Hypothesis Testing with One Sample

1
Chapter 7

• Hypothesis Testing with One Sample

2
Chapter Outline

• 7.1 Introduction to Hypothesis Testing
• 7.2 Hypothesis Testing for the Mean (Large
  Samples)
• 7.3 Hypothesis Testing for the Mean (Small
  Samples)
• 7.4 Hypothesis Testing for Proportions
• 7.5 Hypothesis Testing for Variance and Standard
  Deviation

3
Section 7.1

• Introduction to Hypothesis Testing

4
Section 7.1 Objectives

• State a null hypothesis and an alternative
  hypothesis
• Identify type I and type I errors and interpret
  the level of significance
• Determine whether to use a one-tailed or
  two-tailed statistical test and find a p-value
• Make and interpret a decision based on the
  results of a statistical test
• Write a claim for a hypothesis test

5
Hypothesis Tests

• Hypothesis test
• A process that uses sample statistics to test a
  claim about the value of a population parameter.
  
• For example An automobile manufacturer
  advertises that its new hybrid car has a mean
  mileage of 50 miles per gallon. To test this
  claim, a sample would be taken. If the sample
  mean differs enough from the advertised mean, you
  can decide the advertisement is wrong.

6
Hypothesis Tests

• Statistical hypothesis
• A statement, or claim, about a population
  parameter.
• Need a pair of hypotheses
• one that represents the claim
• the other, its complement
• When one of these hypotheses is false, the other
  must be true.

7
Stating a Hypothesis

• Null hypothesis
• A statistical hypothesis that contains a
  statement of equality such as ?, , or ?.
• Denoted H0 read H subzero or H naught.

• Alternative hypothesis
• A statement of inequality such as gt, ?, or lt.
• Must be true if H0 is false.
• Denoted Ha read H sub-a.

8
Stating a Hypothesis

• To write the null and alternative hypotheses,
  translate the claim made about the population
  parameter from a verbal statement to a
  mathematical statement.
• Then write its complement.

H0 µ k Ha µ gt k
H0 µ k Ha µ lt k
H0 µ k Ha µ ? k

• Regardless of which pair of hypotheses you use,
  you always assume µ k and examine the sampling
  distribution on the basis of this assumption.

9
Example Stating the Null and Alternative
Hypotheses

• Write the claim as a mathematical sentence. State
  the null and alternative hypotheses and identify
  which represents the claim.
• A university publicizes that the proportion of
  its students who graduate in 4 years is 82.

Solution
H0 Ha
Equality condition
(Claim)
p 0.82
Complement of H0
p ? 0.82
10
Example Stating the Null and Alternative
Hypotheses

• Write the claim as a mathematical sentence.
  State the null and alternative hypotheses and
  identify which represents the claim.
• A water faucet manufacturer announces that the
  mean flow rate of a certain type of faucet is
  less than 2.5 gallons per minute.

Solution
H0 Ha
Complement of Ha
µ 2.5 gallons per minute
Inequality condition
(Claim)
µ lt 2.5 gallons per minute
11
Example Stating the Null and Alternative
Hypotheses

• Write the claim as a mathematical sentence.
  State the null and alternative hypotheses and
  identify which represents the claim.
• A cereal company advertises that the mean weight
  of the contents of its 20-ounce size cereal boxes
  is more than 20 ounces.

Solution
H0 Ha
µ 20 ounces
Complement of Ha
Inequality condition
(Claim)
µ gt 20 ounces
12
Types of Errors

• No matter which hypothesis represents the claim,
  always begin the hypothesis test assuming that
  the equality condition in the null hypothesis is
  true.
• At the end of the test, one of two decisions will
  be made
• reject the null hypothesis
• fail to reject the null hypothesis
• Because your decision is based on a sample, there
  is the possibility of making the wrong decision.

13
Types of Errors

• A type I error occurs if the null hypothesis is
  rejected when it is true.
• A type II error occurs if the null hypothesis is
  not rejected when it is false.

14
Example Identifying Type I and Type II Errors

• The USDA limit for salmonella contamination for
  chicken is 20. A meat inspector reports that the
  chicken produced by a company exceeds the USDA
  limit. You perform a hypothesis test to determine
  whether the meat inspectors claim is true. When
  will a type I or type II error occur? Which is
  more serious? (Source United States Department
  of Agriculture)

15
Solution Identifying Type I and Type II Errors

• Let p represent the proportion of chicken that is
  contaminated.

H0 Ha
Hypotheses
p 0.2
p gt 0.2
(Claim)
16
Solution Identifying Type I and Type II Errors
H0 Ha
Hypotheses
p 0.2
(Claim)
p gt 0.2
A type I error is rejecting H0 when it is true.
The actual proportion of contaminated chicken is
less than or equal to 0.2, but you decide to
reject H0.
A type II error is failing to reject H0 when it
is false.
The actual proportion of contaminated chicken is
greater than 0.2, but you do not reject H0.
17
Solution Identifying Type I and Type II Errors
H0 Ha
Hypotheses
p 0.2
(Claim)
p gt 0.2

• With a type I error, you might create a health
  scare and hurt the sales of chicken producers who
  were actually meeting the USDA limits.
• With a type II error, you could be allowing
  chicken that exceeded the USDA contamination
  limit to be sold to consumers.
• A type II error could result in sickness or even
  death.

18
Level of Significance

• Level of significance
• Your maximum allowable probability of making a
  type I error.
• Denoted by ?, the lowercase Greek letter alpha.
• By setting the level of significance at a small
  value, you are saying that you want the
  probability of rejecting a true null hypothesis
  to be small.
• Commonly used levels of significance
• ? 0.10 ? 0.05 ? 0.01
• P(type II error) ß (beta)

19
Statistical Tests

• After stating the null and alternative hypotheses
  and specifying the level of significance, a
  random sample is taken from the population and
  sample statistics are calculated.
• The statistic that is compared with the parameter
  in the null hypothesis is called the test
  statistic.

20
P-values

• P-value (or probability value)
• The probability, if the null hypothesis is true,
  of obtaining a sample statistic with a value as
  extreme or more extreme than the one determined
  from the sample data.
• Depends on the nature of the test.

21
Nature of the Test

• Three types of hypothesis tests
• left-tailed test
• right-tailed test
• two-tailed test
• The type of test depends on the region of the
  sampling distribution that favors a rejection of
  H0.
• This region is indicated by the alternative
  hypothesis.

22
Left-tailed Test

• The alternative hypothesis Ha contains the
  less-than inequality symbol (lt).

H0 µ ? k Ha µ lt k
23
Right-tailed Test

• The alternative hypothesis Ha contains the
  greater-than inequality symbol (gt).

H0 µ k Ha µ gt k
24
Two-tailed Test

• The alternative hypothesis Ha contains the not
  equal inequality symbol (?). Each tail has an
  area of ½P.

H0 µ k Ha µ ? k
25
Example Identifying The Nature of a Test

• For each claim, state H0 and Ha. Then determine
  whether the hypothesis test is a left-tailed,
  right-tailed, or two-tailed test. Sketch a normal
  sampling distribution and shade the area for the
  P-value.
• A university publicizes that the proportion of
  its students who graduate in 4 years is 82.

Solution
p 0.82
H0 Ha
p ? 0.82
Two-tailed test
26
Example Identifying The Nature of a Test

• For each claim, state H0 and Ha. Then determine
  whether the hypothesis test is a left-tailed,
  right-tailed, or two-tailed test. Sketch a normal
  sampling distribution and shade the area for the
  P-value.
• A water faucet manufacturer announces that the
  mean flow rate of a certain type of faucet is
  less than 2.5 gallons per minute.

Solution
µ 2.5 gpm
H0 Ha
µ lt 2.5 gpm
Left-tailed test
27
Example Identifying The Nature of a Test

• For each claim, state H0 and Ha. Then determine
  whether the hypothesis test is a left-tailed,
  right-tailed, or two-tailed test. Sketch a normal
  sampling distribution and shade the area for the
  P-value.
• A cereal company advertises that the mean weight
  of the contents of its 20-ounce size cereal boxes
  is more than 20 ounces.

Solution
µ 20 oz
H0 Ha
µ gt 20 oz
Right-tailed test
28
Making a Decision

• Decision Rule Based on P-value
• Compare the P-value with ?.
• If P ? ?, then reject H0.
• If P gt ?, then fail to reject H0.

There is enough evidence to reject the claim
There is enough evidence to support the claim
There is not enough evidence to reject the claim
There is not enough evidence to support the claim
29
Example Interpreting a Decision

• You perform a hypothesis test for the following
  claim. How should you interpret your decision if
  you reject H0? If you fail to reject H0?
• H0 (Claim) A university publicizes that the
  proportion of its students who graduate in 4
  years is 82.

30
Solution Interpreting a Decision

• The claim is represented by H0.
• If you reject H0 you should conclude there is
  sufficient evidence to indicate that the
  universitys claim is false.
• If you fail to reject H0, you should conclude
  there is insufficient evidence to indicate that
  the universitys claim (of a four-year graduation
  rate of 82) is false.

31
Example Interpreting a Decision

• You perform a hypothesis test for the following
  claim. How should you interpret your decision if
  you reject H0? If you fail to reject H0?
• Ha (Claim) Consumer Reports states that the mean
  stopping distance (on a dry surface) for a Honda
  Civic is less than 136 feet.

• Solution
• The claim is represented by Ha.
• H0 is the mean stopping distanceis greater than
  or equal to 136 feet.

32
Solution Interpreting a Decision

• If you reject H0 you should conclude there is
  enough evidence to support Consumer Reports
  claim that the stopping distance for a Honda
  Civic is less than 136 feet.
• If you fail to reject H0, you should conclude
  there is not enough evidence to support Consumer
  Reports claim that the stopping distance for a
  Honda Civic is less than 136 feet.

33
Steps for Hypothesis Testing

• State the claim mathematically and verbally.
  Identify the null and alternative hypotheses.
• H0 ? Ha ?
• Specify the level of significance.
• a ?
• Determine the standardized sampling distribution
  and draw its graph.
• Calculate the test statisticand its standardized
  value.Add it to your sketch.

34
Steps for Hypothesis Testing

• Find the P-value.
• Use the following decision rule.
• Write a statement to interpret the decision in
  the context of the original claim.

Is the P-value less than or equal to the level of
significance?
Fail to reject H0.
Reject H0.
35
Section 7.1 Summary

• Stated a null hypothesis and an alternative
  hypothesis
• Identified type I and type I errors and
  interpreted the level of significance
• Determined whether to use a one-tailed or
  two-tailed statistical test and found a p-value
• Made and interpreted a decision based on the
  results of a statistical test
• Wrote a claim for a hypothesis test

36
Section 7.2

• Hypothesis Testing for the Mean (Large Samples)

37
Section 7.2 Objectives

• Find P-values and use them to test a mean µ
• Use P-values for a z-test
• Find critical values and rejection regions in a
  normal distribution
• Use rejection regions for a z-test

38
Using P-values to Make a Decision

• Decision Rule Based on P-value
• To use a P-value to make a conclusion in a
  hypothesis test, compare the P-value with ?.
• If P ? ?, then reject H0.
• If P gt ?, then fail to reject H0.

39
Example Interpreting a P-value

• The P-value for a hypothesis test is P 0.0237.
  What is your decision if the level of
  significance is
• 0.05?
• 0.01?

SolutionBecause 0.0237 lt 0.05, you should
reject the null hypothesis.
Solution Because 0.0237 gt 0.01, you should fail
to reject the null hypothesis.
40
Finding the P-value

• After determining the hypothesis tests
  standardized test statistic and the test
  statistics corresponding area, do one of the
  following to find the P-value.
• For a left-tailed test, P (Area in left tail).
• For a right-tailed test, P (Area in right
  tail).
• For a two-tailed test, P 2(Area in tail of test
  statistic).

41
Example Finding the P-value

• Find the P-value for a left-tailed hypothesis
  test with a test statistic of z -2.23. Decide
  whether to reject H0 if the level of significance
  is a 0.01.

SolutionFor a left-tailed test, P (Area in
left tail)
P 0.0129
Because 0.0129 gt 0.01, you should fail to reject
H0
42
Example Finding the P-value

• Find the P-value for a two-tailed hypothesis test
  with a test statistic of z 2.14. Decide whether
  to reject H0 if the level of significance is a
  0.05.

SolutionFor a two-tailed test, P 2(Area in
tail of test statistic)
1 0.9838 0.0162
P 2(0.0162) 0.0324
0.9838
Because 0.0324 lt 0.05, you should reject H0
43
Z-Test for a Mean µ

• Can be used when the population is normal and ?
  is known, or for any population when the sample
  size n is at least 30.
• The test statistic is the sample mean
• The standardized test statistic is z
• When n ? 30, the sample standard deviation s can
  be substituted for ?.

44
Using P-values for a z-Test for Mean µ
In Words In Symbols

• State the claim mathematically and verbally.
  Identify the null and alternative hypotheses.
• Specify the level of significance.
• Determine the standardized test statistic.
• Find the area that corresponds to z.

State H0 and Ha.
Identify ?.
Use Table 4 in Appendix B.
45
Using P-values for a z-Test for Mean µ
In Words In Symbols

• Find the P-value.
• For a left-tailed test, P (Area in left tail).
• For a right-tailed test, P (Area in right
  tail).
• For a two-tailed test, P 2(Area in tail of test
  statistic).
• Make a decision to reject or fail to reject the
  null hypothesis.
• Interpret the decision in the context of the
  original claim.

Reject H0 if P-value is less than or equal to ?.
Otherwise, fail to reject H0.
46
Example Hypothesis Testing Using P-values

• In an advertisement, a pizza shop claims that its
  mean delivery time is less than 30 minutes. A
  random selection of 36 delivery times has a
  sample mean of 28.5 minutes and a standard
  deviation of 3.5 minutes. Is there enough
  evidence to support the claim at ? 0.01? Use a
  P-value.

47
Solution Hypothesis Testing Using P-values

• H0
• Ha
• ?
• Test Statistic

• P-value

0.01

• Decision

0.0051 lt 0.01 Reject H0
At the 1 level of significance, you have
sufficient evidence to conclude the mean delivery
time is less than 30 minutes.
48
Example Hypothesis Testing Using P-values

• You think that the average franchise investment
  information shown in the graph is incorrect, so
  you randomly select 30 franchises and determine
  the necessary investment for each. The sample
  mean investment is 135,000 with astandard
  deviation of 30,000. Is there enough evidence
  to support your claim at ? 0.05? Use a
  P-value.

49
Solution Hypothesis Testing Using P-values

• H0
• Ha
• ?
• Test Statistic

• P-value

P 2(0.0655) 0.1310
0.05

• Decision

0.1310 gt 0.05 Fail to reject H0
At the 5 level of significance, there is not
sufficient evidence to conclude the mean
franchise investment is different from 143,260.
50
Rejection Regions and Critical Values

• Rejection region (or critical region)
• The range of values for which the null hypothesis
  is not probable.
• If a test statistic falls in this region, the
  null hypothesis is rejected.
• A critical value z0 separates the rejection
  region from the nonrejection region.

51
Rejection Regions and Critical Values

• Finding Critical Values in a Normal Distribution
• Specify the level of significance ?.
• Decide whether the test is left-, right-, or
  two-tailed.
• Find the critical value(s) z0. If the hypothesis
  test is
• left-tailed, find the z-score that corresponds to
  an area of ?,
• right-tailed, find the z-score that corresponds
  to an area of 1 ?,
• two-tailed, find the z-score that corresponds to
  ½? and 1 ½?.
• Sketch the standard normal distribution. Draw a
  vertical line at each critical value and shade
  the rejection region(s).

52
Example Finding Critical Values

• Find the critical value and rejection region for
  a two-tailed test with ? 0.05.

Solution
1 a 0.95
½a 0.025
½a 0.025
z0 1.96
-z0 -1.96
The rejection regions are to the left of -z0
-1.96 and to the right of z0 1.96.
53
Decision Rule Based on Rejection Region

• To use a rejection region to conduct a hypothesis
  test, calculate the standardized test statistic,
  z. If the standardized test statistic
• is in the rejection region, then reject H0.
• is not in the rejection region, then fail to
  reject H0.

54
Using Rejection Regions for a z-Test for a Mean µ
In Words In Symbols
State H0 and Ha.

• State the claim mathematically and verbally.
  Identify the null and alternative hypotheses.
• Specify the level of significance.
• Sketch the sampling distribution.
• Determine the critical value(s).
• Determine the rejection region(s).

Identify ?.
Use Table 4 in Appendix B.
55
Using Rejection Regions for a z-Test for a Mean µ
In Words In Symbols

• Find the standardized test statistic.
• Make a decision to reject or fail to reject the
  null hypothesis.
• Interpret the decision in the context of the
  original claim.

If z is in the rejection region, reject H0.
Otherwise, fail to reject H0.
56
Example Testing with Rejection Regions

• Employees in a large accounting firm claim that
  the mean salary of the firms accountants is less
  than that of its competitors, which is 45,000.
  A random sample of 30 of the firms accountants
  has a mean salary of 43,500 with a standard
  deviation of 5200. At a 0.05, test the
  employees claim.

57
Solution Testing with Rejection Regions

• H0
• Ha
• ?
• Rejection Region

• Test Statistic

0.05

• Decision

Fail to reject H0
At the 5 level of significance, there is not
sufficient evidence to support the employees
claim that the mean salary is less than 45,000.
-1.645
-1.58
58
Example Testing with Rejection Regions

• The U.S. Department of Agriculture reports that
  the mean cost of raising a child from birth to
  age 2 in a rural area is 10,460. You believe
  this value is incorrect, so you select a random
  sample of 900 children (age 2) and find that the
  mean cost is 10,345 with a standard deviation of
  1540. At a 0.05, is there enough evidence to
  conclude that the mean cost is different from
  10,460? (Adapted from U.S. Department of
  Agriculture Center for Nutrition Policy and
  Promotion)

59
Solution Testing with Rejection Regions

• H0
• Ha
• ?
• Rejection Region

• Test Statistic

0.05
Reject H0

• Decision

At the 5 level of significance, you have enough
evidence to conclude the mean cost of raising a
child from birth to age 2 in a rural area is
significantly different from 10,460.
-1.96
1.96
-2.24
60
Section 7.2 Summary

• Found P-values and used them to test a mean µ
• Used P-values for a z-test
• Found critical values and rejection regions in a
  normal distribution
• Used rejection regions for a z-test

61
Section 7.3

• Hypothesis Testing for the Mean (Small Samples)

62
Section 7.3 Objectives

• Find critical values in a t-distribution
• Use the t-test to test a mean µ
• Use technology to find P-values and use them with
  a t-test to test a mean µ

63
Finding Critical Values in a t-Distribution

• Identify the level of significance ?.
• Identify the degrees of freedom d.f. n 1.
• Find the critical value(s) using Table 5 in
  Appendix B in the row with n 1 degrees of
  freedom. If the hypothesis test is
• left-tailed, use One Tail, ? column with a
  negative sign,
• right-tailed, use One Tail, ? column with a
  positive sign,
• two-tailed, use Two Tails, ? column with a
  negative and a positive sign.

64
Example Finding Critical Values for t

• Find the critical value t0 for a left-tailed test
  given? 0.05 and n 21.

• Solution
• The degrees of freedom are d.f. n 1 21 1
  20.
• Look at a 0.05 in the One Tail, ? column.
• Because the test is left-tailed, the critical
  value is negative.

65
Example Finding Critical Values for t

• Find the critical values t0 and -t0 for a
  two-tailed test given ? 0.05 and n 26.

• Solution
• The degrees of freedom are d.f. n 1 26 1
  25.
• Look at a 0.05 in the Two Tail, ? column.
• Because the test is two-tailed, one critical
  value is negative and one is positive.

66
t-Test for a Mean µ (n lt 30, ? Unknown)

• t-Test for a Mean
• A statistical test for a population mean.
• The t-test can be used when the population is
  normal or nearly normal, ? is unknown, and n lt
  30.
• The test statistic is the sample mean
• The standardized test statistic is t.
• The degrees of freedom are d.f. n 1.

67
Using the t-Test for a Mean µ(Small Sample)
In Words In Symbols

• State the claim mathematically and verbally.
  Identify the null and alternative hypotheses.
• Specify the level of significance.
• Identify the degrees of freedom and sketch the
  sampling distribution.
• Determine any critical value(s).

State H0 and Ha.
Identify ?.
d.f. n 1.
Use Table 5 in Appendix B.
68
Using the t-Test for a Mean µ(Small Sample)
In Words In Symbols

• Determine any rejection region(s).
• Find the standardized test statistic.
• Make a decision to reject or fail to reject the
  null hypothesis.
• Interpret the decision in the context of the
  original claim.

If t is in the rejection region, reject H0.
Otherwise, fail to reject H0.
69
Example Testing µ with a Small Sample

• A used car dealer says that the mean price of a
  2005 Honda Pilot LX is at least 23,900. You
  suspect this claim is incorrect and find that a
  random sample of 14 similar vehicles has a mean
  price of 23,000 and a standard deviation of
  1113. Is there enough evidence to reject the
  dealers claim at a 0.05? Assume the population
  is normally distributed. (Adapted from Kelley
  Blue Book)

70
Solution Testing µ with a Small Sample

• H0
• Ha
• a
• df
• Rejection Region

• Test Statistic
• Decision

0.05
Reject H0
14 1 13
At the 0.05 level of significance, there is
enough evidence to reject the claim that the mean
price of a 2005 Honda Pilot LX is at least 23,900
-1.771
-3.026
71
Example Testing µ with a Small Sample

• An industrial company claims that the mean pH
  level of the water in a nearby river is 6.8. You
  randomly select 19 water samples and measure the
  pH of each. The sample mean and standard
  deviation are 6.7 and 0.24, respectively. Is
  there enough evidence to reject the companys
  claim at a 0.05? Assume the population is
  normally distributed.

72
Solution Testing µ with a Small Sample

• H0
• Ha
• a
• df
• Rejection Region

• Test Statistic
• Decision

0.05
Fail to reject H0
19 1 18
At the 0.05 level of significance, there is not
enough evidence to reject the claim that the mean
pH is 6.8.
-2.101
2.101
-1.816
73
Example Using P-values with t-Tests

• The American Automobile Association claims that
  the mean daily meal cost for a family of four
  traveling on vacation in Florida is 118. A
  random sample of 11 such families has a mean
  daily meal cost of 128 with a standard deviation
  of 20. Is there enough evidence to reject the
  claim at a 0.10? Assume the population is
  normally distributed. (Adapted from American
  Automobile Association)

74
Solution Using P-values with t-Tests

• H0
• Ha

TI-83/84set up
Calculate
Draw

• Decision

0.1664 gt 0.10
Fail to reject H0. At the 0.10 level of
significance, there is not enough evidence to
reject the claim that the mean daily meal cost
for a family of four traveling on vacation in
Florida is 118.
75
Section 7.3 Summary

• Found critical values in a t-distribution
• Used the t-test to test a mean µ
• Used technology to find P-values and used them
  with a t-test to test a mean µ

76
Section 7.4

• Hypothesis Testing for Proportions

77
Section 7.4 Objectives

• Use the z-test to test a population proportion p

78
z-Test for a Population Proportion

• z-Test for a Population Proportion
• A statistical test for a population proportion.
• Can be used when a binomial distribution is given
  such that np ? 5 and nq ? 5.
• The test statistic is the sample proportion .
  
• The standardized test statistic is z.

79
Using a z-Test for a Proportion p
Verify that np 5 and nq 5
In Words In Symbols

• State the claim mathematically and verbally.
  Identify the null and alternative hypotheses.
• Specify the level of significance.
• Sketch the sampling distribution.
• Determine any critical value(s).

State H0 and Ha.
Identify ?.
Use Table 5 in Appendix B.
80
Using a z-Test for a Proportion p
In Words In Symbols

• Determine any rejection region(s).
• Find the standardized test statistic.
• Make a decision to reject or fail to reject the
  null hypothesis.
• Interpret the decision in the context of the
  original claim.

If z is in the rejection region, reject H0.
Otherwise, fail to reject H0.
81
Example Hypothesis Test for Proportions

• Zogby International claims that 45 of people in
  the United States support making cigarettes
  illegal within the next 5 to 10 years. You decide
  to test this claim and ask a random sample of 200
  people in the United States whether they support
  making cigarettes illegal within the next 5 to 10
  years. Of the 200 people, 49 support this law.
  At a 0.05 is there enough evidence to reject
  the claim?

• Solution
• Verify that np 5 and nq 5.

np 200(0.45) 90 and nq 200(0.55) 110
82
Solution Hypothesis Test for Proportions

• H0
• Ha
• ?
• Rejection Region

• Test Statistic

0.05
Fail to reject H0

• Decision

At the 5 level of significance, there is not
enough evidence to reject the claim that 45 of
people in the U.S. support making cigarettes
illegal within the next 5 to 10 years.
-1.96
1.96
1.14
83
Example Hypothesis Test for Proportions

• The Pew Research Center claims that more than 55
  of U.S. adults regularly watch their local
  television news. You decide to test this claim
  and ask a random sample of 425 adults in the
  United States whether they regularly watch their
  local television news. Of the 425 adults, 255
  respond yes. At a 0.05 is there enough evidence
  to support the claim?

• Solution
• Verify that np 5 and nq 5.

np 425(0.55) 234 and nq 425 (0.45) 191
84
Solution Hypothesis Test for Proportions

• H0
• Ha
• ?
• Rejection Region

• Test Statistic

0.05
Reject H0

• Decision

At the 5 level of significance, there is enough
evidence to support the claim that more than 55
of U.S. adults regularly watch their local
television news.
1.645
2.07
85
Section 7.4 Summary

• Used the z-test to test a population proportion p

86
Section 7.5

• Hypothesis Testing for Variance and Standard
  Deviation

87
Section 7.5 Objectives

• Find critical values for a ?2-test
• Use the ?2-test to test a variance or a standard
  deviation

88
Finding Critical Values for the ?2-Test

• Specify the level of significance ?.
• Determine the degrees of freedom d.f. n 1.
• The critical values for the ?2-distribution are
  found in Table 6 of Appendix B. To find the
  critical value(s) for a
• right-tailed test, use the value that corresponds
  to d.f. and ?.
• left-tailed test, use the value that corresponds
  to d.f. and 1 ?.
• two-tailed test, use the values that corresponds
  to d.f. and ½? and d.f. and 1 ½?.

89
Finding Critical Values for the ?2-Test
Right-tailed
Left-tailed
Two-tailed
90
Example Finding Critical Values for ?2

• Find the critical ?2-value for a left-tailed test
  whenn 11 and ? 0.01.

• Solution
• Degrees of freedom n 1 11 1 10 d.f.
• The area to the right of the critical value is 1
  ? 1 0.01 0.99.

From Table 6, the critical value is
.
91
Example Finding Critical Values for ?2

• Find the critical ?2-value for a two-tailed test
  when n 13 and ? 0.01.

• Solution
• Degrees of freedom n 1 13 1 12 d.f.
• The areas to the right of the critical values are
  

From Table 6, the critical values are
and
92
The Chi-Square Test

• ?2-Test for a Variance or Standard Deviation
• A statistical test for a population variance or
  standard deviation.
• Can be used when the population is normal.
• The test statistic is s2.
• The standardized test statistic
• follows a chi-square distribution with degrees
  of freedom d.f. n 1.

93
Using the ?2-Test for a Variance or Standard
Deviation
In Words In Symbols

• State the claim mathematically and verbally.
  Identify the null and alternative hypotheses.
• Specify the level of significance.
• Determine the degrees of freedom and sketch the
  sampling distribution.
• Determine any critical value(s).

State H0 and Ha.
Identify ?.
d.f. n 1
Use Table 6 in Appendix B.
94
Using the ?2-Test for a Variance or Standard
Deviation
In Words In Symbols

• Determine any rejection region(s).
• Find the standardized test statistic.
• Make a decision to reject or fail to reject the
  null hypothesis.
• Interpret the decision in the context of the
  original claim.

If ?2 is in the rejection region, reject H0.
Otherwise, fail to reject H0.
95
Example Hypothesis Test for the Population
Variance

• A dairy processing company claims that the
  variance of the amount of fat in the whole milk
  processed by the company is no more than 0.25.
  You suspect this is wrong and find that a random
  sample of 41 milk containers has a variance of
  0.27. At a 0.05, is there enough evidence to
  reject the companys claim? Assume the population
  is normally distributed.

96
Solution Hypothesis Test for the Population
Variance

• H0
• Ha
• a
• df
• Rejection Region

• Test Statistic
• Decision

0.05
41 1 40
Fail to Reject H0
At the 5 level of significance, there is not
enough evidence to reject the companys claim
that the variance of the amount of fat in the
whole milk is no more than 0.25.
55.758
43.2
97
Example Hypothesis Test for the Standard
Deviation

• A restaurant claims that the standard deviation
  in the length of serving times is less than 2.9
  minutes. A random sample of 23 serving times has
  a standard deviation of 2.1 minutes. At a
  0.10, is there enough evidence to support the
  restaurants claim? Assume the population is
  normally distributed.

98
Solution Hypothesis Test for the Standard
Deviation

• H0
• Ha
• a
• df
• Rejection Region

• Test Statistic
• Decision

0.10
23 1 22
Reject H0
At the 10 level of significance, there is enough
evidence to support the claim that the standard
deviation for the length of serving times is less
than 2.9 minutes.
14.042
11.536
99
Example Hypothesis Test for the Population
Variance

• A sporting goods manufacturer claims that the
  variance of the strength in a certain fishing
  line is 15.9. A random sample of 15 fishing line
  spools has a variance of 21.8. At a 0.05, is
  there enough evidence to reject the
  manufacturers claim? Assume the population is
  normally distributed.

100
Solution Hypothesis Test for the Population
Variance

• H0
• Ha
• a
• df
• Rejection Region

• Test Statistic
• Decision

0.05
15 1 14
Fail to Reject H0
At the 5 level of significance, there is not
enough evidence to reject the claim that the
variance in the strength of the fishing line is
15.9.
5.629
26.119
19.194
101
Section 7.5 Summary

• Found critical values for a ?2-test
• Used the ?2-test to test a variance or a standard
  deviation