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𝜈: THE ONE-SAMPLE 𝑢-TEST AND SPSS

Business Statistics

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

When to use the one-sample 𝑨-test? ▪ It relies on the fact that the distribution of ത 𝑌 is normal with 𝜈 ത

𝑌 = 𝜈𝑌 and 𝜏 ത 𝑌 = 𝜏𝑌 𝑜, so on the CLT

▪ 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 𝜏𝑌

▪ so knowing 𝑡𝑌 will not work ▪ that is a problem!

THE ONE-SAMPLE 𝑢-TEST FOR 𝜈

Rule of thumb: use of CLT for mean justified when 𝑜 ≥ 30

So, what to do if you don’t know 𝜏𝑌? ▪ Luckily, we can estimate the population variance 𝜏𝑌

2 from the

sample variance 𝑡𝑌

2

▪ calculate the test statistic

ത 𝑌−𝜈ഥ

𝑌

𝑡ഥ

𝑌

=

ത 𝑌−𝜈𝑌 𝑡𝑌/ 𝑜 instead of ത 𝑌−𝜈ഥ

𝑌

𝜏ഥ

𝑌

=

ത 𝑌−𝜈𝑌 𝜏𝑌/ 𝑜?

Sure, we can ... ▪ but instead of

ത 𝑌−𝜈ഥ

𝑌

𝜏ഥ

𝑌 ~𝑂 0,1 , we have

ത 𝑌 − 𝜈 ത

𝑌

𝑡 ത

𝑌

~𝑢𝑜−1 ▪ where 𝑢df 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 ▪ sample variance 𝑡𝑌

2 = 212.4 cm2

We think (or hope, or fear) that 𝜈𝑌 < 181 cm Use the five-step procedure ▪ Steps 1-2: as before (𝐼0: 𝜈 ≥ 181) ▪ Steps 3-5: see below THE ONE-SAMPLE 𝑢-TEST FOR 𝜈

▪ Step 3

▪ under least extreme version of 𝐼0:

ത 𝑌−181 212.4/100 ~𝑢99

▪ no further assumptions required (because 𝑜 ≥ 30)

▪ Step 4

▪ value of test statistic 𝑢calc = −1.3037 ▪ critical value of 𝑢 from the table is 𝑢crit,lower,0.05,df=99 = − 1.660

▪ Step 5

▪ 𝑢calc ∉ 𝑆crit, so do not reject 𝐼0 ▪ conclude that there is no evidence for 𝜈 < 181

THE ONE-SAMPLE 𝑢-TEST FOR 𝜈

−1.660

Summing up: ▪ 𝜏𝑌 known:

▪ use 𝑨-table/perform 𝑨-test

▪ 𝜏𝑌 estimated by 𝑡𝑌:

▪ 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

• ne-sided hypothesis test

THE ONE-SAMPLE 𝑢-TEST FOR 𝜈

We test a hypothesis 𝐼0: 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.

EXERCISE 1

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

• ne-sample 𝑢-test

HYPOTHESES AND SPSS

Using SPSS for the one-sample 𝑢-test ▪ Claim: the average life expectancy is 68 year HYPOTHESES AND SPSS

𝐼0: 𝜈𝑌 = 68 𝑢calc Under 𝐼0: 𝑈~𝑢152 𝑞-value 𝑌=life expectancy

But look carefully: ▪ it is a two-sided test (SPSS says: two-tailed) HYPOTHESES AND SPSS

𝐼0: 𝜈𝑌 = 68 𝑞-value

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 𝐼0: 𝜈 ≥ 68 ▪ sample with 𝑜 = 153 yields ҧ 𝑦 = 64.515 and 𝑡 = 12.7937

HYPOTHESES AND SPSS

64.515 − 68 153 − 1

Two-sided ▪ 𝑄 ത 𝑌 ≤ 64.515 + 𝑄 ത 𝑌 ≥ 71.485 = 𝑄 ቀ ቁ 𝑢 ≤

64.515−68 12.7937/ 153 + 𝑄 𝑢 ≥ 71.485−68 12.7937/ 153 = 𝑄ሺ

ሻ 𝑢 ≤ −3.369 + 𝑄 𝑢 ≥ 3.369 ▪ use SPSS for 𝑢df=152: 𝑞 = 0.001 HYPOTHESES AND SPSS

68 + 68 − 64.515

One-sided (right-sided) and ҧ 𝑦 < 𝜈0 ▪ 𝑄 ത 𝑌 ≤ 64.515 = 𝑄 𝑢 ≤

64.515−68 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 < 𝜈0: ▪ divide by 2 𝑞right = 𝑞two/2 HYPOTHESES AND SPSS

One-sided (left-sided) and ҧ 𝑦 < 𝜈 ▪ 𝑄 ത 𝑌 ≥ 64.515 = 𝑄 𝑢 ≥

64.515−68 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 > 𝜈0: ▪ divide by 2 and subtract the result from 1 𝑞left = 1 − 𝑞two/2 HYPOTHESES AND SPSS

Another example ▪ Suppose ▪ 𝐼0: 𝜈 = 3 ▪ 𝛽 = 0.05 ▪ ҧ 𝑦 = 2.84 ▪ 𝑞−value = 2 × 𝑄

𝜈=3 ത

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 𝐼0: 𝜈 ≥ 3 and which when 𝐼0: 𝜈 ≤ 3? HYPOTHESES AND SPSS

𝑰𝟏: 𝝂 ≤ 𝟒 (right-sided) 𝑰𝟏: 𝝂 ≥ 𝟒 (left-sided) ҧ 𝑦 = 2.84 ( ҧ 𝑦 < 𝜈0) 𝑄

𝜈=3 ത

𝑌 ≥ 2.84 = 1 −

𝑞SPSS 2

𝑄

𝜈=3 ത

𝑌 ≤ 2.84 =

𝑞SPSS 2

ҧ 𝑦 = 3.16 ( ҧ 𝑦 > 𝜈0) 𝑄

𝜈=3 ത

𝑌 ≥ 3.16 =

𝑞SPSS 2

𝑄

𝜈=3 ത

𝑌 ≤ 3.16 = 1 −

𝑞SPSS 2

HYPOTHESES AND SPSS

Suppose we test a hypothesis on the mean 𝐼0: 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 2

Suppose we have the following result:

• A. reject 𝐼0: 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? EXERCISE 3

26 March 2015, Q3b OLD EXAM QUESTION

Doane & Seward 5/E 9.1-9.5 Tutorial exercises week 2

• ne-sided test in SPSS

FURTHER STUDY