Business Statistics CONTENTS The one-sample -test for Hypotheses - PowerPoint PPT Presentation

𝜈 : THE ONE-SAMPLE 𝑢 -TEST AND SPSS Business Statistics

CONTENTS The one-sample 𝑢 -test for 𝜈 Hypotheses and SPSS Old exam question Further study

THE ONE-SAMPLE 𝑢 -TEST FOR 𝜈 When to use the one-sample 𝑨 -test? ▪ It relies on the fact that the distribution of ത 𝑌 is normal with 𝜏 𝑌 𝑌 = 𝜈 𝑌 and 𝜏 ത 𝑜 , so on the CLT 𝜈 ത 𝑌 = ▪ So it can be used to test a hypothesis of the mean only ▪ not a hypothesis on 𝜏 , median, etc. ▪ It works when the population 𝑌 is normal ▪ or when 𝑌 is symmetric and 𝑜 ≥ 15 ▪ or when 𝑜 ≥ 30 ▪ It requires knowledge of 𝜏 𝑌 Rule of thumb: ▪ so knowing 𝑡 𝑌 will not work use of CLT for mean justified when 𝑜 ≥ 30 ▪ that is a problem!

THE ONE-SAMPLE 𝑢 -TEST FOR 𝜈 So, what to do if you don’t know 𝜏 𝑌 ? 2 from the ▪ Luckily, we can estimate the population variance 𝜏 𝑌 2 sample variance 𝑡 𝑌 ത ത ത ത 𝑌−𝜈 ഥ 𝑌−𝜈 ഥ 𝑌−𝜈 𝑌 𝑌−𝜈 𝑌 ▪ calculate the test statistic 𝑡 𝑌 / 𝑜 instead of 𝜏 𝑌 / 𝑜 ? 𝑌 𝑌 = = 𝑡 ഥ 𝜏 ഥ 𝑌 𝑌 Sure, we can ... ത 𝑌−𝜈 ഥ ▪ but instead of 𝑌 ~𝑂 0,1 , we have 𝑌 𝜏 ഥ ത 𝑌 − 𝜈 ത 𝑌 ~𝑢 𝑜−1 𝑡 ത 𝑌 ▪ where 𝑢 df is the 𝑢 -distribution with df degrees of freedom

THE ONE-SAMPLE 𝑢 -TEST FOR 𝜈 Example: ▪ sample of 𝑜 = 100 body heights ▪ sample mean ҧ 𝑦 = 179.1 cm 2 = 212.4 cm 2 ▪ sample variance 𝑡 𝑌 We think (or hope, or fear) that 𝜈 𝑌 < 181 cm Use the five-step procedure ▪ Steps 1-2: as before ( 𝐼 0 : 𝜈 ≥ 181 ) ▪ Steps 3-5: see below

THE ONE-SAMPLE 𝑢 -TEST FOR 𝜈 ▪ Step 3 ത 𝑌−181 ▪ under least extreme version of 𝐼 0 : 212.4/100 ~𝑢 99 ▪ no further assumptions required (because 𝑜 ≥ 30 ) ▪ Step 4 ▪ value of test statistic 𝑢 calc = −1.3037 ▪ critical value of 𝑢 from the table is 𝑢 crit,lower,0.05,df=99 = − 1.660 −1.660 0 ▪ Step 5 ▪ 𝑢 calc ∉ 𝑆 crit , so do not reject 𝐼 0 ▪ conclude that there is no evidence for 𝜈 < 181

THE ONE-SAMPLE 𝑢 -TEST FOR 𝜈 Summing up: ▪ 𝜏 𝑌 known: ▪ use 𝑨 -table/perform 𝑨 -test ▪ 𝜏 𝑌 estimated by 𝑡 𝑌 : ▪ use 𝑢 -table/perform 𝑢 -test with df = 𝑜 − 1 degrees of freedom ▪ In both cases: ▪ use two-tailed critical value ( 𝛽/2 ) to construct a confidence interval ▪ use two-tailed critical value ( 𝛽/2 ) to construct a critical region for a two-sided hypothesis test ▪ use one-tailed critical value ( 𝛽 ) to construct a critical region for a one-sided hypothesis test

EXERCISE 1 We test a hypothesis 𝐼 0 : 𝜈 = 310 with a sample size 𝑜 = 50 and 𝜏 unknown. Which statements are correct? A. We can use the 𝑨 -test. B. We can use the 𝑢 -test. C. We can’t perform this test. D. The sample is normally distributed. E. The population is normally distributed. F. The sampling distribution of the mean is normal.

HYPOTHESES AND SPSS Using SPSS for the one-sample 𝑨 -test ▪ This is not possible! ▪ SPSS (realistically) assumes you don’t know 𝜏 𝑌 ▪ SPSS will always use 𝑡 𝑌 to estimate 𝜏 𝑌 and then do a one-sample 𝑢 -test

HYPOTHESES AND SPSS 𝑌 =life Using SPSS for the one-sample 𝑢 -test expectancy ▪ Claim: the average life expectancy is 68 year 𝐼 0 : 𝜈 𝑌 = 68 Under 𝐼 0 : 𝑈~𝑢 152 𝑞 -value 𝑢 calc

HYPOTHESES AND SPSS But look carefully: ▪ it is a two-sided test (SPSS says: two-tailed) 𝐼 0 : 𝜈 𝑌 = 68 𝑞 -value

HYPOTHESES AND SPSS Doing a one-sided test with SPSS ▪ How to do a one-sided 𝑢 -test in SPSS? ▪ we could do it by hand with the intermediate results ▪ we could do it by hand with the two-sided results ▪ you must be able to do both ▪ Example: ▪ null hypothesis 𝐼 0 : 𝜈 ≥ 68 ▪ sample with 𝑜 = 153 yields ҧ 𝑦 = 64.515 and 𝑡 = 12.7937 64.515 − 68 153 − 1

HYPOTHESES AND SPSS 68 + 68 − 64.515 Two-sided ▪ 𝑄 ത 𝑌 ≤ 64.515 + 𝑄 ത 𝑌 ≥ 71.485 = 𝑄 ቀ 𝑢 ≤ 64.515−68 71.485−68 12.7937/ 153 = 𝑄ሺ 12.7937/ 153 + 𝑄 𝑢 ≥ ቁ 𝑢 ≤ ሻ −3.369 + 𝑄 𝑢 ≥ 3.369 ▪ use SPSS for 𝑢 df=152 : 𝑞 = 0.001

HYPOTHESES AND SPSS One-sided (right-sided) and ҧ 𝑦 < 𝜈 0 64.515−68 ▪ 𝑄 ത 𝑌 ≤ 64.515 = 𝑄 𝑢 ≤ 12.7937/ 153 = 𝑄 𝑢 ≤ −3.369 ▪ this is exactly half of the reported two-sided 𝑞 -value ▪ 𝑞−value = 0.0005 So, to move from the SPSS-reported two-sided 𝑞 -value to right-sided 𝑞 -value, in case of ҧ 𝑦 < 𝜈 0 : ▪ divide by 2 𝑞 right = 𝑞 two /2

HYPOTHESES AND SPSS One-sided (left-sided) and ҧ 𝑦 < 𝜈 64.515−68 ▪ 𝑄 ത 𝑌 ≥ 64.515 = 𝑄 𝑢 ≥ 12.7937/ 153 = 𝑄 𝑢 ≥ −3.369 ▪ this is exactly 1 − 𝑄 𝑢 ≤ −3.369 ▪ 𝑞−value = 0.9995 So, to move from the SPSS-reported two-sided 𝑞 -value to left- sided 𝑞 -value, in case of ҧ 𝑦 > 𝜈 0 : ▪ divide by 2 and subtract the result from 1 𝑞 left = 1 − 𝑞 two /2

ҧ HYPOTHESES AND SPSS Another example ▪ Suppose ▪ 𝐼 0 : 𝜈 = 3 ▪ 𝛽 = 0.05 ▪ 𝑦 = 2.84 𝜈=3 ത ▪ 𝑞−value = 2 × 𝑄 𝑌 ≤ 2.84 = 0.0456

HYPOTHESES AND SPSS Both ҧ 𝑦 = 2.84 and ҧ 𝑦 = 3.16 give the same two-sided 𝑞 - value! ▪ so which is true when 𝐼 0 : 𝜈 ≥ 3 and which when 𝐼 0 : 𝜈 ≤ 3 ?

ҧ ҧ HYPOTHESES AND SPSS 𝑰 𝟏 : 𝝂 ≤ 𝟒 𝑰 𝟏 : 𝝂 ≥ 𝟒 (right-sided) (left-sided) 𝑞 SPSS 𝑞 SPSS 𝑦 = 2.84 𝜈=3 ത 𝜈=3 ത 𝑄 𝑌 ≥ 2.84 = 1 − 𝑄 𝑌 ≤ 2.84 = 2 2 ( ҧ 𝑦 < 𝜈 0 ) 𝑞 SPSS 𝑞 SPSS 𝜈=3 ത 𝜈=3 ത 𝑦 = 3.16 𝑄 𝑌 ≥ 3.16 = 𝑄 𝑌 ≤ 3.16 = 1 − 2 2 ( ҧ 𝑦 > 𝜈 0 )

EXERCISE 2 Suppose we test a hypothesis on the mean 𝐼 0 : 𝜈 ≥ 310 with significance level 𝛽 = 0.05 . We sample data, and perform the test and find a sample mean ҧ 𝑦 = 307 and a two- sided 𝑞 -value 0.08 . What do we conclude?

EXERCISE 3 Suppose we have the following result: A. reject 𝐼 0 : 𝜈 ≤ 6 . B. 𝑞−value < 𝛽 . C. probably 𝜈 > 6 . What is a correct sequence? A→B→C , C→B→A , B→A→C , etc?

OLD EXAM QUESTION 26 March 2015, Q3b

FURTHER STUDY Doane & Seward 5/E 9.1-9.5 Tutorial exercises week 2 one-sided test in SPSS