Chapter 10 – Hypothesis Testing

• What is a hypothesis?

A statement about a population that may or may not be true.

• What is hypothesis testing?A statistical test to prove or disprove a hypothesis. At the end of the test, either the hypothesis is rejected or not rejected.

Steps in hypothesis testing

• There are 5 steps in hypothesis testing:

• Step 1 – pattern of population distribution (normal)

• Step 2 – Formulation of hypothesis

Null Hypothesis (H0)Alternate Hypothesis (H1)

Steps in hypothesis testing

• Step 3 - Level of significance (α )

Alpha is the probability of rejecting a null hypothesis when it is true. Alpha is also known as the level of risk. It is described in terms of percent or decimals (5% or 0.05, 1% or 0.01, and so on.)

• Step 4 - Test statisticIt is a quantity that is used to compare with critical Z from appendix D to determine if a null hypothesis is to be rejected or not.

Steps in hypothesis testing• This quantity is the z-statistic _ Z = (x – ) / ( n)

. Step 5 - Decision Rule:

It is a rule by which a null hypothesis is rejected or not rejected.

It always takes the form:

Reject H0 if Z-statistic is > Critical Z or < - Critical Z

Critical Z-value is obtained from appendix D.

Steps in hypothesis testing

Decision: Reject H0 or Do not reject H0

Hypothesis tests can be one-tailed or two-tailed depending on how the alternate hypothesis is written.

Let’s take an example• Suppose:

H : μ = 100 0 H : μ > 100 1 This is a one tailed test.

• Another example: H : μ = 100 0 H : μ < 100 1

An example of a two tailed test

• Suppose:

H : μ = 100 0 H : μ ≠ 100 1

• This is a two tailed test.• Exercises from book: Problem 7(P age 239),

Problem 8 (Page 240), Problem 9 (Page 240)

Exercises from Book

• Problem 7 (pg 283):• a. ONE-TAILED b. ONE-TAILED • c. ONE-TAILED d. ONE-TAILED• e. TWO-TAILED • Problem 8 (pg 283):• a. Ho : μ = 500, H1 : μ ≠ 500• b. Ho : μ = 500, H1 : μ > 500• c. Ho : μ = 500, H1 : μ < 500

Exercises from book

• Problem 9 (pg 283)

• Ho : μ = 60; H1 : μ > 60

• Problem 10 (pg 283)

• Ho : μ = 1000; H1 : μ ≠ 1000

• Ho : μ ≤ 1000; H1 : μ > 1000

Testing for population mean

• Such testing can be made under 2 situations: (1) When population standard deviation is known and (2) when population standard deviation is unknown. This chapter discusses the first.Refer to example problem 10-2 (page 241)

• Two hypotheses are: H : μ = 10 0 H : μ < 10 1

Testing for population mean

• Alpha ( α ) is 5% (or 0.05).

Critical Z from appendix D is –1.645

• Reject H0 if Z-statistic < - 1.645

Test statistic (Z-statistic) _ X – μ 8.8 - 10

Z = -------- = ---------- = - 2.0 σ / √ n 2.4 / 4

Testing for population mean

Decision:

Reject H0

• What does it mean? The new manufacturing process reduces tar content of cigarettes.

• Example problem 10-4.

Testing for population mean

H : μ = 1000 0H : μ > 1000 1α = 0.025

Critical Z = 1.96 (From Appendix D) _ X – μ 1050 - 1000Test Statistic Z = --------- = -------------- = 2.5 σ / n 100 / 5Decision: Reject H 0What does it mean? The new manufacturing process

increases life of fuses.

This is a two tailed test Example Problem 10-7 (page 248) H : μ = 100 0

H : μ ≠ 100 1

Critical Z’s are – 2.575 and + 2.575

Decision Rule:

Reject H0 if Z-statistic > + 2.575 or < - 2.575 _ X - 100 Z-statistic = -------------- = 1.667 Decision: Do not reject H √ (225 / 25) 0 IQs of the school district students are no different from the national average

Example problem 13• Problem 13 (page 251)

H : μ ≥ 30 0 H : μ < 30 1Critical Z = - 2.326

Decision Rule: Reject H0 if Z-statistic ≤ - 2.326 27 - 30Z-Statistic = ------------- = - 5.0 6 / √ 100

Decision: Reject H0

• What does it mean? Vendor’s claim that mean weight of his chickens is at least 30 ounces is rejected.

Testing for difference between 2 means

• These tests are performed when one wants to compare 2 groups of populations.

• For example, is these a difference between average GPAs of students of two universities? Or, are average monthly incomes the same between two groups of people?

• The testing procedure is similar to what has been presented in the previous section.

Testing for difference between 2 means

• The null hypothesis is always of the form:

H : μ = μ • 0 1 2•

The alternative hypothesis can be either of the 3 forms:

• μ > μ or μ < μ or μ ≠ μ

1 2 1 2 1 2

Testing for difference between 2 means

Formula 10-1, p. 252

The decision rule is the same as before

Example problem 10-8 (page 252-253)

H : μ = μ H : μ ≠ μ

0 1 2 1 1 2

Testing for difference between 2 means

0.8 – 1.0• Z- statistic = --------------------------- = - 1.0 √ (0.36/25) + (0.64/25)

• At α = 0.05, critical Z = -1.96 and +1.96

Decision: Do not reject H 0• Interpretation: The avg. nicotine contents of two

brands of cigarettes are equal.

Problem 4, pg 255

• Is average hourly output of male workers less than that of female workers?

• This is a test involving difference of means between two population groups.

• Data are given in the table

Ho: μ1 = μ2; H1: μ1 < μ2

(μ1 indicates output of males)

• Z-statistic =

(150-153)/√[(70/36) + (74/36)]

= -1.5

• At α = 0.05, critical Z is -1.645

• Decision: Do not reject Ho

• Interpretation: Average outputs of male and female workers are the same.

Testing for population proportion

• In some cases, we are interested in population proportions rather than population means. For example, are more than 50% of CSULB students females? Or, is a new medicine more than 60% effective?

• Two hypothesis are:

H : p = a given value 0 H : p > a given value OR 1 p < a given value OR

p ≠ a given value

Example 10-11 (Page 257-258)

The test statistic

• Z-statistic = (X – n.p)/√[(n.p(1-p)]

• The decision rule is the same as before.

Example problem 10-12 (pg 259)

• A TV program attracts 50% audience. A new anchorperson has been hired. Does the new hire increase audience level? Use 5% level of significance.

• Ho: p = 0.5; H1: p > 0.5

• Given n = 100; critical Z = 1.645

• P = 55/100

Problem 10-12, cont.

• Z-statistic = (55-50)/√[(100)(0.5)(0.5)] = 1

• Decision: Do not reject Ho

• Interpretation:– The audience level has remained the same

(at 50%) after hiring the new anchorperson.

Problem 6, pg 261

• An auto manufacturer claims that 20% of customers prefer his products. In a sample of 100 customers, 15 indicate their preferences for his products. At 5% level of significance, can we support his claim?

• Ho: p = 0.2; H1: p ≠ 0.2

Problem 6, cont.

• Z-statistic = (15-20)/√[(20)(0.08)] = -1.25

• Decision: Do not reject Ho.

• Interpretation – The auto manufacturer’s claim cannot be

disputed.

Hypothesis Testing

• Strength of rejection (p-value)– When we reject a null hypothesis, there is a

quantity that describes the strength of rejection. This quantity is called the p-value.

– The lower the p-value, the greater the strength of refection. In other words, the lower the p-value, the greater the strength in the rejection.

What is p-value?

• It is the area to right right of the calculated value of Z-statistic when the area of rejection is on the right side of the normal curve. When the area of rejection is on the left side of the normal curve, the p-value is the area to the left of the calculated value of the Z-statistic. If there are two area of rejection, add the area to the right of Z-statistic and the area to the left of Z-statistic.

How do we find the P-value?

• Let’s take an example problem:• Test if the mean waiting time is less than 3

minutes.• α = 0.05

• μ = 3, σ = 1, n=50, X=2.75• Ho: μ = 3; H1: μ < 3• Decision rule: Reject Ho if z-statistic < critical Z• Z-statistic = (2.75 - 3)/(1/√50) = -1.768• Critical Z - -1.645 so we reject Ho.• Interpretation: Mean waiting time is < 3 minutes.

What is the p-value?

• Find the area to the left of -1.768.

• This area is the p-value.

• P-value = 1.0 – 0.9616 = 0.0384