statistics for business and economics chapter 7 inferences based on a single sample: tests of...

Statistics for Business and Economics

Chapter 7Inferences Based on a Single Sample: Tests

of Hypotheses

Content

1. The Elements of a Test of Hypothesis2. Formulating Hypotheses and Setting Up the Rejection

Region3. Observed Significance Levels: p-Values4. Test of Hypothesis about a Population Mean: Normal

(z) Statistic5. Test of Hypothesis about a Population Mean: Student’s

t-Statistic6. Large-Sample Test of Hypothesis about a Population

Proportion

7.1

The Elements ofa Test of Hypothesis

Hypothesis Testing

Population

I believe the population mean age is 50 (hypothesis).

Mean X = 20

Random sample

Reject hypothesis! Not close.

Reject hypothesis! Not close.

What’s a Hypothesis?

A statistical hypothesis is a statement about the numerical value of a population parameter.

I believe the mean GPA of this class is 3.5!

© 1984-1994 T/Maker Co.

Determining theTarget Parameter

Parameter Key Words or Phrases Type of Data

µ Mean; average Quantitative

p Proportion; percentage; fraction; rate

Qualitative

2

(Will not covered

in this course)

Variance; variability; spread

Quantitative

Null Hypothesis• The null hypothesis, denoted H0, represents the hypothesis that will

be “retained” unless the data provide convincing evidence that it is false. This usually represents the “status quo” or some claim about the population parameter that the researcher wants to test.

• You may think of null hypothesis as the “favored” hypothesis; we reject it in favor of the alternative hypothesis Ha if and only if the evidence provided by the sample data are strong against H0 and in favor of Ha .

• “retain H0” is commonly referred to as “do not reject”.• Stated in one of the following forms

H0: = some value)

H0: ≤ some value)

H0: some value)

Alternative Hypothesis

•The alternative (research) hypothesis, denoted Ha, represents the hypothesis that will be accepted only if the data provide convincing evidence of its truth. •This usually represents the values of a population parameter for which the researcher wants to gather evidence to support.

Alternative Hypothesis1. Opposite of null hypothesis2. The hypothesis that will be accepted only if

the data provide convincing evidence of its truth

3. Designated Ha 4. Stated in one of the following forms

Ha: some value)

Ha: some value)

Ha: some value)

Identifying Hypotheses

Example 1: If the hypothesis of a researcher is that the population mean is not 3, set-up the hypotheses to be tested.Steps:

• State the question statistically ≠ 3

• State the opposite statistically = 3

• State the null hypothesis statistically H0: = 3

• State the alternative hypothesis statistically Ha: ≠ 3


Example 2: If the hypothesis of a researcher is that the population mean is greater than 3, set-up the hypotheses to be tested.Steps:

• State the question statistically > 3

• State the opposite statistically ≤ 3

• State the null hypothesis statistically H0: ≤ 3

• State the alternative hypothesis statistically Ha: > 3


• State the question statistically: = 12

• State the opposite statistically: 12

• Select the alternative hypothesis: Ha: 12

• State the null hypothesis: H0: = 12

Example 3: Is the population average amount of TV viewing 12 hours?


• State the question statistically: 12

• State the opposite statistically: = 12


• State the null hypothesis: H0: = 12

Example 4: Is the population average amount of TV viewing different from 12 hours?





• State the null hypothesis: H0: 20

Example 5: Is the average cost per hat less than or equal to $20?





• State the null hypothesis: H0: 25

Example 6: Is the average amount spent in the bookstore greater than $25?

Test Statistic

The test statistic is a sample statistic, computed from information provided in the sample, that the researcher uses to decide between the null and alternative hypotheses.

Type I Error

• A Type I error occurs if the researcher rejects the null hypothesis in favor of the alternative hypothesis when, in fact, H0 is true.

• The probability of committing a Type I error is denoted by .

Example• Lets assume we would like to test

H0 : = 2400 versus Ha : > 2400 based on large sample of size n.• By the CLT, sample mean will follow an approximately normal distribution.• We will reject the null hypothesis in favor of Ha : > 2400 if the sample mean falls far

above 2400. • As we always did, by standardizing the value of we will have a random variable z which

has standard normal distribution so that we will use standard normal distribution (z) table.• So if we standardize , we will have

.• This statistic is our test statistic.• Then we will reject H0 : = 2400 in favor of Ha : > 2400 if z > critical value. • This area where z > critical value will be called “rejection region”.• Critical value will be determined from z-table based on the value of type I error.

Rejection RegionThe rejection region of a statistical test is the set of possible values of the test statistic for which the researcher will reject H0 in favor of Ha.

Type II Error

A Type II error occurs if the researcher retains the null hypothesis when, in fact, H0 is false. The probability of committing a Type II error is denoted by .

Conclusions and Consequences for a Test of Hypothesis

True State of Nature

Conclusion H0 True Ha True

Do not reject H0 (Assume H0 True)

Correct decision Type II error (probability )

Reject H0 (Assume Ha True)

Type I error (probability )

Correct decision

Elements of a Test of Hypothesis

1. Null hypothesis (H0): A theory about the specific values of one or more population parameters. The theory generally represents the status quo, which we adopt until it is proven false.

2. Alternative (research) hypothesis (Ha): A theory that contradicts the null hypothesis. The theory generally represents that which we will adopt only when sufficient evidence exists to establish its truth.


3. Test statistic: A sample statistic used to decide whether to reject the null hypothesis.

4. Rejection region: The numerical values of the test statistic for which the null hypothesis will be rejected. The rejection region is chosen so that the probability is that it will contain the test statistic when the null hypothesis is true, thereby leading to a Type I error. The value of is usually chosen to be small (e.g., .01, .05, or .10) and is referred to as the level of significance of the test.


5. Assumptions: Clear statement(s) of any assumptions made about the population(s) being sampled.

6. Experiment and calculation of test statistic: Performance of the sampling experiment and determination of the numerical value of the test statistic.


7. Conclusion:

a. If the numerical value of the test statistic falls in the rejection region, we reject the null hypothesis and conclude that the alternative hypothesis is true.

b. We know that the hypothesis-testing process will lead to this conclusion incorrectly (Type I error) only 100% of the time when H0 is true.


7. Conclusion:

b. If the test statistic does not fall in the rejection region, we do not reject H0. Thus, we reserve judgment about which hypothesis is true.

c. We do not conclude that the null hypothesis is true because we do not (in general) know the probability that our test procedure will lead to an incorrect acceptance of H0 (Type II error).

7.2

Setting Up the Rejection Region

Steps for Selecting the Null and Alternative Hypotheses

1. The rejection region is determined by alternative hypothesis.a. One-tailed, upper-tailed (e.g., Ha: µ > 2,400)

b. One-tailed, lower-tailed (e.g., Ha: µ < 2,400)

c. Two-tailed (e.g., Ha: µ ≠ 2,400)

One-Tailed Test

A one-tailed test of hypothesis is one in which the alternative hypothesis is directional and includes the symbol “ < ” or “ >.”

Two-Tailed Test

A two-tailed test of hypothesis is one in which the alternative hypothesis does not specify departure from H0 in a particular direction and is written with the symbol “ ≠.”

Rejection Region (One-Tail Test-lower tailed)

Ho

ValueCriticalValue

Sample Statistic

RejectionRegion

Fail to RejectRegion

Sampling Distribution

1 –

Level of Confidence

P(Z< )=

Rejection Regions (One-Tailed Test-Upper-tailed)

Ho

Value CriticalValue

Sample Statistic

RejectionRegion



1 –

Level of Confidence

P(Z> )=

Rejection Regions (Two-Tailed Test)

Ho

Value CriticalValue

CriticalValue

Sample Statistic

RejectionRegion

RejectionRegion



1 –

Level of Confidence

P(Z< )=/2 P(Z> )=/2

Rejection Regions (Two-Tailed level =0.05 Test)

Ho

Value CriticalValue=1.96

CriticalValue

/2=0.025a

Sample Statistic

RejectionRegion

RejectionRegion



1 – =0.95

Level of Confidence

/2=0.025a

Rejection Regions

Alternative Hypotheses

Lower-Tailed

Upper-Tailed

Two-Tailed

= .10 z < –1.282 z > 1.282 z < –1.645 or z > 1.645

= .05 z < –1.645 z > 1.645 z < –1.96 or z > 1.96

= .01 z < –2.326 z > 2.326 z < –2.575 or z > 2.575

7.3

Observed Significance Levels:p-Values

p-Value• Probability of obtaining a test statistic more extreme

(or than actual sample value, given H0 is true • Can be thought of as a measure of the “credibility” of

the null hypothesisH0 .

• is the nominal level of significance. This value is assumed by an analyst.

• p-value is also probability for making type-I error. • But, p-value is called “observed level of significance”.• It is used to make rejection decision

• If p-value , do not reject H0

• If p-value < , reject H0

• The p-value shows our confidence to reject null hypothesis.

• If this value is smaller than , then the probability that we will reject null hypothesis when it is true is even smaller than the maximum tolerated error probability, .

• So we can conclude that null hypothesis is wrong and can be rejected in favor of alternative hypothesis.

• The smaller the p-value is, the more confident we are with our decision to reject H0 .

Steps for Calculating the p-Value for a Test of Hypothesis

1. Determine the value of the test statistic z corresponding to the result of the sampling experiment.


2a. If the test is one-tailed, the p-value is equal to the tail area beyond z in the same direction as the alternative hypothesis. Thus, if the alternative hypothesis is of the form > , the p-value is the area to the right of, or above, the observed z-value. Conversely, if the alternative is of the form < , the p-value is the area to the left of, or below, the observed z-value.


2b. If the test is two-tailed, the p-value is equal to twice the tail area beyond the observed z-value in the direction of the sign of z – that is, if z is positive, the p-value is twice the area to the right of, or above, the observed z-value. Conversely, if z is negative, the p-value is twice the area to the left of, or below, the observed z-value.

Reporting Test Results asp-Values: How to Decide Whether to

Reject H0

1. Choose the maximum value of that you are willing to tolerate.

2. If the observed significance level (p-value) of the test is less than the chosen value of , reject the null hypothesis. Otherwise, do not reject the null hypothesis.

3. Decision is always the same as found with the level , critical value, rejection region approach.

4. Typical values for are 0.01, 0.05, 0.10.

Two-Tailed z Test p-Value Example

Does an average box of cereal contain 368 grams of cereal? A random sample of 25 boxes showed x = 372.5. The company has specified to be 15 grams. Find the p-value. How does it compare to = .05? 368 gm.

Two-Tailed z Test p-Value Solution

z0 1.50

z value of sample statistic (observed)

z x n

372.5 368

15

25

1.50

Two-Tailed Z Test p-Value Solution

1/2 p-Value1/2 p-Value

z value of sample statistic (observed)

p-Value is P(z –1.50 or z 1.50)

z0 1.50–1.50

From z table: lookup 1.50

.4332

.5000– .4332

.0668


1/2 p-Value.0668

1/2 p-Value.0668

p-Value is P(z –1.50 or z 1.50) = .1336

z0 1.50–1.50


0 1.50–1.50 z

Reject H0Reject H0

1/2 p-Value = .06681/2 p-Value = .0668

1/2 = .0251/2 = .025

p-Value = .1336 = .05 Do not reject H0.

Test statistic is in ‘Do not reject’ region

One-Tailed z Test p-Value Example

Does an average box of cereal contain more than 368 grams of cereal? A random sample of 25 boxes showed x = 372.5. The company has specified to be 15 grams. Find the p-value. How does it compare to = .05?

368 gm.

One-Tailed z Test p-Value Solution

z0 1.50

z value of sample statistic

z x n

372.5 368

15

25

1.50


Use alternative hypothesis to find direction

p-Value is P(z 1.50)


p-Value

z0 1.50


.4332

.5000– .4332

.0668


p-Value.0668




.5000– .4332

.0668

p-Value is P(z 1.50) = .0668

z0 1.50

.4332

= .05


0 1.50 z

Reject H0

p-Value = .0668

(p-Value = .0668) ( = .05). Do not reject H0.

Test statistic is in ‘Do not reject’ region

p-Value Thinking Challenge

You’re an analyst for Ford. You want to find out if the average miles per gallon of Escorts is less than 32 mpg. Similar models have a standard deviation of 3.8 mpg. You take a sample of 60 Escorts & compute a sample mean of 30.7 mpg. What is the p-value? How does it compare to = .01?


p-Value Solution*

z0–2.65z value of sample statistic From z table:

lookup 2.65

.4960

p-Value.004

.5000– .4960

.0040

p-Value is P(z -2.65) = .004.p-Value < ( = .01). Reject H0.

7.4

Test of Hypotheses about a Population Mean:Normal (z) Statistic

Large-Sample Test of Hypothesis about µ

One-Tailed Test Two-Tailed TestH0: µ = µ0 H0: µ = µ0

Ha: µ < µ0 Ha: µ ≠ µ0

(or Ha: µ > µ0)

Test Statistic: Test Statistic:z

x µ0

x

x µ0

s nz

x µ0

x

x µ0

s n


One-Tailed TestRejection region:

z < –z(or z > zwhen Ha: µ > µ0)

where z is chosen so that

P(z > z) =

You may also use p-value to give your decision.• If p-value , do not reject H0

• If p-value < , reject H0


Two-Tailed TestRejection region:

|z| > zwhere z is chosen so that

P(|z| > z) = /2

Note: µ0 is the symbol for the numerical value assigned to µ under the null hypothesis.

Conditions Required for a Valid Large-Sample Hypothesis Test for

µ

1. A random sample is selected from the target population.

2. The sample size n is large (i.e., n ≥ 30). (Due to the Central Limit Theorem, this condition guarantees that the test statistic will be approximately normal regardless of the shape of the underlying probability distribution of the population.)

Possible Conclusions for a Test of Hypothesis

1. If the calculated test statistic falls in the rejection region, reject H0 and conclude that the alternative hypothesis Ha is true. State that you are rejecting H0 at the level of significance. Remember that the confidence is in the testing process, not the particular result of a single test.

Possible Conclusions for a Test of Hypothesis

2. If the test statistic does not fall in the rejection region, conclude that the sampling experiment does not provide sufficient evidence to reject H0 at the level of significance. [Generally, we will not “accept” the null hypothesis unless the probability of a Type II error has been calculated.]

Two-Tailed z Test Example

Does an average box of cereal contain 368 grams of cereal? A random sample of 25 boxes had x = 372.5. The company has specified to be 25 grams. Test at the .05 level of significance.

368 gm.

Two-Tailed z Test Solution

• H0: • Ha: • • n • Critical Value(s):

Test Statistic:

Decision:

Conclusion:

= 368

368

.0525

z0 1.96–1.96

.025

Reject H0 Reject H0

.025

z x n

372.5 368

25

25

0.9

Do not reject at = .05

No evidence average is not 368

Two-Tailed z Test Thinking Challenge

You’re a Q/C inspector. You want to find out if a new machine is making electrical cords to customer specification: average breaking strength of 70 lb. with = 3.5 lb. You take a sample of 36 cords & compute a sample mean of 69.7 lb. At the .05 level of significance, is there evidence that the machine is not meeting the average breaking strength?

Two-Tailed z Test Solution*

• H0: • Ha: • = • n = • Critical Value(s):

Test Statistic:

Decision:

Conclusion:

= 70

70

.05

36

z0 1.96–1.96

.025

Reject H0 Reject H0

.025

z x n

69.7 70

3.5

36

.51


No evidence average is not 70

One-Tailed z Test Example

Does an average box of cereal contain more than 368 grams of cereal? A random sample of 25 boxes showed x = 372.5. The company has specified to be 25 grams. Test at the .05 level of significance.

368 gm.

One-Tailed z Test Solution

• H0:

• Ha: • = • n = • Critical Value(s):

Test Statistic:

Decision:

Conclusion:

= 368

> 368

.05

25

z0 1.645

.05

Reject

z x n

372.5 368

15

25

1.50


No evidence average is more than 368

One-Tailed z Test Thinking Challenge

You’re an analyst for Ford. You want to find out if the average miles per gallon of Escorts is at least 32 mpg. Similar models have a standard deviation of 3.8 mpg. You take a sample of 60 Escorts & compute a sample mean of 30.7 mpg. At the .01 level of significance, is there evidence that the miles per gallon is less than 32?

One-Tailed z Test Solution*

• H0:

• Ha: • = • n =• Critical Value(s):

Test Statistic:

Decision:

Conclusion:

= 32 < 32

.01

60

z0-2.33

.01

Reject

z x n

30.7 32

3.8

60

2.65

Reject at = .01

There is evidence average is less than 32

7.5

Test of Hypothesis about a Population Mean:

Student’s t-Statistic

Small-Sample Test of Hypothesis about µ

One-Tailed TestH0: µ = µ0

Ha: µ < µ0 (or Ha: µ > µ0)

Test statistic:

Rejection region: t < –t (or t > t when Ha: µ > µ0) where t and t are based on (n – 1) degrees of freedom

t x s n

Small-Sample Test of Hypothesis about µ

Two-Tailed TestH0: µ = µ0

Ha: µ ≠ µ0

Test statistic:

Rejection region: |t| > t

t x s n

Conditions Required for a Valid Small-Sample Hypothesis Test for

µ

1. A random sample is selected from the target population.

2. The population from which the sample is selected has a distribution that is approximately normal.

Two-Tailed t Test Example

Does an average box of cereal contain 368 grams of cereal? A random sample of 36 boxes had a mean of 372.5 and a standard deviation of 12 grams. Test at the .05 level of significance.

368 gm.

Two-Tailed t Test Solution

• H0: • Ha: • = • df = • Critical Value(s):

Test Statistic:

Decision:

Conclusion:

= 368 368

.05

36 – 1 = 35

t0 2.030-2.030

.025

Reject H0 Reject H0

.025

t x

s

n

372.5 368

12

36

2.25

Reject at = .05

There is evidence population average is not 368

Two-Tailed t TestThinking Challenge

You work for the FTC. A manufacturer of detergent claims that the mean weight of detergent is 3.25 lb. You take a random sample of 64 containers. You calculate the sample average to be 3.238 lb. with a standard deviation of .117 lb. At the .01 level of significance, is the manufacturer correct?

3.25 lb.

Two-Tailed t Test Solution*

• H0: • Ha: • • df • Critical Value(s):

Test Statistic:

Decision:

Conclusion:

= 3.25

3.25

.01

64 – 1 = 63

t0 2.656-2.656

.005

Reject H0 Reject H0

.005

t x

s

n

3.238 3.25

.117

64

.82


There is no evidence average is not 3.25

One-Tailed t TestExample

Is the average capacity of batteries less than 140 ampere-hours? A random sample of 20 batteries had a mean of 138.47 and a standard deviation of 2.66. Assume a normal distribution. Test at the .05 level of significance.

One-Tailed t Test Solution

• H0:

• Ha: • =• df =• Critical Value(s):

Test Statistic:

Decision:

Conclusion:

= 140

< 140

.0520 – 1 = 19

t0-1.729

.05

Reject H0

t x

s

n

138.47 140

2.66

20

2.57

Reject at = .05

There is evidence population average is less than 140

One-Tailed t Test Thinking Challenge

You’re a marketing analyst for Wal-Mart. Wal-Mart had teddy bears on sale last week. The weekly sales ($ 00) of bears sold in 10 stores was: 8 11 0 4 7 8 10 5 8 3 At the .05 level of significance, is there evidence that the average bear sales per store is more than 5 ($ 00)?

One-Tailed t Test Solution*

• H0:

• Ha: • = • df = • Critical Value(s):

Test Statistic:

Decision:

Conclusion:

= 5 > 5

.05

10 – 1 = 9

t0 1.833

.05

Reject H0

t x

s

n

6.4 53.373

10

1.31


There is no evidence average is more than 5

7.6

Large-Sample Test of Hypothesis about a Population Proportion

Large-Sample Test of Hypothesis about p

One-Tailed TestH0: p = p0

Ha: p < p0 (or Ha: p > p0)

Test statistic:

Rejection region: z < –z(or z > z when Ha: p > p0)Note: p0 is the symbol for the numerical value of p assigned in the null hypothesis

z p̂ p0

p̂

where p̂ p0q0 n

q0 1 p0

Large-Sample Test of Hypothesis about p

Two-Tailed TestH0: p = p0

Ha: p ≠ p0

Test statistic:

Rejection region: |z| < z

Note: p0 is the symbol for the numerical value of p assigned in the null hypothesis

z p̂ p0

p̂

where p̂ p0q0 n

q0 1 p0

Conditions Required for a Valid Large-Sample Hypothesis Test for

p

1. A random sample is selected from a binomial population.

2. The sample size n is large. (This condition will be satisfied if both np0 ≥ 15 and nq0 ≥ 15.)

One-Proportion z Test Example

The present packaging system produces 10% defective cereal boxes. Using a new system, a random sample of 200 boxes had11 defects. Does the new system produce fewer defects? Test at the .05 level of significance.

One-Proportion z Test Solution

• H0:

• Ha: • = • n =• Critical Value(s):

Test Statistic:

Decision:

Conclusion:

p = .10

p < .10

.05

200

z0-1.645

.05

Reject H0

0

0 0

11.10ˆ 200 2.12

.10 .90

200

p pz

p q

n

Reject at = .05

There is evidence new system < 10% defective

One-Proportion z Test Thinking Challenge

You’re an accounting manager. A year-end audit showed 4% of transactions had errors. You implement new procedures. A random sample of 500 transactions had 25 errors. Has the proportion of incorrect transactions changed at the .05 level of significance?

One-Proportion z Test Solution*

• H0:

• Ha: • = • n = • Critical Value(s):

Test Statistic:

Decision:

Conclusion:

p = .04

p .04

.05

500

z0 1.96-1.96

.025

Reject H 0 Reject H0

.025

0

0 0

25.04ˆ 500 1.14

.04 .96

500

p pz

p q

n


There is evidence proportion is not 4%

statistics for business and economics chapter 7 inferences based on a single sample: tests of...

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