1. 7 Chapter Hypothesis Testing with One Sample
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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
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3. Introduction to Hypothesis Testing
Section 7.1
Introduction to Hypothesis Testing
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4. Section 7.1 Objectives
State a null hypothesis and an alternative hypothesis
Identify type I and type II 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
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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.
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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.
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7. complementary statements
Stating a Hypothesis
Null hypothesis
A statistical hypothesis that contains a statement of equality such as ≤, =, or ≥.
Denoted H0 read “H sub-zero” or “H naught.”
Alternative hypothesis
A statement of strict inequality such as >, ≠, or <.
Must be true if H0 is false.
Denoted Ha read “H sub-a.”
complementary statements
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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 > k
H0: μ ≥ k
Ha: μ < 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.
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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 school publicizes that the proportion of its students who are involved in at least one extracurricular activity is 61%.
Solution:
H0:
Ha:
p = 0.61
Equality condition
(Claim)
p ≠ 0.61
Complement of H0
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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 car dealership announces that the mean time for an oil change is less than 15 minutes.
Solution:
H0:
Ha:
μ ≥ 15 minutes
Complement of Ha
μ < 15 minutes
Inequality condition
(Claim)
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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 company advertises that the mean life of its furnaces is more than 18 years
Solution:
H0:
Ha:
μ ≤ 18 years
Complement of Ha
μ > 18 years
Inequality condition
(Claim)
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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.
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13. Types of Errors
Actual Truth of H0
Decision
H0 is true
H0 is false
Do not reject H0
Correct Decision
Type II Error
Reject H0
Type I Error
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.
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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 inspector’s claim is true. When will a type I or type II error occur? Which is more serious? (Source: United States Department of Agriculture)
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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 > 0.2
(Claim)
0.16
0.18
0.20
0.22
0.24 p
H0: p ≤ 0.20
H0: p > 0.20
Chicken meets USDA limits.
Chicken exceeds USDA limits.
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16. Solution: Identifying Type I and Type II Errors
Hypotheses:
H0:
Ha:
p ≤ 0.2
p > 0.2
(Claim)
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.
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17. Solution: Identifying Type I and Type II Errors
Hypotheses:
H0:
Ha:
p ≤ 0.2
p > 0.2
(Claim)
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.
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18. Level of Significance Level of significance
Your maximum allowable probability of making a type I error.
Denoted by α, 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:
α = 0.10 α = 0.05 α = 0.01
P(type II error) = β (beta)
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19. Standardized test statistic
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. σ2
χ2 (Section 7.5) s2
z (Section 7.4) p
t (Section 7.3 n < 30)
z (Section 7.2 n ≥ 30) μ
Standardized test statistic
Test statistic
Population parameter
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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.
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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.
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22. Left-tailed Test
The alternative hypothesis Ha contains the less-than inequality symbol (<).
H0: μ ≥ k
Ha: μ < k
P is the area to the left of the standardized test statistic. z 1 2 3 –3 –2 –1
Test statistic
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23. Right-tailed Test
The alternative hypothesis Ha contains the greater-than inequality symbol (>).
H0: μ ≤ k
Ha: μ > k
P is the area to the right of the standardized test statistic. z 1 2 3 –3 –2 –1
Test statistic
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24. Two-tailed Test
The alternative hypothesis Ha contains the not-equal-to symbol (≠). Each tail has an area of ½P.
H0: μ = k
Ha: μ ≠ k
P is twice the area to the right of the positive standardized test statistic.
P is twice the area to the left of the negative standardized test statistic. z 1 2 3 –3 –2 –1
Test statistic
Test statistic
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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 school publicizes that the proportion of its students who are involved in at least one extracurricular activity is 61%.
Solution:
H0:
Ha:
p = 0.61
p ≠ 0.61
Two-tailed test
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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 car dealership announces that the mean time for an oil change is less than 15 minutes.
Solution: z -z
P-value area
H0:
Ha:
μ ≥ 15 min
μ < 15 min
Left-tailed test
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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 company advertises that the mean life of its furnaces is more than 18 years.
Solution: z
P-value area
H0:
Ha:
μ ≤ 18 yr
μ > 18 yr
Right-tailed test
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28. Making a Decision Decision Rule Based on P-value
Compare the P-value with α.
If P ≤ α , then reject H0.
If P > α, then fail to reject H0.
Claim
Decision
Claim is H0
Claim is Ha
Reject H0
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
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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 school publicizes that the proportion of its students who are involved in at least one extracurricular activity is 61%.
Solution:
The claim is represented by H0.
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30. Solution: Interpreting a Decision
If you reject H0, then you should conclude “there is enough evidence to reject the school’s claim that the proportion of students who are involved in at least one extracurricular activity is 61%.”
If you fail to reject H0, then you should conclude “there is not enough evidence to reject the school’s claim that proportion of students who are involved in at least one extracurricular activity is 61%.”
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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): A car dealership announces that the mean time for an oil change is less than 15 minutes.
Solution:
The claim is represented by Ha.
H0 is “the mean time for an oil change is greater than or equal to 15 minutes.”
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32. Solution: Interpreting a Decision
If you reject H0, then you should conclude “there is enough evidence to support the dealership’s claim that the mean time for an oil change is less than 15 minutes.”
If you fail to reject H0, then you should conclude “there is not enough evidence to support the dealership’s claim that the mean time for an oil change is less than 15 minutes.”
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33. Steps for Hypothesis Testing
State the claim mathematically and verbally. Identify the null and alternative hypotheses.
H0: ? Ha: ?
Specify the level of significance.
α = ?
Determine the standardized sampling distribution and sketch its graph.
Calculate the test statistic and its corresponding
standardized test statistic. Add it to your sketch.
This sampling distribution is based on the assumption that H0 is true. z z
Standardized test statistic
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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? No
Fail to reject H0.
Yes
Reject H0.
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35. Section 7.1 Summary
Stated a null hypothesis and an alternative hypothesis
Identified type I and type II 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
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36. Hypothesis Testing for the Mean (Large Samples)
Section 7.2
Hypothesis Testing for the Mean (Large Samples)
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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
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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 α.
If P ≤ α, then reject H0.
If P > α, then fail to reject H0.
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39. Example: Interpreting a P-value
The P-value for a hypothesis test is P = What is your decision if the level of significance is
α = 0.05?
α = 0.01?
Solution: Because < 0.05, you should reject the null hypothesis.
Solution:
Because > 0.01, you should fail to reject the null hypothesis.
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40. Finding the P-value
After determining the hypothesis test’s standardized test statistic and the test statistic’s 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).
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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 α = 0.01.
Solution: For a left-tailed test, P = (Area in left tail) z
-2.23
P =
Because > 0.01, you should fail to reject H0.
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42. Example: Finding the P-value
Find the P-value for a two-tailed hypothesis test with a test statistic of z = Decide whether to reject H0 if the level of significance is α = 0.05.
Solution: For a two-tailed test, P = 2(Area in tail of test statistic) z
2.14
1 – =
P = 2(0.0162) =
0.9838
Because < 0.05, you should reject H0.
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43. Z-Test for a Mean μ
Can be used when the population is normal and σ 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 σ.
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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 α.
Use Table 4 in Appendix B.
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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 α. Otherwise, fail to reject H0.
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46. Example: Hypothesis Testing Using P-values
In auto racing, a pit crew claims that its mean pit stop time (for 4 new tires and fuel) is less than 13 seconds. A random selection of 32 pit stop times has a sample mean of 12.9 seconds and a standard deviation of 0.19 second. Is there enough evidence to support the claim at α = 0.01? Use a P-value.
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47. Solution: Hypothesis Testing Using P-values
Ha:
α =
Test Statistic:
μ ≥ 13 sec
μ < 13 sec (Claim)
0.01
< 0.01
Reject H0 .
Decision:
At the 1% level of significance, you have sufficient evidence to support the claim that the mean pit stop time is less than 13 seconds.
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48. Example: Hypothesis Testing Using P-values
The National Institute of Diabetes and Digestive and Kidney Diseases reports that the average cost of bariatric (weight loss) surgery is $22,500. You think this information is incorrect. You randomly select 30 bariatric surgery patients and find that the average cost for their surgeries is $21,545 with a standard deviation of $3015. Is there enough evidence to support your claim at α = 0.05? Use a P-value. (Adapted from National Institute of Diabetes and Digestive and Kidney Diseases)
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49. Solution: Hypothesis Testing Using P-values
Ha:
α =
Test Statistic:
μ = $22,500
μ ≠ 22,500 (Claim)
0.05
Decision:
> 0.05
Fail to reject H0 .
At the 5% level of significance, there is not sufficient evidence to support the claim that the mean cost of bariatric surgery is different from $22,500.
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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.
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51. Rejection Regions and Critical Values
Finding Critical Values in a Normal Distribution
Specify the level of significance α.
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 α,
right-tailed, find the z-score that corresponds to an area of 1 – α,
two-tailed, find the z-score that corresponds to ½α and 1 – ½α.
Sketch the standard normal distribution. Draw a vertical line at each critical value and shade the rejection region(s).
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52. Example: Finding Critical Values
Find the critical value and rejection region for a two-tailed test with α = 0.05.
Solution:
1 – α = 0.95 z z0
½α = 0.025
½α = 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.
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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. z z0
Fail to reject H0.
Reject H0.
Left-Tailed Test
z < z0 z z0
Reject Ho.
Fail to reject Ho.
z > z0
Right-Tailed Test z
–z0
Two-Tailed Test z0
z < –z0
z > z0
Reject H0
Fail to reject H0
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54. Using Rejection Regions for a z-Test for a 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 critical value(s).
Determine the rejection region(s).
State H0 and Ha.
Identify α.
Use Table 4 in Appendix B.
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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.
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56. Example: Testing with Rejection Regions
Employees at a construction and mining company claim that the mean salary of the company’s mechanical engineers is less than that of the one of its competitors, which is $68,000. A random sample of 30 of the company’s mechanical engineers has a mean salary of $66,900 with a standard deviation of $5500. At α = 0.05, test the employees’ claim.
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57. Solution: Testing with Rejection Regions
Ha:
α =
Rejection Region:
μ ≥ $68,000
μ < $68,000 (Claim)
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 $68,000.
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58. Example: Testing with Rejection Regions
The U.S. Department of Agriculture claims that the mean cost of raising a child from birth to age 2 by husband-wife families in the U.S. is $13,120. A random sample of 500 children (age 2) has a mean cost of $12,925 with a standard deviation of $1745. At α = 0.10, is there enough evidence to reject the claim? (Adapted from U.S. Department of Agriculture Center for Nutrition Policy and Promotion)
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59. Solution: Testing with Rejection Regions
Ha:
α =
Rejection Region:
μ = $13,120 (Claim)
μ ≠ $13,120
Test Statistic
0.10
Decision:
Reject H0 .
At the 10% level of significance, you have enough evidence to reject the claim that the mean cost of raising a child from birth to age 2 by husband-wife families in the U.S. is $13,120.
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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
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61. Hypothesis Testing for the Mean (Small Samples)
Section 7.3
Hypothesis Testing for the Mean (Small Samples)
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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
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63. Finding Critical Values in a t-Distribution
Identify the level of significance α.
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, α ” column with a negative sign,
right-tailed, use “One Tail, α ” column with a positive sign,
two-tailed, use “Two Tails, α ” column with a negative and a positive sign.
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64. Example: Finding Critical Values for t
Find the critical value t0 for a left-tailed test given α = 0.05 and n = 21.
Solution:
The degrees of freedom are d.f. = n – 1 = 21 – 1 = 20.
Look at α = 0.05 in the “One Tail, α” column.
Because the test is left-tailed, the critical value is negative. t
t0 = –1.725
0.05
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65. Example: Finding Critical Values for t
Find the critical values –t0 and t0 for a two-tailed test given α = 0.10 and n = 26.
Solution:
The degrees of freedom are d.f. = n – 1 = 26 – 1 = 25.
Look at α = 0.10 in the “Two Tail, α” column.
Because the test is two-tailed, one critical value is negative and one is positive.
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66. t-Test for a Mean μ (n < 30, σ Unknown)
A statistical test for a population mean.
The t-test can be used when the population is normal or nearly normal, σ is unknown, and n < 30.
The test statistic is the sample mean
The standardized test statistic is t.
The degrees of freedom are d.f. = n – 1.
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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.
Determine the critical value(s).
Determine the rejection region(s).
State H0 and Ha.
Identify α.
d.f. = n – 1.
Use Table 5 in Appendix B.
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68. Using the t-Test for a Mean μ (Small Sample)
In Words In Symbols
Find the standardized test statistic and sketch the sampling distribution
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.
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69. Example: Testing μ with a Small Sample
A used car dealer says that the mean price of a 2008 Honda CR-V is at least $20,500. You suspect this claim is incorrect and find that a random sample of 14 similar vehicles has a mean price of $19,850 and a standard deviation of $1084. Is there enough evidence to reject the dealer’s claim at α = 0.05? Assume the population is normally distributed. (Adapted from Kelley Blue Book)
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70. Solution: Testing μ with a Small Sample
Test Statistic:
Decision:
H0:
Ha:
α =
df =
Rejection Region:
μ ≥ $20,500 (Claim)
μ < $20,500
0.05
Reject H0 .
14 – 1 = 13
At the 5% level of significance, there is enough evidence to reject the claim that the mean price of a 2008 Honda CR-V is at least $20,500.
t ≈ –2.244
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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 company’s claim at α = 0.05? Assume the population is normally distributed.
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72. Solution: Testing μ with a Small Sample
α =
df =
Rejection Region:
μ = 6.8 (Claim)
μ ≠ 6.8
Test Statistic:
Decision:
0.05
19 – 1 = 18
Fail to reject H0 .
At the 5% level of significance, there is not enough evidence to reject the claim that the mean pH is 6.8. t
–2.101
0.025
2.101
–2.101
2.101
–1.816
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73. Example: Using P-values with t-Tests
A Department of Motor Vehicles office claims that the mean wait time is less than 14 minutes. A random sample of 10 people has a mean wait time of 13 minutes with a standard deviation of 3.5 minutes. At α = 0.10, test the office’s claim. Assume the population is normally distributed.
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74. Solution: Using P-values with t-Tests
Ha:
μ ≥ 14 min
μ < 14 min (Claim)
TI-83/84 setup:
Calculate:
Draw:
P ≈
Decision:
Since > 0.10, fail to reject H0.
At the 10% level of significance, there is not enough evidence to support the office’s claim that the mean wait time is less than 14 minutes.
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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
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76. Hypothesis Testing for Proportions
Section 7.4
Hypothesis Testing for Proportions
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77. Section 7.4 Objectives
Use the z-test to test a population proportion p
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78. z-Test for a Population Proportion
z-Test for a Population Proportion p
A statistical test for a population proportion p.
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.
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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.
Determine the critical value(s).
Determine the rejection region(s).
State H0 and Ha.
Identify α.
Use Table 4 in Appendix B.
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80. Using a z-Test for a Proportion p
In Words In Symbols
Find the standardized test statistic and sketch the sampling distribution.
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.
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81. Example: Hypothesis Test for Proportions
A research center claims that less than 50% of U.S. adults have accessed the Internet over a wireless network with a laptop computer. In a random sample of 100 adults, 39% say they have accessed the Internet over a wireless network with a laptop computer. At α = 0.01, is there enough evidence to support the researcher’s claim? (Adopted from Pew Research Center)
Solution:
Verify that np ≥ 5 and nq ≥ 5.
np = 100(0.50) = 50 and nq = 100(0.50) = 50
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82. Solution: Hypothesis Test for Proportions
Test Statistic
H0:
Ha:
α =
Rejection Region:
p ≥ 0.5
p ≠ 0.45
0.01
Decision:
Fail to reject H0 .
At the 1% level of significance, there is not enough evidence to support the claim that less than 50% of U.S. adults have accessed the Internet over a wireless network with a laptop computer.
z = –2.2
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83. Example: Hypothesis Test for Proportions
A research center claims that 25% of college graduates think a college degree is not worth the cost. You decide to test this claim and ask a random sample of 200 college graduates whether they think a college degree is not worth the cost. Of those surveyed, 21% reply yes. At α = 0.10 is there enough evidence to reject the claim?
Solution:
Verify that np ≥ 5 and nq ≥ 5.
np = 200(0.25) = 50 and nq = 200 (0.75) = 150
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84. Solution: Hypothesis Test for Proportions
Test Statistic
H0:
Ha:
α =
Rejection Region:
p = (Claim)
p ≠ 0.25
0.10
Decision:
Fail to reject H0 .
At the 10% level of significance, there is enough evidence to reject the claim that 25% of college graduates think a college degree is not worth the cost.
z ≈ –1.31
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85. Section 7.4 Summary Used the z-test to test a population proportion p
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86. Hypothesis Testing for Variance and Standard Deviation
Section 7.5
Hypothesis Testing for Variance and Standard Deviation
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87. Section 7.5 Objectives Find critical values for a χ2-test
Use the χ2-test to test a variance or a standard deviation
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88. Finding Critical Values for the χ2-Test
Specify the level of significance α.
Determine the degrees of freedom d.f. = n – 1.
The critical values for the χ2-distribution are found in Table 6 in Appendix B. To find the critical value(s) for a
right-tailed test, use the value that corresponds to d.f. and α.
left-tailed test, use the value that corresponds to d.f. and 1 – α.
two-tailed test, use the values that corresponds to d.f. and ½α, and d.f. and 1 – ½α.
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89. Finding Critical Values for the χ2-Test
1 –
Right-tailed
Left-tailed χ2
Two-tailed
1 –
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90. Example: Finding Critical Values for χ2
Find the critical χ2-value for a left-tailed test when n = 11 and α = 0.01.
Solution:
Degrees of freedom: n – 1 = 11 – 1 = 10 d.f.
The area to the right of the critical value is 1 – α = 1 – 0.01 = 0.99.
χ0 = 2.558
From Table 6, the critical value is
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91. Example: Finding Critical Values for χ2
Find the critical χ2-value for a two-tailed test when n = 9 and α = 0.05.
Solution:
Degrees of freedom: n – 1 = 9 – 1 = 8 d.f.
The areas to the right of the critical values are
From Table 6, the critical values are and .
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92. The Chi-Square Test χ2-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.
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93. Using the χ2-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.
Determine the critical value(s).
State H0 and Ha.
Identify α.
d.f. = n – 1
Use Table 6 in Appendix B.
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94. Using the χ2-Test for a Variance or Standard Deviation
In Words In Symbols
Determine the rejection region(s).
Find the standardized test statistic and sketch the sampling distribution.
Make a decision to reject or fail to reject the null hypothesis.
Interpret the decision in the context of the original claim.
If χ2 is in the rejection region, reject H0. Otherwise, fail to reject H0.
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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 You suspect this is wrong and find that a random sample of 41 milk containers has a variance of At α = 0.05, is there enough evidence to reject the company’s claim? Assume the population is normally distributed.
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96. Solution: Hypothesis Test for the Population Variance
Ha:
α =
df =
Rejection Region:
σ2 ≤ 0.25 (Claim)
σ2 > 0.25
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 company’s claim that the variance of the amount of fat in the whole milk is no more than 0.25.
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97. Example: Hypothesis Test for the Standard Deviation
A company claims that the standard deviation of the lengths of time it takes an incoming telephone call to be transferred to the correct office is less than 1.4 minutes. A random sample of 25 incoming telephone calls has a standard deviation of 1.1 minutes. At α = 0.10, is there enough evidence to support the company’s claim? Assume the population is normally distributed.
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98. Solution: Hypothesis Test for the Standard Deviation
Ha:
α =
df =
Rejection Region:
σ ≥ 1.4 min.
σ < 1.4 min. (Claim)
Test Statistic:
Decision:
0.10
25 – 1 = 24
Reject H0 .
At the 10% level of significance, there is enough evidence to support the claim that the standard deviation of the lengths of time it takes an incoming telephone call to be transferred to the correct office is less than 1.4 minutes.
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99. Example: Hypothesis Test for the Population Variance
A sporting goods manufacturer claims that the variance of the strengths of a certain fishing line is A random sample of 15 fishing line spools has a variance of At α = 0.05, is there enough evidence to reject the manufacturer’s claim? Assume the population is normally distributed.
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100. Solution: Hypothesis Test for the Population Variance
Ha:
α =
df =
Rejection Region:
σ2 = 15.9 (Claim)
σ2 ≠ 15.9
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 strengths of the fishing line is 15.9.
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101. Section 7.5 Summary Found critical values for a χ2-test
Used the χ2-test to test a variance or a standard deviation
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101 of 101