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Unit 7 Chapter 7 Summary

Chapter 6 presented the method of interval estimation used in inferential statistic to draw a conclusion about a population parameter. Chapter 7 gives another method, hypothesis testing, to test a specific value of the parameter. A hypothesis test is a process that uses sample statistics to test a claim about the population parameter. A hypothesis test has five steps. Each will be explained individually then examples for complete tests will be given.

Step 1: State working hypothesis. This consists of a null hypothesis H0 and an alternative hypothesis H1.

Null Hypothesis H0 contains a statement about the population parameter that includes some condition of equality, such as =, ≥ , ≤. This hypothesis can be reject during testing but never accepted as true.

Alternative Hypothesis Ha is the complement of the null hypothesis. It is a statement that must be true if H0 is false (or H0 is rejected). It contains a statement of strict inequality, such as ≠, >, k |

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|H0 : µ ≤ k |Ha: > k |

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|H0 : µ ≥ k |Ha: < k |

Examples:

1. A light bulb manufacturer’s competitor announces that the mean life time of its competitor’s bulbs are less than 1000 hours. The claim “the mean …is less than 1000 hours” can be written as µ < 1000. Its complement is µ ≥ 1000. Because µ ≥ 1000 contains the statement of equality it becomes the null hypothesis. In this case, the alternative hypothesis represents the claim.

H0 : µ ≥ 1000

Ha : µ < 1000 (claim)

2. A snack company advertises that the mean weigh of the contents of a 20-ounce size bag is more than 20 ounces.

H0 : µ ≤ 20 ounces

Ha : µ > 20 ounces (claim)

3. A consumer analyst says that the mean life of Hot Shot automobile battery is not 60 months.

H0 : µ = 60 months

H1 : µ ≠ 60 months (claim)

4. A company claims that its brand of paint has a mean drying time of less than 45 minutes.

H0 : µ ≥ 45

Ha : µ < 45 (claim)

5. A researcher claims that 65% of voters favor gun control.

H0 : µ = 65% (claim)

Ha : µ ≠ 65%

6. Hancock Motors claims that its new truck will average equal to or better than 25 miles per gallons in the city.

H0 : µ ≥ 25 (claim)

H1 : µ < 25

Step 2: Determine and calculate the test statistic. This will be determined by if distribution is normal, if σ known or unknown or if sample size is less than 30 or greater than or equal to 30.

If x has a normal distribution with known σ, the test statistic is

[pic]

Where [pic] = sample mean

µ = value stated in H0

n = sample size

Step 3: P-value is the probability of chance. The P-value of a hypothesis test depends on the nature of the test: Left-tail, right-tail, or two-tailed. The tail follows the H1 condition. Left-tail test will contain . Two-tail tests will contain ≠.

|Left-tail test |Right-tailed test |Two-tailed test |

|[pic] |[pic] |[pic] |

|[pic] |

|[pic] |

|[pic] |

Step 4: Develop the decision rule. A risk or error probability, denoted α (alpha),must be set. The beginning assumption in the hypothesis test is that the equality condition in the null hypothesis H0 is true. Thus,when you perform a test, you make one of two decisions:

1. reject H0

2. fail to reject H0

This level of significance α (alpha) is the probability of rejecting H0 when it is true. Another name for this is Type I error.

Examples:

1. For α = 5% or 0.05. There is a 5% chance of error which equates to a 95% level of confidence.

2. For α = 10% or 0.10. There is a 10% chance of error which equates to a 90% level of confidence.

Right or left tail test as shown above have all their error in one of the tails. A two-tailed test has its error split evenly in both tails. For α = 5% or 0.05, each tail would have 2.5% error in each tail.

Decision Rules:

1) Using P-values

This α and the P-value are used to define the decision rule.

To use a P-value to make a conclusion, compare the P-value with α :

1. If P ≤ α, then reject H0

2. If P > α, then fail to reject H0.

Note: If H0 is rejected, H1 can be concluded as true; but failing to reject H0 does not mean H0 is accepted as true.

How to find P-value

|Find the P-value for a left-tailed test with a test statistic of z = -2.23.|[pic] |

|Decide whether to reject H0 if the level of significance is α = 0.01. | |

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|From the normal table z = -2.23 is P= 0.0129 which is the area in the left | |

|tail. | |

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|Since 0.0129 > 0.01, you should fail to reject H0. | |

|Find the P-value for a two-tailed test with test statistic of z = 2.14. |[pic] |

|Decide whether to reject H0 if the level of significance α = 0.05. The | |

|graph shows a standard normal curve with the shaded areas to the left of z | |

|= -2.14 and to the right of z = 2.14. For a two-tailed test, P = 2(area in | |

|tail). | |

| | |

|From the normal table the area corresponding to z = 2.14, the area in the | |

|right tail is 1 – 0.9838 = 0.0162.So the P-value for the two-tail test with| |

|z =2.14 is P = 2(0.0162) = 0.0324. | |

| | |

|Since 0.0324 < 0.05, then reject H0. | |

Examples:

1. Find the P-value for a left-tailed test with a test statistic of z = -1.62. Decide whether to reject H0 if the level of significance is α = 0.05.

From the table, z = -1.62 gives P = 0.0526. Since 0.0526 > 0.05, fail to reject H0.

2. Find the P-value for a two-tailed test with test statistic of z = 2.31. Decide whether to reject H0 if α = 0.01

From the table z = 2.31, the area in the right tail is 1 – 0.9896 = 0.0104. So the P-value for the two-tail test with z = 2.31 is P = 2(0.0104) = 0.0208. Since 0.0208 > 0.01, the conclusion is to fail to reject H0.

1) Using Rejections Region

|[pic] |

|[pic] |

Hypothesis Testing Steps using P-value

1. Define H0 and Ha and set level of significance α

2. Specify test statistic and calculate the value

3. Find the P-value

4. Test conclusion with α and P-value

5. Interpret the test result

Testing µ when σ known

Examples

1. A manufacturer of sprinkler systems designed for fire protection claims that the average activating temperature is at least 135o F. To test this claim, you randomly select a sample of 32 systems and find the mean activation temperature to be 133o F. Assume σ = 3.3o F. At α = 0.10 do you have enough evidence to support the manufacturer’s claim.

Given: µ = 135, n = 32, [pic] = 133, σ = 3.3, α = 0.10

1. H0 : µ ≥ 135 (claim) α = 0.10

Ha : µ < 135 one tail test

2. Test statistic [pic]

3. From the table z = -3.43, the area in the left tail is P = 0.0003.

4. Since 0.0003 < 0.10, reject H0.

5. At the 10% significance level, there is enough evidence to reject the manufacturer’s claim that the average activating temperature is at least 135oF.

2. A nutritionist claims that the mean tuna consumption by a person in the US is 3.1 pounds per year. A random sample of 60 people show a mean tuna consumption per person is 2.9 pounds per year. Assume σ = 0.94 pounds. At that α = 0.08, can you reject the nutritionist’s claim?

Given: µ = 3.1, n = 60, [pic] = 2.9, σ = 0.94, α = 0.08

1. H0 : µ = 3.1 (claim) α = 0.08

Ha : µ ≠ 3.1 one tail test

2. Test statistic [pic]

3. From the table z = -1.65 give 0.0495. Since this is a two-tailed test P = 2(0.0495) = 0.099.

4. Since 0.099 > 0.08, fail to reject H0

5. At the 8% significance level, there is not enough information to reject the nutrition’s claim that the mean tuna consumption by a person is 3.1 pounds per year.

Hypothesis Testing Steps using Rejection Region(s)

6. Define H0 and Ha and set level of significance α

7. Specify test statistic and calculate the value

8. Find the Rejections Region

9. Test conclusion with the critical value(s) and the test statistic

10. Interpret the test result

Testing µ when σ known

Examples

3. A manufacturer of sprinkler systems designed for fire protection claims that the average activating temperature is at least 135o F. To test this claim, you randomly select a sample of 32 systems and find the mean activation temperature to be 133o F. Assume σ = 3.3o F. At α = 0.10 (level of confidence 90%) do you have enough evidence to support the manufacturer’s claim.

Given: µ = 135, n = 32, [pic] = 133, σ = 3.3, α = 0.10

1. H0 : µ ≥ 135 (claim) α = 0.10

Ha : µ < 135 one tail test

2. Test statistic [pic]

3. Rejections region: z < -1.28

4. Since – 3.43 < - 1.28, reject H0.

5. At the 10% significance level, there is enough evidence to reject the manufacturer’s claim that the average activating temperature is at least 135oF.

4. A nutritionist claims that the mean tuna consumption by a person in the US is 3.1 pounds per year. A random sample of 60 people show a mean tuna consumption per person is 2.9 pounds per year. Assume σ = 0.94 pounds. At that α = 0.08 (level of confidence 98%), can you reject the nutritionist’s claim?

Given: µ = 3.1, n = 60, [pic] = 2.9, σ = 0.94, α = 0.08

1. H0 : µ = 3.1 (claim) α = 0.08

Ha : µ ≠ 3.1 two tail test

2. Test statistic [pic]

3. Rejection Regions: z < -2.33 or z > 2.33

4. Since -1.65 ( -2.33, fail to reject H0

5. At the 8% significance level, there is not enough information to reject the nutrition’s claim that the mean tuna consumption by a person is 3.1 pounds per year.

Testing µ when σ unknown (It is recommended that the rejection region method is used with t.

If x has a normal distribution ,then any sample size will work. If you do not have this, then use sample size greater than or equal to 30. The hypothesis test will use the Student’s t distribution.

[pic] with degrees of freedom df = n – 1.

Example:

1. A used car dealer says that the mean price of a 2005 SUV is at least $23,900. It is suspected that this claim is incorrect and find that a random sample of 14 similar Nimbus SUVs has a mean price of $23,000 and a sample standard deviation of $1113. Is there enough evidence to reject the dealer’s claim at α = 0.05? Assume the population is normally distributed.

Given: µ = 23900, n = 14, [pic] = 23000, s = 1113, α = 0.05

1. H0 : µ ≥ 23,900 (claim)

Ha : µ < 23,900 α = 0.05 df = 14 – 1 = 13

2. Test statistics [pic]

3. In the t table on the row for df = 13 and one-tail (, t =1.771. The rejection region is t < - 1.771

4. Since – 3.026 < -1.771 H0 is rejected.

5. There is enough evidence at the 5% level of significance to reject the claim that the mean price of a Nimbus SUV is at least $23,900.

2. An industrial firm claims that the mean pH level of the water in a nearby river is 6.8. A random sample of 19 water samples are taken and measured for the pH. The sample mean and standard deviation are 6.7 and 0.24 respectively. Is there enough evidence to reject the firm’s claim at α = 0.05 (level of confidence 95%)? Assume the population is normally distributed.

Given: µ = 6.8, n = 19, [pic] = 6.7, s = 0.24, α = 0.05

1. H0 : µ = 6.8 (claim) α = 0.03 df = 19 – 1 =18

Ha : µ ≠ 6.8

2. Test statistics [pic]

3. In the t table on the row for df = 18 and two tail (, t is ± 2.101. Rejection regions : t < - 2.101 or t > 2.101

4. Since -1.816 ( -2.101, fail to reject H0

5. There is not enough evidence at the 5% level of significance to reject the claim that the mean pH is 6.8.

Testing a proportion p

Hypothesis tests for proportion occur when a person wants to be able to draw some conclusions the proportion or percentage of a population.

The population parameter for the proportion is p. The statistic has a foundation in the binomial distribution with given values for p, q and n and r as the number of successes. Therefore np > 5 and nq > 5 must be checked. The sample proportion is (p hat) [pic][pic] or it may be given in the problem statement.

The test statistic is [pic]

Where n = number of trials, p = proportion specified in H0, q = 1 – p.

Examples: Using rejection region method

1. A research center claim that less than 20% of Internet user in the US have a wireless network in their home. In a random sample of 100 adults, 15% say they have a wireless network in their home. At α = 0.01 (level of confidence 99%), is there enough evidence to support the researcher’s claim?

Given: p = 0.20, n = 100, [pic]= 0.15, α = 0.01

1. H0 : p ≥ 0.2 α = 0.01

Ha : p < 0.2 (claim) one-tail test

2. p = 0.2 and q = 1 - 0.2 = 0.8

Check np = 20 > 5 and nq = 80 > 5, a z test can be used.

test statistic is [pic]

3. Rejection region: z < - 2.33

4. Since -1.25 ( - 2.33, H0 cannot be rejected.

5. At the 1% significance level, there is not enough evidence to support the claim that less than 20% of Internet users in the US have a wireless network at home. (Note the claim is in H1.)

2. Zogby International claims that 45% of people in the US support making cigarettes illegal within the next 5 to 10 years. Test the claim with a random sample of 200 people in the US asking 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 (level of confidence 95%), is there enough evidence to reject the claim?

Given: p = 0.45, n = 200, [pic]= 0.49, α = 0.05

1. H0 : p = 0.45 (claim) α = 0.05

Ha : p ≠ 0.45 two-tail test

2. p = 0.45 and q = 0.55. Check np = 90 and nq = 110, both are greater than 5, so the z test can be used.

test statistic is [pic]

3. Rejection regions: z < -1.96 or z > 1.96

4. Since 1,14 ( 1.96P= 0.2542 > α = 0.05, you must fail to reject H0.

5. At the 5% significance level, there is not enough evidence to reject the claim that 45% of people in the US support making cigarettes illegal within the next 5 to 10 years.

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