Business Statistics CONTENTS A hypothesis test Hypotheses - - PowerPoint PPT Presentation

HYPOTHESES: LOGIC AND FRAMEWORK

Business Statistics

A hypothesis test Hypotheses Rejection region and significance level Five-step procedure for hypothesis tests More on hypotheses Old exam question Further study CONTENTS

▪ Suppose a beverage company wants to test if its bottles are filled with 1 liter

▪ more than 1 liter: not competitive ▪ less than 1 liter: trouble with the consumers association

▪ They take a random sample of 9 bottles

▪ and find ҧ 𝑦 = 1.02 liter

▪ Can they claim 𝜈 = 1 liter? ▪ Assume:

▪ population is normally distributed ▪ population has standard deviation 𝜏 = 0.003 liter ▪ so: 𝑌~𝑂 𝜈 =? , 𝜏 = 0.003

A HYPOTHESIS TEST

If all assumptions (including 𝜈 = 1!) are true: ▪ The sampling distribution of the mean ( ത 𝑌)

▪ is normal ▪ has mean 𝜈 ത

𝑌 = 𝜈𝑌 = 1

▪ has standard deviation 𝜏 ത

𝑌 = 𝜏𝑌 9 = 0.003 3

= 0.001

▪ So, there is a probability of finding a sample mean ത 𝑌 = 1.02 or even larger, given by

▪ 𝑄𝑂 ത 𝑌 ≥ 1.02 = 𝑄𝑎

ത 𝑌−1 0.001 ≥ 1.02−1 0.001

= 𝑄𝑎 𝑎 ≥ 20 = 0.000 ≈ 0% ▪ very very unlikely!

▪ So, you can reject the claim 𝜈𝑌 = 1 with high confidence! A HYPOTHESIS TEST

Or even larger? We’ll go into that soon.

Now, suppose you had found ҧ 𝑦 = 1.002 liter ▪ There is a probability of finding a sample mean ത 𝑌 = 1.002

• r even larger, given by

▪ 𝑄𝑂 ത 𝑌 ≥ 1.002 = 𝑄𝑎

ത 𝑌−1 0.001 ≥ 1.002−1 0.001

= 𝑄𝑎 𝑎 ≥ 2 = 0.02275 ≈ 2.3% ▪ not very likely, but it may certainly happen now and then

▪ So, you can reject the claim 𝜈𝑌 = 1 with some confidence

▪ but you know that there is some chance to make the wrong decision

▪ Or: you can decide to not reject the claim 𝜈𝑌 = 1

▪ because you know that it may still be true, despite the data

A HYPOTHESIS TEST

In general a hypothesis is an unproven assertion ▪ In statistics:

▪ a hypothesis is a claim about a (population!) parameter

▪ Examples:

▪ the mean monthly cell phone bill of this city is 42$ is (𝜈 = $42) ▪ the proportion of adults in this city with an iPhone is at least 0.68 (𝜌 ≥ 0.68) ▪ the variance of spending on fashion for men is not smaller than that for women (𝜏men

2

≥ 𝜏women

2

) ▪ the median life expectancy is the same for all three income groups (𝑁1 = 𝑁2 = 𝑁3)

HYPOTHESES

Statistical hypotheses have the following aspects: ▪ A (population!) parameter

▪ 𝜈 , 𝜌 , 𝜏2 , etc.

▪ In case of one-sample: a benchmark

▪ 𝜈 = 181 , 𝜌 ≤ 0.2 , etc.

▪ In case of several samples: a comparison

▪ 𝜈1 = 𝜈2 , 𝜌1 − 𝜌2 ≤ 0.2 , 𝜏1

2 = 𝜏2 2 = 𝜏3 2 , etc.

A hypothesis test is a decision between two competing mutually exclusive and collectively exhaustive hypotheses about the value(s) of the parameter(s) HYPOTHESES

▪ Examples of a hypothesis test:

▪ 𝐼0: 𝜈 = 181 versus 𝐼1: 𝜈 ≠ 181 ▪ 𝐼0: 𝜌 ≤ 0.2 versus 𝐼1: 𝜌 > 0.2

▪ Terminology

▪ 𝐼0 is the null hypothesis (on which the test focuses) ▪ 𝐼1 is the alternative hypothesis

▪ We focus on 𝐼0

▪ so if we reject 𝐼0, we automatically accept 𝐼1 ▪ while if we do not reject 𝐼0, we “maintain” 𝐼0 (but do not reject 𝐼1 and do not accept 𝐼0)

HYPOTHESES

A government official wants to proudly announce that unemployment is under 4%. Which hypothesis should he test? EXERCISE 1

▪ Example:

▪ 𝐼0: 𝜈 = 181 versus 𝐼1: 𝜈 ≠ 181

▪ We collect data and perform the hypothesis test ▪ Two possible outcomes:

▪ reject 𝐼0, so accept 𝐼1, and conclude 𝜈 ≠ 181 ▪ do not reject 𝐼0, and conclude that there is no evidence to reject 𝜈 = 181

▪ Whatever the decision is, you may be wrong

▪ there is sampling variation ▪ you may always have an exceptional sample ▪ example: if you want to test if a coin is fair, it may happen that you have only “heads” in your sample, even if the coin is fair!

REJECTION REGION AND SIGNIFICANCE LEVEL

▪ Between rejecting and not rejecting, there is a boundary ▪ This boundary defines the risk you are prepared to take

▪ if you want to test if a coin is fair, and you use a sample of size 20, how many “heads” will induce you to reject the null hypothesis (𝜌 = 0.5)?

▪ You will determine a rejection region

▪ for instance: you will reject the null hypothesis (𝜌 = 0.5) when you obtain 5 heads or fewer, or 15 heads or more

▪ You use a pre-established significance level to determine the boundaries of the rejection region REJECTION REGION AND SIGNIFICANCE LEVEL

▪ So, you define a significance level

▪ conventional symbol 𝛽 ▪ often taken to be 0.05 ▪ but also 0.1, 0.01, 0.005, 0.001, etc are used often

▪ There is a close link between

▪ the confidence level (1 − 𝛽, as used in a confidence interval) ▪ and a significance level (𝛽, as used in a hypothesis test) ▪ confidence level+significance level=1

REJECTION REGION AND SIGNIFICANCE LEVEL

▪ Suppose we have a sample and want to see if it comes from a distribution with mean 𝜈0

▪ assuming normality of the population ▪ assuming a known value for 𝜏 ▪ testing 𝜈 = 𝜈0 at a significance level 𝛽 = 5%

▪ We want to determine boundary values for ത 𝑌 such that the claim 𝜈 = 𝜈0 becomes unlikely

▪ upper boundary: 𝑄 ത 𝑌 ≥ ҧ 𝑦upper = 0.025 ▪ lower boundary: 𝑄 ത 𝑌 ≤ ҧ 𝑦lower = 0.025

▪ If the value of the test statistic is in the rejection region

▪ so if ҧ 𝑦data ≤ 𝜈0 − 1.96𝜏 ത

𝑌 or ҧ

𝑦data ≥ 𝜈0 + 1.96𝜏 ത

𝑌

▪ we reject 𝐼0: 𝜈 = 𝜈0 and accept 𝐼1: 𝜈 ≠ 𝜈0

REJECTION REGION AND SIGNIFICANCE LEVEL

So, we distribute the 𝛽 = 5% equally at both sides

Rejection region for non-standardized statistic ത 𝑌 (𝛽 = 0.05) REJECTION REGION AND SIGNIFICANCE LEVEL

Reject H0 Do not reject H0

𝛽 2 = 0.025 ҧ 𝑦crit = 𝜈0 − 1.96𝜏 ത

𝑌

Reject H0

𝛽 2 = 0.025 𝑨crit = 𝜈0 + 1.96𝜏 ത

𝑌

1 − 𝛽 = 0.95 𝜈0

▪ The rejection region in this test is defined by the boundary values 𝜈0 − 1.96𝜏 ത

𝑌 and 𝜈0 + 1.96𝜏 ത 𝑌

▪ But we can also standardize the test statistic, and focus on 𝑎 =

ത 𝑌−𝜈0 𝜏ഥ

𝑌 rather than on ത

𝑌 ▪ The rejection region in this test is defined by the boundary values −1.96 and 1.96 ▪ If the value of your standardized test statistic is in the rejection region

▪ so if 𝑨data ≤ −1.96 or 𝑨data ≥ 1.96 ▪ reject 𝐼0: 𝜈 = 𝜈0 and accept 𝐼1: 𝜈 ≠ 𝜈0

REJECTION REGION AND SIGNIFICANCE LEVEL

Rejection region for standardized statistic 𝑎 =

ത 𝑌−𝜈0 𝜏ഥ

𝑌

(𝛽 = 0.05) REJECTION REGION AND SIGNIFICANCE LEVEL

Reject H0 Do not reject H0

𝛽 2 = 0.025 𝑨crit = −1.96

Reject H0

𝛽 2 = 0.025 𝑨crit = +1.96 1 − 𝛽 = 0.95

Suppose we test a hypothesis on the mean 𝐼0: 𝜈 = 310 with significance level 𝛽 = 0.05. We sample data, and calculate a test statistic 𝑢 = −2.13. What do we conclude? EXERCISE 2

Five-step procedure ▪ step 1: state the hypotheses and the significance level ▪ step 2: choose a sample statistic and determine the rejection region (qualitatively) ▪ step 3: determine the null distribution, and state and/or check the requirements needed ▪ step 4: calculate the value of the test statistic and its critical value(s) ▪ step 5: draw conclusions FIVE-STEP PROCEDURE FOR HYPOTHESIS TESTS

These steps are done somewhat differently in every book and course. Never mind, all elements reappear.

Using an example about the mean body height 𝜈𝑌 of a population with 𝜏𝑌

2 = 225 cm2

▪ On the basis of a sample of size 𝑜 = 100 with ҧ 𝑦 = 179.1 cm Step 1 ▪ State the hypotheses and the significance level

▪ null hypothesis 𝐼0: 𝜈𝑌 = 181 ▪ alternative hypothesis 𝐼1: 𝜈𝑌 ≠ 181 ▪ significance level 𝛽 = 0.05

FIVE-STEP PROCEDURE FOR HYPOTHESIS TESTS

Step 2 ▪ Choose a sample statistic and determine the rejection region (qualitatively)

▪ sample statistic: sample mean ത 𝑌 ▪ because the hypothesis is about 𝜈𝑌 ▪ rejection region: reject 𝐼0 when ҧ 𝑦 is “too small” or “too large” ▪ because both situations suggest that 𝐼0 is probably wrong

FIVE-STEP PROCEDURE FOR HYPOTHESIS TESTS

Step 3 ▪ Determine the null distribution, and state and/or validate the requirements needed

• A. sampling distribution of ത

𝑌 under 𝐼0: ത 𝑌~𝑂 181,

225 100

▪ or even better: 𝑎 =

ത 𝑌−181 225/100 ~𝑂 0,1

▪ where the sample statistic ത 𝑌 is transformed into a standardized test statistic 𝑎

• B. requirements: because 𝑜 = 100 ≥ 30, the sampling distribution
• f ത

𝑌 will indeed be approximately normal ▪ no additional assumptions are needed

FIVE-STEP PROCEDURE FOR HYPOTHESIS TESTS

Step 4 ▪ Calculate the value of the test statistic and its critical value(s)

▪ value of 𝑎 calculated from the data is

179.1−181 225/100 = −1.267

▪ we write this as 𝑨calc = −1.267 ▪ critical values of 𝑎 from the table are 𝑨crit,lower,0.025 = −1.96 and 𝑨crit,upper,0.025 = 1.96 ▪ rejection region for 𝑎 is 𝑆crit = −∞, −1.96 ∪ [1.96, ∞)

FIVE-STEP PROCEDURE FOR HYPOTHESIS TESTS

−1.96 +1.96 −1.267

Step 5 ▪ Draw conclusions

▪ 𝑨calc ∉ 𝑆crit ▪ the probability of obtaining a sample mean of 179.1 (or even more extreme) while the population mean is 181 exceeds 0.05 ▪ do not reject 𝐼0 ▪ the data sampled are compatible with a population with 𝜈𝑌 = 181 ▪ this does not mean that you reject 𝐼1, and neither that you accept 𝐼0 ▪ it is the weak conclusion

FIVE-STEP PROCEDURE FOR HYPOTHESIS TESTS

Suppose you want to test 𝐼0: 𝜏 ≤ 3. What would be step 2

• f the 5-step procedure?

▪ remember: choose the sample statistic and determine the rejection region (qualitatively)

EXERCISE 3

Alternative five-step procedure, using 𝑞-values ▪ step 1: identical ▪ step 2: identical ▪ step 3: identical ▪ step 4’: calculate the 𝑞-value of the test statistic ▪ step 5’: almost identical FIVE-STEP PROCEDURE FOR HYPOTHESIS TESTS

Step 4’ ▪ Calculate the 𝑞-value of the test statistic

▪ probability of obtaining ҧ 𝑦 = 179.1 or even smaller: 𝑄ሺ ) ത 𝑌 ≤ 179.1 = 𝑄 𝑎 ≤

179.1−181 225/100

= 𝑄 𝑎 ≤ −1.267 = 0.1026 ▪ probability of obtaining ҧ 𝑦 = 182.9 or even larger: 𝑄ሺ ) ത 𝑌 ≥ 182.9 = 𝑄 𝑎 ≥

182.9−181 225/100

= 𝑄 𝑎 ≥ 1.267 = 0.1026 ▪ 𝑞-value: 0.1026 + 0.1026 = 0.2052

FIVE-STEP PROCEDURE FOR HYPOTHESIS TESTS

Why 182.9? It is equally far from 𝜈0 = 181, so equally “extreme” ... So, 𝑞-value is 𝑄൫ ൯ ሺ ) ത 𝑌 ≤ 179.1 ∪ ത 𝑌 ≥ 182.9 |𝐼0

Step 5’ ▪ Draw conclusions

▪ 𝑞−value = 0.2052 > 𝛽 = 0.05 ▪ do not reject 𝐼0 ▪ the data sampled are compatible with a population with 𝜈𝑌 = 181 ▪ this does not mean that you reject 𝐼1, and neither that you accept 𝐼0 ▪ it is the weak conclusion

FIVE-STEP PROCEDURE FOR HYPOTHESIS TESTS

So: ▪ Two possible implementations of the testing procedure

▪ standardize on the basis of the hypothesized distribution, compute 𝑨calc and a rejection region 𝑆calc, and determine if 𝑨calc ∈ 𝑆crit (this was the first procedure using a critical region) ▪ calculate the probability of finding 𝑨calc or even more extreme, and determine if it is smaller than or equal to the significance level 𝛽 (this is the second procedure, using 𝑞-values)

▪ Both methods yield the same conclusion ▪ You should be able to apply both methods FIVE-STEP PROCEDURE FOR HYPOTHESIS TESTS

There is always a pair of hypotheses ▪ Example 1:

▪ you doubt if a coin is fair ▪ 𝐼0: 𝜌 = 0.5 ▪ 𝐼1: 𝜌 ≠ 0.5

▪ Example 2:

▪ you are designing a building and you think that the mean body height is less than 181 cm ▪ 𝐼0: 𝜈 ≥ 181 ▪ 𝐼1: 𝜈 < 181

MORE ON HYPOTHESES

General rules: ▪ 𝐼0 and 𝐼1 are complementary (𝐼1

′ = 𝐼0)

▪ 𝐼0 contains an =-sign (=, ≥, or ≤) ▪ 𝐼1 is the statement we would like to establish (hmmm...) Note: some books and some programs use a different strategy ▪ 𝐼0 containing only an =-sign ▪ 𝐼1 being either ≠, <, or > ▪ so 𝐼0 and 𝐼1 not necessarily complementary MORE ON HYPOTHESES

▪ Testing a two-sided hypothesis

▪ also called an undirected hypothesis ▪ example: 𝐼0: 𝜈 = 181 ▪ two-tailed rejection region (reject for “too small” and for “too large” values of ത 𝑌)

▪ Testing a one-sided hypothesis

▪ also called a directed hypothesis ▪ example: 𝐼0: 𝜈 ≥ 181 ▪ two types: left-sided and right-sided ▪ one-tailed rejection region (reject for “too small” or for “too large” values of ത 𝑌, depending on the direction)

MORE ON HYPOTHESES

Possible one-sided hypotheses: Right-sided hypothesis ▪ 𝐼0: 𝜈 ≤ 𝜈0 vs. 𝐼1: 𝜈 > 𝜈0 ▪ rejection region is at the right Left-sided hypothesis ▪ 𝐼0: 𝜈 ≥ 𝜈0 vs. 𝐼1: 𝜈 < 𝜈0 ▪ rejection region is at the left MORE ON HYPOTHESES

It is easiest to remember this by thinking of 𝐼1: if 𝐼1 contains “<” the rejection region is on the left We often write 𝜈0 for the mean according to the null hypothesis

Where to locate the rejection region (grey area) ▪ for 𝐼0: 𝜈 ≥ 𝜈𝑌 (left-sided) ▪ for 𝐼0: 𝜈 = 𝜈𝑌 (two-sided) ▪ for 𝐼0: 𝜈 ≤ 𝜈𝑌 (right-sided) MORE ON HYPOTHESES

Not precise enough in the book:

• ne-sided and two-sided

What about one-sided vs. one-tailed and two-sided vs. two-tailed? ▪ Suppose the hypothesis is 𝐼0: 𝜈 = 𝜈0

▪ two-sided, because we will reject when ത 𝑌 is too small and when ത 𝑌 is too large ▪ we might do a test on

ത 𝑌−𝜈 something and then reject when the test statistic

falls in either side of the distribution ▪ but we might alternatively do a test on

ത 𝑌−𝜈 2 something and then reject when

the test statistic falls in the right side of the distribution

▪ So, a two-sided hypothesis might lead to a one-tailed rejection region

▪ we will see this later when we study the so-called 𝜓2-test ▪ no need to bother now

MORE ON HYPOTHESES

Example of one-sided hypothesis test Context:

▪ you are designing a building and you think that the mean body height of the population is less than 181 cm ▪ do a hypothesis test at 𝛽 = 5% ▪ sample of size 𝑜 = 100 has ҧ 𝑦 = 179.1 cm ▪ it is known that 𝜏𝑌

2 = 225 cm2

▪ can you be (statistically speaking) confident that 𝜈 < 181?

▪ Use five-step procedure MORE ON HYPOTHESES

▪ Step 1

▪ 𝐼0: 𝜈 ≥ 181, 𝐼1: 𝜈 < 181, 𝛽 = 0.05

▪ Step 2

▪ sample statistic: ത 𝑌 ▪ reject for “too small values”

▪ Step 3

▪ under least extreme version of 𝐼0:

179.1−181 225/100 ~𝑂 0,1

▪ no further requirements (because 𝑜 ≥ 30)

▪ Step 4

▪ value of test statistic 𝑨calc = −1.267 ▪ 𝑨crit,0.05 = −1.645, so critical region 𝑆crit = ሺ−∞, −1.645]

▪ Step 5

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

MORE ON HYPOTHESES

The same problem, using 𝑞-values ▪ Steps 1-3

▪ identical

▪ Step 4’

▪ 𝑞-value: 𝑄 ത 𝑌 ≤ 179.1 = 𝑄

ത 𝑌−𝜈ഥ

𝑌

𝜏ഥ

𝑌

≤

179.1−181 225/100

= 𝑄 𝑎 ≤ −1.267 = 0.1026

▪ Step 5’

▪ 𝑞−value=0.1026 > 𝛽 = 0.05, so do not reject 𝐼0 ▪ conclude that there is no evidence against 𝜈 ≥ 181

MORE ON HYPOTHESES

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 𝑞-value 0.37. What do we conclude? EXERCISE 4

27 May 2014, Q1n OLD EXAM QUESTION

Doane & Seward 5/E 9.1-9.5 Tutorial exercises week 2 hypothesis tests null and alternative testing one sample mean FURTHER STUDY