Fundamentals of Hypothesis Testing: One-Sample Tests

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Fundamentals of Hypothesis Testing: One-Sample Tests Prof. dr. Siswanto Agus Wilopo, M.Sc., Sc.D. Department of Biostatics, Epidemiology and Population Health Faculty of Medicine Universitas Gadjah Mada

9/22/2017

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Learning Objectives In this lecture, you learn: •

The basic principles of hypothesis testing



How to use hypothesis testing to test a mean or proportion



The assumptions of each hypothesis-testing procedure, how to evaluate them, and the consequences if they are seriously violated



How to avoid the pitfalls involved in hypothesis testing



The ethical issues involved in hypothesis testing

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What is a Hypothesis? • A hypothesis is a claim (assumption) about a population parameter: – population mean Example: The mean of systolic blood pressure among adults of this city is μ = 120 mmHg

– population proportion Example: The proportion of adults in this city with hypertension is π = 0.118 9/22/2017

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The Null Hypothesis, H0 • States the claim or assertion to be tested Example: The average number of children in NTT woman is equal to three ( H0 : μ  3 )

• Is always about a population parameter, not about a sample statistic H0 : X  3

H0 : μ  3 9/22/2017

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The Null Hypothesis, H0 (continued)

• Begin with the assumption that the null hypothesis is true – Similar to the notion of innocent until proven guilty • Refers to the status quo • Always contains “=” , “≤” or “” sign • May or may not be rejected 9/22/2017

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The Alternative Hypothesis, H1 • Is the opposite of the null hypothesis – e.g., The average number of children in NTT woman is not equal to 3 ( H1: μ ≠ 3 )

• • • • 9/22/2017

Challenges the status quo Contains the “=” , “≤” or “” sign May or may not be proven Is generally the hypothesis that the researcher is trying to prove Biostatistics I: 2017-18

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Hypothesis Testing Process Claim: the population mean age is 50. (Null Hypothesis: H0: μ = 50 )

Population

Is X 20 likely if μ = 50? If not likely, REJECT Null Hypothesis 9/22/2017

Suppose the sample mean age is 20: X = 20 Biostatistics I: 2017-18

Now select a random sample

Sample 7

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Reason for Rejecting H0 Sampling Distribution of X

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If it is unlikely that we would get a sample mean of this value ... 9/22/2017

μ = 50 If H0 is true

... if in fact this were the population mean… Biostatistics I: 2017-18

X

... then we reject the null hypothesis that μ = 50. 8

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Level of Significance,  • Defines the unlikely values of the sample statistic if the null hypothesis is true – Defines rejection region of the sampling distribution

• Is designated by  , (level of significance) – Typical values are 0.01, 0.05, or 0.10

• Is selected by the researcher at the beginning • Provides the critical value(s) of the test 9/22/2017

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Level of Significance and the Rejection Region Level of significance =

H 0: μ = 3 H 1: μ ≠ 3



/2 Two-tail test

/2

 0

Upper-tail test

H0: μ ≥ 3 H1: μ < 3

 Lower-tail test

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Rejection region is shaded

0

H 0: μ ≤ 3 H 1: μ > 3

Represents critical value

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Errors in Making Decisions • Type I Error – Reject a true null hypothesis – Considered a serious type of error The probability of Type I Error is  The probability of Type I Error is  • Called level of significance of the test

• Set by the researcher in advance 9/22/2017

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Errors in Making Decisions (continued)

• Type II Error – Fail to reject a false null hypothesis The TypeIIIIError Errorisisβ β The probability of Type

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Outcomes and Probabilities Possible Hypothesis Test Outcomes Actual situation

Key: Outcome (Probability)

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Decision

H0 True

Do Not Reject H0

No error (1 -  )

Type II Error (β)

Reject H0

Type I Error ()

No Error (1-β)

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H0 False

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Type I & II Error Relationship  Type I and Type II errors cannot happen at the same time 

Type I error can only occur if H0 is true



Type II error can only occur if H0 is false If Type I error probability (  )

, then

Type II error probability ( β ) 9/22/2017

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Factors Affecting Type II Error • All else equal, – β when the difference between hypothesized parameter and its true value

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– β

when 

– β

when σ

– β

when n Biostatistics I: 2017-18

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Hypothesis Tests for the Mean Hypothesis Tests for 

 Known (Z test)

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 Unknown (t test)

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Z Test of Hypothesis for the Mean (σ Known) • Convert sample statistic ( X ) to a Z test statistic Hypothesis Tests for 

Known σKnown (Z test)

Unknown σUnknown (t test)

The test statistic is: Z 

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X μ σ n Biostatistics I: 2017-18

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Critical Value Approach to Testing • For a two-tail test for the mean, σ known: • Convert sample statistic ( X ) to test statistic (Z statistic )

• Determine the critical Z values for a specified level of significance  from a table or computer • Decision Rule: If the test statistic falls in the rejection region, reject H0 ; otherwise do not reject H0 9/22/2017

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Two-Tail Tests  There are two cutoff values (critical values), defining the regions of rejection

H0: μ = 3 H1: μ  3

/2

/2 X

3 Reject H0

-Z

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Lower critical value

Do not reject H0

0

Reject H0

+Z

Upper critical value

Z

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6 Steps in Hypothesis Testing 1. State the null hypothesis, H0 and the alternative hypothesis, H1 2. Choose the level of significance, , and the sample size, n 3. Determine the appropriate test statistic and sampling distribution 4. Determine the critical values that divide the rejection and nonrejection regions 9/22/2017

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6 Steps in Hypothesis Testing

(continued)

5. Collect data and compute the value of the test statistic 6. Make the statistical decision and state the managerial conclusion.    9/22/2017

If the test statistic falls into the non-rejection region, do not reject the null hypothesis H0. If the test statistic falls into the rejection region, reject the null hypothesis. Express the managerial conclusion in the context of the problem 21 Biostatistics I: 2017-18

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Hypothesis Testing Example Test the claim that the true mean of children in NTT woman is # equal to 3. (Assume σ = 0.8) 1. State the appropriate null and alternative hypotheses  H0: μ = 3 H1: μ ≠ 3 (This is a two-tail test) 2. Specify the desired level of significance and the sample size  Suppose that  = 0.05 and n = 100 are chosen for this test 9/22/2017

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Hypothesis Testing Example (continued)

3. Determine the appropriate technique  σ is known so this is a Z test. 4. Determine the critical values  For  = 0.05 the critical Z values are ±1.96 5. Collect the data and compute the test statistic  Suppose the sample results are n = 100, X = 2.84 (σ = 0.8 is assumed known) So the test statistic is: Z 

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Xμ 2.84  3  .16    2.0 σ 0.8 .08 n 100 Biostatistics I: 2017-18

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Hypothesis Testing Example (continued)

6. Is the test statistic in the rejection region?  = 0.05/2

Reject H0 if Z < -1.96 or Z > 1.96; otherwise do not reject H0 9/22/2017

 = 0.05/2

Reject H0

Do not reject H0

-Z= -1.96

0

Reject H0

+Z= +1.96

Here, Z = -2.0 < -1.96, so the test statistic is in the rejection region Biostatistics I: 2017-18

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Hypothesis Testing Example (continued)

6(continued). Reach a decision and interpret the result

 = 0.05/2

 = 0.05/2

Reject H0

Do not reject H0

-Z= -1.96

0

Reject H0

+Z= +1.96

-2.0

Since Z = -2.0 < -1.96, we reject the null hypothesis and conclude that there is sufficient evidence that the mean number of children of NTT woman is not equal to 3 9/22/2017

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p-Value Approach to Testing • p-value: Probability of obtaining a test statistic more extreme ( ≤ or  ) than the observed sample value given H0 is true – Also called observed level of significance – Smallest value of  for which H0 can be rejected 9/22/2017

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p-Value Approach to Testing (continued)

• Convert Sample Statistic (e.g., X ) to Test Statistic (e.g., Z statistic ) • Obtain the p-value from a table or computer • Compare the p-value with  – If p-value <  , reject H0 – If p-value   , do not reject H0 9/22/2017

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p-Value Example • Example:

How likely is it to see a sample mean of 2.84 (or something further from the mean, in either direction) if the true mean is  = 3.0?

X = 2.84 is translated to a Z score of Z = -2.0 P(Z  2.0)  0.0228 P(Z  2.0)  0.0228

/2 = 0.025

/2 = 0.025

0.0228

0.0228

p-value = 0.0228 + 0.0228 = 0.0456 9/22/2017

-1.96 -2.0

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1.96 2.0

Z 28

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p-Value Example • Compare the p-value with 

(continued)

– If p-value <  , reject H0 – If p-value   , do not reject H0 Here: p-value = 0.0456  = 0.05 Since 0.0456 < 0.05, we reject the null hypothesis

/2 = 0.025 0.0228

0.0228

-1.96 9/22/2017

/2 = 0.025

-2.0

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1.96 2.0

Z 29

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Connection to Confidence Intervals  For X = 2.84, σ = 0.8 and n = 100, the 95% confidence interval is: 0.8 2.84 - (1.96) to 100

0.8 2.84  (1.96) 100

2.6832 ≤ μ ≤ 2.9968  Since this interval does not contain the hypothesized mean (3.0), we reject the null hypothesis at  = 0.05 9/22/2017

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One-Tail Tests • In many cases, the alternative hypothesis focuses on a particular direction H0: μ ≥ 3 H1: μ < 3 H0: μ ≤ 3 H1: μ > 3 9/22/2017

This is a lower-tail test since the alternative hypothesis is focused on the lower tail below the mean of 3 This is an upper-tail test since the alternative hypothesis is focused on the upper tail above the mean of 3 Biostatistics I: 2017-18

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Lower-Tail Tests H0: μ ≥ 3  There is only one critical value, since the rejection area is in only one tail

H1: μ < 3

 Reject H0

-Z

Do not reject H0

Z

0

μ

X

Critical value 9/22/2017

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Upper-Tail Tests  There is only one critical value, since the rejection area is in only one tail

Z _ X

H0: μ ≤ 3 H1: μ > 3

 Do not reject H0

0



Reject H0

μ

Critical value 9/22/2017

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Example: Upper-Tail Z Test for Mean ( Known) An investigator thinks that annual health expenditure have increased, and now average over $52 per year. The investigator wishes to test this claim. (Assume  = 10 is known) Form hypothesis test: H0: μ ≤ 52 the average is not over $52 per month H1: μ > 52

the average is greater than $52 per year (i.e., sufficient evidence exists to support the invitagotor’s claim)

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Example: Find Rejection Region (continued)

• Suppose that  = 0.10 is chosen for this test Find the rejection region:

Reject H0

 = 0.10

Do not reject H0

0

1.28

Reject H0

Reject H0 if Z > 1.28 9/22/2017

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Review: One-Tail Critical Value Standardized Normal Distribution Table (Portion)

What is Z given  = 0.10?

0.90

0.10

 = 0.10 0.90

Z

.07

.08

.09

1.1 .8790 .8810 .8830 1.2 .8980 .8997 .9015

z

0 1.28

1.3 .9147 .9162 .9177

Critical Value = 1.28 9/22/2017

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Example: Test Statistic (continued)

Obtain sample and compute the test statistic Suppose a sample is taken with the following results: n = 64, X = 53.1 (=10 was assumed known) – Then the test statistic is: Xμ 53.1  52 Z   0.88 σ 10 n 64

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Example: Decision (continued)

Reach a decision and interpret the result: Reject H0

 = 0.10

Do not reject H0

1.28

0

Reject H0

Z = 0.88

Do not reject H0 since Z = 0.88 ≤ 1.28 9/22/2017

i.e.: there is not sufficient evidence that the mean bill is over $52 Biostatistics I: 2017-18

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p -Value Solution (continued)

Calculate the p-value and compare to  (assuming that μ = 52.0)

p-value = 0.1894 Reject H0  = 0.10

0 Do not reject H0

1.28

Reject H0

Z = 0.88

P( X  53.1) 53.1 52.0    P Z   10/ 64    P(Z  0.88)  1 0.8106  0.1894

Do not reject H0 since p-value = 0.1894 >  = 0.10 9/22/2017

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t Test of Hypothesis for the Mean (σ Unknown) • Convert sample statistic ( X ) to a t test statistic Hypothesis Tests for 

Known σKnown (Z test)

Unknown σUnknown (t test) The test statistic is:

t n-1 9/22/2017

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X μ  S n

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Example: Two-Tail Test ( Unknown) The average height of male in Indonesia is said to be 168 cm. A random sample of 25 male resulted in X = 172.50 cm and S = 15.40 cm. Test at the  = 0.05 level.

H0: μ = 168 H1: μ  168

(Assume the population distribution is normal) 9/22/2017

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Example Solution: Two-Tail Test H0: μ = 168 H1: μ  168 •  = 0.05

/2=.025

Reject H0

-t n-1,α/2 -2.0639

• n = 25 •  is unknown, so use a t statistic • Critical Value: t24 = ± 2.0639 9/22/2017

t n1 

/2=.025

Do not reject H0

Reject H0

0

1.46

t n-1,α/2 2.0639

X μ 172.50  168   1.46 S 15.40 n 25

Do not reject H0: not sufficient evidence that true mean height is different than 168 cm Biostatistics I: 2017-18

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Connection to Confidence Intervals  For X = 172.5, S = 15.40 and n = 25, the 95% confidence interval is: 172.5 - (2.0639) 15.4/ 25

to 172.5 + (2.0639) 15.4/ 25

166.14 ≤ μ ≤ 178.86  Since this interval contains the Hypothesized mean (168), we do not reject the null hypothesis at  = 0.05 9/22/2017

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Hypothesis Tests for Proportions • Involves categorical variables • Two possible outcomes – “Success” (possesses a certain characteristic) – “Failure” (does not possesses that characteristic)

• Fraction or proportion of the population in the “success” category is denoted by π 9/22/2017

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Proportions (continued)

• Sample proportion in the success category is denoted by p X number of successesin sample – p n  samplesize

• When both nπ and n(1-π) are at least 5, p can be approximated by a normal distribution with mean and standard deviation  (1   ) – σp  μp   n 9/22/2017

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Hypothesis Tests for Proportions • The sampling distribution of p is approximately normal, so the test statistic is a Z value:

Z

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pπ π(1 π ) n

Hypothesis Tests for p nπ  5 and n(1-π)  5

nπ < 5 or n(1-π) < 5 Not discussed in this chapter

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Z Test for Proportion in Terms of Number of Successes • An equivalent form to the last slide, but in terms of the number of successes, X:

X  n Z n (1  ) 9/22/2017

Hypothesis Tests for X X5 and n-X  5

X<5 or n-X < 5 Not discussed in this chapter

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Example: Z Test for Proportion A researcher claims that it receives 8% responses from its mailing. To test this claim, a random sample of 500 were surveyed with 25 responses. Test at the  = 0.05 significance level. 9/22/2017

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Check: n π = (500)(.08) = 40 n(1-π) = (500)(.92) = 460

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Z Test for Proportion: Solution Test Statistic: p  .05  .08 Z   2.47  (1  ) .08(1 .08)

H0: π = 0.08 H1: π  0.08  = 0.05 n = 500, p = 0.05

n

Decision:

Critical Values: ± 1.96 Reject

500

Reject

Reject H0 at  = 0.05

Conclusion: .025

.025 -1.96

-2.47 9/22/2017

0

1.96

There is sufficient evidence to reject the z researcher claim of 8% response rate. Biostatistics I: 2017-18

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p-Value Solution (continued)

Calculate the p-value and compare to  (For a two-tail test the p-value is always two-tail) Do not reject H0

Reject H0 /2 = .025

Reject H0

p-value = 0.0136:

/2 = .025

P(Z  2.47)  P(Z  2.47) 0.0068

0.0068 -1.96

Z = -2.47

0

 2(0.0068)  0.0136

1.96 Z = 2.47

Reject H0 since p-value = 0.0136 <  = 0.05 9/22/2017

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Potential Pitfalls and Ethical Considerations • Use randomly collected data to reduce selection biases • Do not use human subjects without informed consent • Choose the level of significance, α, and the type of test (one-tail or two-tail) before data collection • Do not employ “data snooping” to choose between one-tail and two-tail test, or to determine the level of significance • Do not practice “data cleansing” to hide observations that do not support a stated hypothesis • Report all pertinent findings 9/22/2017

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Chapter Summary • Addressed hypothesis testing methodology • Performed Z Test for the mean (σ known) • Discussed critical value and p–value approaches to hypothesis testing • Performed one-tail and two-tail tests 9/22/2017

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Chapter Summary (continued)

• Performed t test for the mean (σ unknown)

• Performed Z test for the proportion • Discussed pitfalls and ethical issues

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