View
219
Download
1
Category
Preview:
Citation preview
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.
Recommended