Fundamentals of Hypothesis Testing: One-Sample Tests

İŞL 276

What is a Hypothesis?

A hypothesis is a claim (assumption) about a population parameter:

population mean

population proportion

Example: The mean monthly cell phone bill of this city is μ = $42

Example: The proportion of adults in this city with cell phones is π = 0.68

The Null Hypothesis, H0

States the claim or assertion to be tested

Example: Average number of mobile phone in a

Turkish family is equal to 3 ( )

Is always about a population parameter, not about a sample statistic

3μ:H0

3μ:H0 3X:H0

The Null Hypothesis, H0

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

(continued)

The Alternative Hypothesis, H1

Is the opposite of the null hypothesis e.g., The average number of TV sets in U.S.

homes is not equal to 3 ( H1: μ ≠ 3 )

Challenges the status quo Never contains the “=” , “≤” or “” sign May or may not be proven Is generally the hypothesis that the researcher is

trying to prove

Population

Claim: thepopulationmean age is 50.(Null Hypothesis:

REJECT

Supposethe samplemean age

is 20: X = 20Sample

Null Hypothesis

20 likely if μ = 50?Is

Hypothesis Testing Process

If not likely,

Now select a random sample

H0: μ = 50 )

X

Sampling Distribution of X

μ = 50If H0 is true

If it is unlikely that we would get a sample mean of

this value ...

... then we reject the null

hypothesis that μ = 50.

Reason for Rejecting H0

20

... if in fact this were the population mean…

X

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

Level of Significance and the Rejection Region

H0: μ ≥ 3

H1: μ < 30

H0: μ ≤ 3 H1:

μ > 3

Represents critical value

Lower-tail test

Level of significance =

0Upper-tail test

Two-tail test

Rejection region is shaded

/2

0

/2H0: μ = 3

H1: μ ≠ 3

Hypothesis Tests for the Mean

Known Unknown

Hypothesis Tests for

(Z test) (t test)

Z Test of Hypothesis for the Mean (σ Known)

Convert sample statistic ( ) to a Z test statistic X

The test statistic is:

n

σμX

Z

σ Known σ Unknown

Hypothesis Tests for

Known Unknown(Z test) (t test)

Critical Value Approach to Testing

For a two-tail test for the mean, σ known:

Convert sample statistic ( ) to test statistic (Z statistic )

Determine the critical Z values for a specifiedlevel 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

X

Do not reject H0 Reject H0Reject H0

There are two cutoff values

(critical values), defining the regions of rejection

Two-Tail Tests

/2

-Z 0

H0: μ = 3

H1: μ 3

+Z

/2

Lower critical value

Upper critical value

3

Z

X

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

6 Steps in Hypothesis Testing

5. Collect data and compute the value of the test statistic

6. Make the statistical decision and state the managerial conclusion. If the test statistic falls into the nonrejection 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

(continued)

Hypothesis Testing Example

Test the claim that the true mean of mobile phone in a Turkish family is equal to 3. Assume σ = 0.8, α=0.05 and sample results

are n = 100, = 2.84

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

X

2.0.08

.16

100

0.832.84

n

σμX

Z

Hypothesis Testing Example

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:

(continued)

Reject H0 Do not reject H0

6. Is the test statistic in the rejection region?

= 0.05/2

-Z= -1.96 0Reject H0 if Z < -1.96 or Z > 1.96;

otherwise do not reject H0

Hypothesis Testing Example

(continued)

= 0.05/2

Reject H0

+Z= +1.96

Here, Z = -2.0 < -1.96, so the test statistic is in the rejection region

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

-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 phone in Turkish family is not equal to 3

Hypothesis Testing Example

(continued)

Reject H0 Do not reject H0

= 0.05/2

-Z= -1.96 0

= 0.05/2

Reject H0

+Z= +1.96

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

p-Value Approach to Testing

Convert Sample Statistic (e.g., ) 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

X

(continued)

0.0228

/2 = 0.025

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?

-1.96 0

-2.0

0.02282.0)P(Z

0.02282.0)P(Z

Z1.96

2.0

X = 2.84 is translated to a Z score of Z = -2.0

p-value

= 0.0228 + 0.0228 = 0.0456

0.0228

/2 = 0.025

Compare the p-value with

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

(continued)

p-Value Example

0.0228

/2 = 0.025

-1.96 0

-2.0

Z1.96

2.0

0.0228

/2 = 0.025

Connection to Confidence Intervals

• For sample mean is 2.84 and σ = 0.8 and n = 100, the 95% confidence interval is:

2.6832 ≤ μ ≤ 2.9968

Since this interval does not contain the hypothesized mean (3.0), we reject the null hypothesis at = 0.05

100

0.8 (1.96) 2.84 to

100

0.8 (1.96) - 2.84

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

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

Reject H0 Do not reject H0

There is only one

critical value, since the

rejection area is in

only one tail

Lower-Tail Tests

-Z 0

μ

H0: μ ≥ 3

H1: μ < 3

Z

X

Critical value

Reject H0Do not reject H0

Upper-Tail Tests

Zα0

μ

H0: μ ≤ 3

H1: μ > 3 There is only one

critical value, since the

rejection area is in

only one tail

Critical value

Z

X_

Ex:Upper-Tail Z Test for Mean ( Known)

A phone industry manager thinks that customer monthly cell phone bills have increased, and now average over $52 per month. The company wishes to test this claim. (Assume = 10 is known and α=0.10)

H0: μ ≤ 52 the average is not over $52 per month

H1: μ > 52 the average is greater than $52 per month(i.e., sufficient evidence exists to support the

manager’s claim)

Form hypothesis test:

Reject H0Do not reject H0

Suppose that = 0.10 is chosen for this test

Find the rejection region:

= 0.10

1.280

Reject H0

Reject H0 if Z > 1.28

Example: Find Rejection Region

(continued)

Review:One-Tail Critical Value

Z .07 .09

1.1 .8790 .8810 .8830

1.2 .8980 .9015

1.3 .9147 .9162 .9177z 0 1.28

.08

Standardized Normal Distribution Table (Portion)What is Z given = 0.10?

= 0.10

Critical Value = 1.28

0.90

.8997

0.10

0.90

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:

0.88

64

105253.1

n

σμX

Z

Example: Test Statistic

(continued)

Reject H0Do not reject H0

Example: Decision

= 0.10

1.280

Reject H0

Do not reject H0 since Z = 0.88 ≤ 1.28

i.e.: there is not sufficient evidence that the mean bill is over $52

Z = 0.88

Reach a decision and interpret the result:(continued)

Reject H0

= 0.10

Do not reject H0 1.28

0

Reject H0

Z = 0.88

Calculate the p-value and compare to

(assuming that μ = 52.0)

(continued)

0.1894

0.810610.88)P(Z

6410/

52.053.1ZP

53.1)XP(

p-value = 0.1894

p -Value Solution

Do not reject H0 since p-value = 0.1894 > = 0.10

t Test of Hypothesis for Mean (σ Unknown)

Convert sample statistic ( ) to a t test statistic X

The test statistic is:

n

SμX

t 1-n

Hypothesis Tests for

σ Known σ Unknown Known Unknown(Z test) (t test)

Example: Two-Tail Test( Unknown)

The average cost of a hotel room in New York is said to be $168 per night. A random sample of 25 hotels resulted in X = $172.50 and

s = $15.40. Test at the

= 0.05 level.(Assume the population distribution is normal)

H0: μ= 168

H1: μ

168

= 0.05

n = 25

is unknown, so use a t statistic

Critical Value:

t24 = ± 2.0639

Example Solution: Two-Tail Test

Do not reject H0: not sufficient evidence that true mean cost is different than $168

Reject H0Reject H0

/2=.025

-t n-1,α/2

Do not reject H0

0

/2=.025

-2.0639 2.0639

1.46

25

15.40168172.50

n

SμX

t 1n

1.46

H0: μ= 168

H1: μ

168t n-1,α/2

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

Proportions

Sample proportion in the success category is denoted by p

When both nπ and n(1-π) are at least 5, p can be approximated by a normal distribution with mean and standard deviation

sizesample

sampleinsuccessesofnumber

n

Xp

pμn

)(1σ

p

(continued)

The sampling distribution of p is approximately normal, so the test statistic is a Z value:

Hypothesis Tests for Proportions

n)(1

pZ

ππ

π

nπ 5and

n(1-π) 5

Hypothesis Tests for p

nπ < 5or

n(1-π) < 5

Not discussed in this chapter

An equivalent form to the last slide, but in terms of the number of successes, X:

Z Test for Proportion

)(1n

nXZ

X 5and

n-X 5

Hypothesis Tests for X

X < 5or

n-X < 5

Not discussed in this chapter

Example: Z Test for Proportion

A marketing company 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.

Check:

n π = (500)(.08) = 40

n(1-π) = (500)(.92) = 460

Z Test for Proportion: Solution

= 0.05

n = 500, p = 0.05

Reject H0 at = 0.05

H0: π = 0.08

H1: π 0.08

Critical Values: ± 1.96

Test Statistic:

Decision:

Conclusion:

z0

Reject Reject

.025.025

1.96

-2.47

There is sufficient evidence to reject the

company’s claim of 8% response rate.

2.47

500.08).08(1

.08.05

n)(1

pZ

-1.96

Do not reject H0

Reject H0Reject H0

/2 = .025

1.960

Z = -2.47

Calculate the p-value and compare to (For a two-tail test the p-value is always two-tail)

(continued)

0.01362(0.0068)

2.47)P(Z2.47)P(Z

p-value = 0.0136:

p-Value Solution

Reject H0 since p-value = 0.0136 < = 0.05

Z = 2.47

-1.96

/2 = .025

0.00680.0068

44

Example 1 Price/earnings ratios for stocks.

Theory: stable rate of P/E in market = 13.If P/E (market) < 13, you should invest in the

stock market.If P/E (market) > 13, you should take your

money out. We have a sample of 50

= 12.1. Historical σ = 3.0456

Can we estimate if the population P/E is 13 or not?

45

Estimate Steps Common steps:

Set hypothesis:H0: μ = 13HA: μ ≠ 13

Select = .05.

46

Calculating Test Statistic

4307.050

0456.3Error Standard x

n

0896.24307.0

131.12-x value

x

0

z

47

p-value Approach Calculate the p-value.

We will calculate for the lower tail→ then make an adjustment for the upper tail.

48

p-value, Two-Tailed Test

0 Z=2.09 zZ=–2.09

p(z < –2.09) = ??

p(z > 2.09) = ??

We can just calculate one value, and double it.

49

Calculating the p-value cont’d Find 2.090 on the Standard Normal Distribution tables:

z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09

…

1.9

2.0 .4772 .4778 .4783 .4788 .4793 .4798 .4803 .4808 .4812 .4817

2.1

…

50

p-value, Two-Tailed Test

0 Z=2.09 zZ=–2.09

p(z < –2.09) = ??

p(z > 2.09) = 0.5 -.4817

= 0.0183

Doubling the value, we find the p-value = 0.0366

0.4817

51

Should We Reject the Null Hypothesis? Yes!

p-value = 0.0366 < = 0.05. There is only a 3.66% chance that the measured

price/earnings ratio sample mean of 12.1 is not equal to the stable rate of 13 by random chance.

Example #2:

You want to test the hypothesis that a treatment conducted at the ward you are working on in a hospital is more beneficial than the average treatment used in other wards of the hospital.Given that the mean wellness score of the people on your ward is 87 (N=20), and that the mean for the entire hospital is 76 and the standard deviation of population is 15, is your ward significantly better?

State Ho and H1

State if you’re using a one- or two-tailed test and why. State the p-value that supports your claim.

Hypothesis Testing w/ One Sample

Smaller Portion = p < .0006

3.27 3.36

11

20

157687

z

Example A family therapist states that parents talk to their

teenagers an average of 27 minutes per week. Surprised by that claim, a psychologist decided to collect some data on the amount of time parents spend in conversation with their teenage children. For the n = 12 parents, the study revealed that following times (in minutes) devoted to conversation in a week:

Example Do the psychologists findings differ significantly

from the therapist’s claim? If so, is the family expert’s claim an overestimate or underestimate of actual time spent talking to children? Use the 0.05 level of significance with two tails.

Mean = 24.58, s2 = 12.24, (s/√n) = 1.00

29 22 19 25 27 28

21 22 24 26 30 22

Hypothesis Testing w/ One Sample Example #2:

H0 = μ = 27

Critical t (df = 11) = ±2.201 Reject H0, and conclude that the data are

significantly different from the therapist’s claim

57

J) Hypothesis Tests, 2 Means: ’s Unknown Two datasets –> is the mean value of one larger than

the other? Is it larger by a specific amount?

μ1 vs. μ2 –> μ1 – μ2 vs. D0.

Often set D0 = 0 –> is μ1 = μ2?

58

Example: Female vs. Male Salaries Saskatchewan 2001 Census data:

- only Bachelor’s degrees- aged 21-64- work full-time- not in school

Men: M = $46,452.48, sM = 36,260.1, nM = 557.

Women: W = $35,121.94, sW = 20,571.3, nW = 534.

M – W = $11,330.44 } our point estimate. Is this an artifact of the sample, or do men

make significantly more than women?

59

Hypothesis, Significance Level, Test Statistic Research hypothesis: men get paid more:

1. H0: μM – μW < 0H1: μM – μW > 0

2. Select = 0.05

3. Compute test t-statistic:

375.6

534426178351

5571314794928

0)94.3512148.46452()(22

0

W

W

M

M

WM

ns

ns

Dxxt

60

4. a. Compute the Degrees of Freedom Can compute by hand, or get from Excel: Critical value of t=1.65

22 21 2

1 22 22 2

1 2

1 1 2 2

1 11 1

s sn n

dfs s

n n n n

22 21 2

1 22 22 2

1 2

1 1 2 2

1 11 1

s sn n

dfs s

n n n n

= 888

61

4. b. Computing the p-value Degrees of Freedom 0.20 0.10 0.02 0.025 0.01 0.005

…

100 .845 1.290 1.660 1.984 2.364 2.626

.842 1.282 1.645 1.960 2.326 2.576

6.375 up here somewhere

The p-value <<< 0.005.

62

5. Check the Hypothesis Since the p-value is <<< 0.05, we reject H0.

We conclude that we can accept the alternative hypothesis that men get paid more than women at a very high level of confidence (greater than 99%).

63

Excel t-Test: Two-Sample Assuming Unequal Variances

  Male Female

Mean 46452.47935 35121.94195

Variance 1314794928 423178351.3

Observations 557 534

Hypothesized Mean Diff. 0

df 888

t Stat 6.381034789

P(T<=t) one-tail 1.41441E-10 .00000000014

t Critical one-tail 1.646571945

P(T<=t) two-tail 2.82883E-10

t Critical two-tail 1.962639544