Fα/2 where Fα/2 is based on (n1 - 1) and (n2 - 1) degrees of freedom. • Assumptions. 23
24. 24. Example • Investment risk is generally measured by the volatility of possible outcomes of the investment. The most common method for measuring investment volatility is by computing the variance ( or standard deviation) of possible outcomes. Returns over the past 10 years for first alternative and 8 years for the second alternative produced the following data: • Data Summary: • Investment 1: n1 = 10, x¯1 = 17.8%; s2 1 = 3.21 • Investment 2: n2 = 8, x¯2 = 17.8%; s2 2 = 7.14 • Both populations are assumed to be normally distributed. Varsha Varde 24
25. 25. Solution • Q1: Do the data present suffcient evidence to indicate that the risks for investments1 and 2 are unequal ? • Solution. • Test:H0 : σ2 1 = σ2 2 • Ha : σ2 1 ≠ σ2 2 (two-tailed test). • T.S. :F = S2 1/S2 2 =7.14/3.21= 2.22 • .RR: Reject H0 if F > Fα/2 where • Fα/2,n2-1,n1-1 = F.025,7,9 = 4.20 • Decision: Do not reject H0 • Conclusion: At 5% significance level there is insuffcient statistical evidence to indicate that the risks for investments 1 and 2 are unequal. • Exercise. Do the upper tail test. That is • Ha : σ2 1 > σ2 2. 25
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