1. Hypothesis Tests Regarding a Parameter
Chapter 10
Hypothesis Tests Regarding a Parameter
10.1
10.2
10.3
10.4

2. The Language of Hypothesis Testing
Section
10.1
The Language of Hypothesis Testing

3. 10.1 Objectives Determine the null and alternative hypotheses
Explain Type I and Type II errors
State conclusions to hypothesis tests 3

4. Objective 1
Determine the Null and Alternative Hypotheses
A hypothesis is a statement regarding a characteristic of one or more populations.
In this chapter, we look at hypotheses regarding a single population parameter. 4

5. Examples of Claims Regarding a Characteristic of a Single Population
In 2008, 62% of American adults regularly volunteered their time for charity work. A researcher believes that this percentage is different today.
According to a study published in March, 2006 the mean length of a phone call on a cellular telephone was 3.25 minutes. A researcher believes that the mean length of a call has increased since then.
Using an old manufacturing process, the standard deviation of the amount of wine put in a bottle was 0.23 ounces. With new equipment, the quality control manager believes the standard deviation has decreased.
CAUTION!
We test these types of statements using sample data because it is usually impossible or impractical to gain access to the entire population. If population data are available, there is no need for inferential statistics. 5

6. Steps in Hypothesis Testing
Hypothesis testing is a procedure, based on sample evidence and probability, used to test statements regarding a characteristic of one or more populations.
Steps in Hypothesis Testing
Make a statement regarding the nature of the population.
Collect evidence (sample data) to test the statement.
Analyze the data to assess the plausibility of the statement. 6

7. The null hypothesis, denoted H0, is a statement to be tested
The null hypothesis, denoted H0, is a statement to be tested. The null hypothesis is a statement of no change, no effect or no difference and is assumed true until evidence indicates otherwise.
The alternative hypothesis, denoted H1, is a statement that we are trying to find evidence to support. 7

8. In this chapter, there are three ways to set up the null and alternative hypotheses:
Equal versus not equal hypothesis (two-tailed test)
H0: parameter = some value
H1: parameter ≠ some value
Equal versus less than (left-tailed test)
H1: parameter < some value
Equal versus greater than (right-tailed test)
H1: parameter > some value 8

9. “In Other Words”
The null hypothesis is a statement of status quo or no difference and always contains a statement of equality. The null hypothesis is assumed to be true until we have evidence to the contrary. We seek evidence that supports the statement in the alternative hypothesis. 9

10. Parallel Example 2a: Forming Hypotheses
For each of the following claims, determine the null and alternative hypotheses. State whether the test is two-tailed, left-tailed or right-tailed.
In 2008, 62% of American adults regularly volunteered their time for charity work. A researcher believes that this percentage is different today.
H0:
H1:
p=0.62
p≠0.62 claim
alternative hypothesis is a two-tailed hypothesis. 10

11. Parallel Example 2b: Forming Hypotheses
For each of the following claims, determine the null and alternative hypotheses. State whether the test is two-tailed, left-tailed or right-tailed.
b) According to a study published in March, 2006 the mean length of a phone call on a cellular telephone was 3.25 minutes. A researcher believes that the mean length of a call has increased since then.
H0:
H1:
μ = 3.25.
μ > claim
right-tailed test 11

12. Parallel Example 2c: Forming Hypotheses
For each of the following claims, determine the null and alternative hypotheses. State whether the test is two-tailed, left-tailed or right-tailed.
c) Using an old manufacturing process, the standard deviation of the amount of wine put in a bottle was 0.23 ounces. With new equipment, the quality control manager believes the standard deviation has decreased.
H0:
H1:
σ = 0.23
σ < claim
left-tailed test. 12

13. Objective 2 Explain Type I and Type II Errors
Type I error = P(rejecting H0 when H0 is true)
Type II error = P(not rejecting H0 when H1 is true) 13

14. Parallel Example 3a: Type I and Type II Errors
For each of the following claims, explain what it would mean to make a Type I error. What would it mean to make a Type II error?
In 2008, 62% of American adults regularly volunteered their time for charity work. A researcher believes that this percentage is different today.
H0:
H1:
p = 0.62
p ≠ 0.62 claim
Type I error = P(rejecting H0 when H0 is true)
Type II error = P(not rejecting H0 when H1 is true)
A Type I error is made if the researcher concludes that p ≠ 0.62 when the true proportion of Americans 18 years or older who participated in some form of charity work is currently 62%.
A Type II error is made if the sample evidence leads the researcher to believe that the current percentage of Americans 18 years or older who participated in some form of charity work is still 62% when, in fact, this percentage differs from 62%. 14

15. Parallel Example 3b: Type I and Type II Errors
For each of the following claims, explain what it would mean to make a Type I error. What would it mean to make a Type II error?
b) According to a study published in March, 2006 the mean length of a phone call on a cellular telephone was 3.25 minutes. A researcher believes that the mean length of a call has increased since then.
H0:
H1:
μ = 3.25
μ > 3.25 claim
Type I error = P(rejecting H0 when H0 is true)
Type II error = P(not rejecting H0 when H1 is true)
A Type I error occurs if the sample evidence leads the researcher to conclude that μ > 3.25 when, in fact, the actual mean call length on a cellular phone is still 3.25 minutes.
A Type II error occurs if the researcher fails to reject the hypothesis that the mean length of a phone call on a cellular phone is 3.25 minutes when, in fact, it is longer than 3.25 minutes. 15

16. = P(rejecting H0 when H0 is true) β = P(Type II Error)
α = P(Type I Error)
= P(rejecting H0 when H0 is true)
β = P(Type II Error)
= P(not rejecting H0 when H1 is true)
The probability of making a Type I error, α, is chosen by the researcher before the sample data is collected.
The level of significance, α, is the probability of making a Type I error. 16

17. “In Other Words”
As the probability of a Type I error increases, the probability of a Type II error decreases, and vice-versa. 17

18. Objective 3 State Conclusions to Hypothesis Tests CAUTION!
We never “accept” the null hypothesis, because, without having access to the entire population, we don’t know the exact value of the parameter stated in the null. Rather, we say that we do not reject the null hypothesis. This is just like the court system. We never declare a defendant “innocent”, but rather say the defendant is “not guilty”. 18

19. Making a Decision - Based on the P-value
H0: claim
Ha:
P-value < α
Reject the null
There is enough evidence to reject the claim
H0: claim
Ha:
P-value > α
Do NOT reject the null
There is not enough evidence to reject the claim
H0:
Ha: claim
P-value < α
Reject the null
There is enough evidence to support the claim
H0:
Ha: claim
P-value > α
Do NOT Reject the null
There is not enough evidence to support the claim
Larson/Farber 4th ed.

20. Parallel Example 4a: Stating the Conclusion
According to a study published in March, 2006 the mean length of a phone call on a cellular telephone was 3.25 minutes. A researcher believes that the mean length of a call has increased since then.
Suppose the sample evidence indicates that the null hypothesis should be rejected. State the wording of the conclusion.
H0:
H1:
μ = 3.25
μ > 3.25 claim
Reject the H0
Since the null hypothesis is rejected, there is sufficient evidence to conclude that the mean length of a phone call on a cell phone is greater than 3.25 minutes. 20

21. Parallel Example 4b: Stating the Conclusion
According to a study published in March, 2006 the mean length of a phone call on a cellular telephone was 3.25 minutes. A researcher believes that the mean length of a call has increased since then.
b) Suppose the sample evidence indicates that the null hypothesis should not be rejected. State the wording of the conclusion.
μ = 3.25
μ > 3.25 claim
H0:
H1:
Fail to reject the H0
There is insufficient evidence to conclude that the mean length of a phone call on a cell phone is greater than 3.25 minutes.
In other words, the sample evidence is consistent with the mean call length equaling 3.25 minutes. 21

22. 10.1 Summary Determine the null and alternative hypotheses
Null always = sign
Explain Type I and Type II errors
State conclusions to hypothesis tests
Type I error = P(rejecting H0 when H0 is true)
Type II error = P(not rejecting H0 when H1 is true)
Since the Ho is rejected, there is sufficient evidence to support the H1
Since we fail to reject the Ho, there is NOT sufficient evidence to support the H1 22

23. Hypothesis Tests for a Population Proportion
Section
10.2
Hypothesis Tests for a Population Proportion

24. 10.2 Objectives Explain the logic of hypothesis testing
Test the hypotheses about a population proportion
Test hypotheses about a population proportion using the binomial probability distribution. 24

25. Objective 1 Explain the Logic of Hypothesis Testing
When observed results are unlikely under the assumption that the null hypothesis is true, we say the result is statistically significant.
When results are found to be statistically significant, we reject the null hypothesis. 25

26. A researcher obtains a random sample of 1000 people and finds that 534 are in favor of the banning cell phone use while driving, so = 534/1000. Does this suggest that more than 50% of people favor the policy? Or is it possible that the true proportion of registered voters who favor the policy is some proportion less than 0.5 and we just happened to survey a majority in favor of the policy? In other words, would it be unusual to obtain a sample proportion of or higher from a population whose proportion is 0.5? What is convincing, or statistically significant, evidence?
H0:
Ha:
p = 0.5
p > 0.5 26

27. We would reject the null hypothesis.
x= 534, n = 1000
To determine if a sample proportion of = is statistically significant, we could build a probability model, use p = 0.5
Since np(1 – p) = 100(0.5)(1 – 0.5) = 250 ≥ 10
The sample size (n = 1000) is sufficiently smaller than the population size, we can use the normal model to describe the variability in .
What is the probability of a sample falling that far away from the center?
P-value = P(z0 > 2.15)
= 0.016
Z0 = 2.15
We would reject the null hypothesis. 27

28. Hypothesis Testing Using the Classical Approach
If the sample proportion is too many standard deviations from the proportion stated in the null hypothesis, we reject the null hypothesis.
Hypothesis Testing Using the P-value Approach
If the probability of getting a sample proportion as extreme or more extreme than the one obtained is small under the assumption the statement in the null hypothesis is true, reject the null hypothesis.
Basically, reject null hypothesis if the P-value < α 28

29. Objective 2 Test hypotheses about a population proportion.
provided that the following requirements are satisfied:
The sample is a simple random sample.
np(1-p) ≥ 10.
The sampled values are independent of each other.
Hypothesis-Testing
State the hypothesis and identify the claim
Find the critical value(s) (table)
Compute the test value
Find the P-value
Make a decision to reject or not reject the null hypothesis
Summarize the results
Test statistic:
Point estimate: 29

30. Parallel Example 1: Testing a Hypothesis about a
Parallel Example 1: Testing a Hypothesis about a Population Proportion: Large Sample Size
In 1997, 46% of Americans said they did not trust the media “when it comes to reporting the news fully, accurately and fairly”. In a 2007 poll of 1010 adults nationwide, 525 stated they did not trust the media. At the α = 0.05 level of significance, is there evidence to support the claim that the percentage of Americans that do not trust the media to report fully and accurately has increased since 1997? Source: Gallup Poll
We want to know if p > First, we must verify the requirements to perform the hypothesis test:
This is a simple random sample.
np0(1 – p0) = 1010(0.46)(1 – 0.46) = > 10
Since the sample size is less than 5% of the population size, the assumption of independence is met. 30

31. The level of significance is α = 0.05, right tailed, zα= 1.645
In 1997, 46% of Americans said they did not trust the media “when it comes to reporting the news fully, accurately and fairly”. In a 2007 poll of 1010 adults nationwide, 525 stated they did not trust the media. At the α = 0.05 level of significance, is there evidence to support the claim that the percentage of Americans that do not trust the media to report fully and accurately has increased since 1997? z
Zc =
1.645
H0: p = 0.46
H1: p > claim
The level of significance is
α = 0.05, right tailed, zα= 1.645
The test statistic is
Z = 3.83
P-value = P(Z > 3.83) ≈
P-value < α, reject the null hypothesis.
There is sufficient evidence at the α = 0.05 level of significance to conclude that the percentage of Americans that do not trust the media to report fully and accurately has increased since 1997. 31

32. Test Statistic: p = q = n = x = .45 .55 200 0.49∙200 = 98 .49 .51 H0:
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 α = 0.05 is there enough evidence to reject the claim?
p =
q =
n =
x =
.45
.55
200
0.49∙200 = 98
.49
.51
Verify that np(1 – p) > 10
np(1 – p) = 200(0.45)(0.55) = 49.5
H0:
Ha:
 =
Test Statistic:
P-value:
p = claim
p ≠ 0.45
Rejection Region: z
0.05, 2-tailed, zα/2 = ±1.960
Zc =
-1.96
1.96
Z = 1.14
Decision:
Fail to reject H0
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.
2•P(z > 1.14)
= 2•normalcdf(1.14,100) ≈
Larson/Farber 4th ed.

33. The Pew Research Center claims that more than 55% of U. S
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 α = 0.05 is there enough evidence to support the claim?
We want to know if p > First, we must verify the requirements to perform the hypothesis test:
This is a simple random sample.
np0(1 – p0) = 425(0.55)(1 – 0.55) = > 10
Since the sample size is less than 5% of the population size, the assumption of independence is met.
p = .55
q = .45
n = 425
x = 255
255/425
170/425
H0:
Ha: 
p = 0.55
p > claim
= 0.05 , right tailed, zα = 1.645
Larson/Farber 4th ed.

34. p = .55 q = .45 n = 425 x = 255 H0: Ha:  = Test Statistic: p = 0.55
255/425
170/425
H0:
Ha:
 =
Test Statistic:
p = 0.55
p > claim
Rejection Region z
0.05, right tailed, zα = 1.645
zc =
1.645
Z = 2.07
Decision:
Reject H0
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.
P-value:
P(z > 2.07)
= normalcdf(2.07,100) ≈

35. For the sampling distribution of to be approximately normal, we require np(1– p) be at least 10. What if this requirement is not met?
We stated that an event was unusual if the probability of observing the event was less than This criterion is based on the P-value approach to testing hypotheses; the probability that we computed was the P-value. We use this same approach to test hypotheses regarding a population proportion for small samples using the Binomial Distribution. 35

36. p = .489 q = .511 n = 35 x = 21 Decision: Fail to reject H0 H0: Ha: 
According to the US Department of Agriculture, 48.9% of males aged 20 to 39 years consume the recommended daily requirement of calcium. After an aggressive “Got Milk” campaign, the USDA conducts a survey of 35 randomly selected males aged 20 to 39 and finds that 21 of them consume the recommended daily allowance of calcium. At the α = 0.10 level of significance, is there evidence to conclude that the percentage of males aged 20 to 39 who consume the recommended amount of calcium has increased?
We want to know if p >
First, we must verify the requirements to perform the hypothesis test:
This is a simple random sample.
np(1 – p) = 35(0.489)(1 – 0.489) = is NOT > 10
Since the sample size is less than 5% of the population size, the assumption of independence is met.
p = .489
q = .511
n = 35
x = 21
H0:
Ha: 
p = 0.489
p > claim
Decision:
Fail to reject H0
= (no critical value since we have to use Binomcdf)
There is NOT sufficient evidence to support the claim ….
P-value = P( x > 21)
= 1 – P( x < 20)
= 1 – Bincdf (35,0.489,20) =

37. Summary Hypothesis Testing about Proportions:
provided that the following requirements are satisfied:
The sample is a simple random sample.
np(1-p) ≥ 10.
The sampled values are independent of each other.
Hypothesis-Testing
State the hypothesis and identify the claim
Find the critical value(s) (table)
Compute the test value
Find the P-value
Make a decision to reject or not reject the null hypothesis
Summarize the results
Test statistic:
Point estimate: 37

38. Hypothesis Tests for a Population Mean
Section
10.3
Hypothesis Tests for a Population Mean

39. 10.3 Objectives Test hypotheses about a mean
Understand the difference between statistical significance and practical significance. 39

40. Test hypotheses about a population mean.
provided that the following requirements are satisfied:
The sample is a simple random sample.
The sample has no outliers, and the population from which the sample is drawn is normally distributed or the sample size is large (n ≥ 30).
The sampled values are independent of each other.
Hypothesis-Testing
State the hypothesis and identify the claim
Find the critical value(s) (table)
Compute the test value
Find the P-value
Make a decision to reject or not reject the null hypothesis
Summarize the results
Test statistic:
Point estimate: 40

41. Left-Tailed Two-Tailed Right-Tailed
Classical Approach
Left-Tailed Two-Tailed Right-Tailed
Compare the critical value with the test statistic: 41

42. Left-Tailed Two-Tailed Right-Tailed
P-Value Approach
Left-Tailed Two-Tailed Right-Tailed
If the P-value < α, reject the null hypothesis. 42

43. The procedure is robust, which means that minor departures from normality will not adversely affect the results of the test.
However, for small samples, if the data have outliers, the procedure should not be used. 43

44. Parallel Example 1: Testing a Hypothesis about a
Parallel Example 1: Testing a Hypothesis about a Population Mean, Large Sample
Assume the resting metabolic rate (RMR) of healthy males in complete silence is 5710 kJ/day. Researchers measured the RMR of 45 healthy males who were listening to calm classical music and found their mean RMR to be with a standard deviation of
At the α = 0.05 level of significance, is there evidence to conclude that the mean RMR of males listening to calm classical music is different than 5710 kJ/day?
The sample is a simple random sample.
The sample has no outliers, and the population from which the sample is drawn is normally distributed or the sample size is large (n ≥ 30).
The sampled values are independent of each other.
Since the sample size is large, we follow the steps for testing hypotheses about a population mean for large samples. 44

45. H0: H1: α = test statistic: H0: μ = 5710 H1: μ ≠ 5710 claim α = 0.05
Assume the resting metabolic rate (RMR) of healthy males in complete silence is 5710 kJ/day. Researchers measured the RMR of 45 healthy males who were listening to calm classical music and found their mean RMR to be with a standard deviation of
At the α = 0.05 level of significance, is there evidence to conclude that the mean RMR of males listening to calm classical music is different than 5710 kJ/day?
H0:
H1:
α =
test statistic:
H0: μ = 5710
H1: μ ≠ claim
α = 0.05
test statistic: t =
s =
n = 45
tc =
2.015
to =
Find the critical values:
two-tailed test, α = 0.05, df = n –1 = 44
tα/2 = ±2.015
Fail to reject H0
Find the P-value:
two-tailed,
2•P(t0 < ,df=44) = 0.990
There is NOT sufficient evidence to support the claim 45

46. Solution
There is insufficient evidence at the α = 0.05 level of significance to conclude that the mean RMR of males listening to calm classical music differs from 5710 kJ/day. 46

47. Parallel Example 3: Testing a Hypothesis about a
Parallel Example 3: Testing a Hypothesis about a Population Mean, Small Sample
According to the United States Mint, quarters weigh 5.67 grams. A researcher is interested in determining whether the “state” quarters have a weight that is different from 5.67 grams. He randomly selects 18 “state” quarters, weighs them and obtains the following data.
At the α = 0.05 level of significance, is there evidence to conclude that state quarters have a weight different than grams?
Since the sample size is small, we must verify that the data come from a population that is normally distributed with no outliers before proceeding
Put the data values into L1 47

48. NPP approximately linear. No outliers.
Assumption of normality appears reasonable. 48

49. H0: μ = 5.67 H1: μ ≠ 5.67 claim α = 0.05 test statistic: H0: H1: α =
According to the United States Mint, quarters weigh 5.67 grams. A researcher is interested in determining whether the “state” quarters have a weight that is different from 5.67 grams. He randomly selects 18 “state” quarters, weighs them and obtains the following data.
At the α = 0.05 level of significance, is there evidence to conclude that state quarters have a weight different than grams? t
H0: μ = 5.67
H1: μ ≠ claim
α = 0.05
test statistic:
H0:
H1:
α =
test statistic: =
s =
n = 18
tc = -2.11
2.11
to = 2.75
Find the critical values:
two-tailed test, α = 0.05, df = n –1 =17
tα/2 = ±2.11
Reject H0
Find the P-value:
two-tailed,
2•P(t0 > 2.75,df=17) = 0.014
There is sufficient evidence to support the claim 49

50. Solution
There is sufficient evidence at the α = 0.05 level of significance to conclude that the mean weight of the state quarters differs from 5.67 grams. 50

51. Objective 2
Understand the Difference between Statistical Significance and Practical Significance
When a large sample size is used in a hypothesis test, the results could be statistically significant even though the difference between the sample statistic and mean stated in the null hypothesis may have no practical significance.
Practical significance refers to the idea that, while small differences between the statistic and parameter stated in the null hypothesis are statistically significant, the difference may not be large enough to cause concern or be considered important. 51

52. H0: H1: α = test statistic: H0: μ = 25.2 H1: μ > 25.2 claim
Parallel Example 7: Statistical versus Practical Significance
In 2003, the average age of a mother at the time of her first childbirth was To determine if the average age has increased, a random sample of 1200 mothers is taken and is found to have a sample mean age of 25.5 with a standard deviation of 4.8, determine whether the mean age has increased using a significance level of α = 0.05. t
= 25.5
s = 4.8
n = 1200
H0:
H1:
α =
test statistic:
H0: μ = 25.2
H1: μ > claim
α = 0.05
test statistic:
tc =
1.646
to = 2.17
Find the critical values:
right-tailed test, α = 0.05, df = n –1 =1199
tα = 1.646
Reject H0
Find the P-value:
Right-tailed,
P(t0 > 2.17,df=1199) = 0.015
There is sufficient evidence to support the claim 52

53. Solution
There is sufficient evidence at the 0.05 significance level to conclude that the mean age of a mother at the time of her first childbirth is greater than 25.2.
Although we found the difference in age to be significant, there is really no practical significance in the age difference (25.2 versus 25.5). 53

54. Large Sample Sizes
Large sample sizes can lead to statistically significant results while the difference between the statistic and parameter is not enough to be considered practically significant. 54

55. Test hypotheses about a population mean.
provided that the following requirements are satisfied:
The sample is a simple random sample.
The sample has no outliers, and the population from which the sample is drawn is normally distributed or the sample size is large (n ≥ 30).
The sampled values are independent of each other.
Hypothesis-Testing
State the hypothesis and identify the claim
Find the critical value(s) (table)
Compute the test value
Find the P-value
Make a decision to reject or not reject the null hypothesis
Summarize the results
10.3 Summary
Test statistic:
Point estimate: 55

56. Putting It Together: Which Method Do I Use?
Section
10.4
Putting It Together: Which Method Do I Use?

57. 57

58. This example of 10.2 not covered 2014
Parallel Example 4: Hypothesis Test for a Population Proportion: Small Sample Size
In 2006, 10.5% of all live births in the United States were to mothers under 20 years of age. A sociologist claims that births to mothers under 20 years of age is decreasing. She conducts a simple random sample of 34 births and finds that 3 of them were to mothers under 20 years of age. Test the sociologist’s claim at the α = 0.01 level of significance.
This example of 10.2 not covered 2014
We want to know if p > First, we must verify the requirements to perform the hypothesis test:
This is a simple random sample.
np0(1 – p0) = 34(0.105)(1 – 0.105) = < 10
Thus, the sampling distribution of is not approximately normal. 58

59. This example of 10.2 not covered 2014
In 2006, 10.5% of all live births in the United States were to mothers under 20 years of age. A sociologist claims that births to mothers under 20 years of age is decreasing. She conducts a simple random sample of 34 births and finds that 3 of them were to mothers under 20 years of age. Test the sociologist’s claim at the α = 0.01 level of significance.
Let X represent the number of live births in the United States to mothers under 20 years of age. We have x = 3 successes in n = 34 trials so = 3/34= We want to determine whether this result is unusual, if the population truly has a p = 0.105
H0: p = 0.105
H1: p < claim
This example of 10.2 not covered 2014
P-value = P(X ≤ 3 assuming p=0.105 )
= P(X = 0) + P(X =1) + P(X =2) + P(X = 3)
= 0.51
Binomcdf(n,p,x)
Binomcdf(34,0.015,3)
P-value > α, fail to reject the null hypothesis.
There is insufficient evidence to conclude that the percentage of live births in the United States to mothers under the age of 20 has decreased below the 2006 level of 10.5%. 59