1. Defining the null and alternative hypotheses
For the purposes of this class, the null hypothesis represents the status quo and will always be of the form
Ho: μ = μo
The choice of the alternative hypothesis depends on the purpose of the hypothesis test.
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2. Defining the null and alternative hypotheses
If the primary concern is deciding whether the population mean is different from some specified value, then the alternative takes the form:
Ha : μ ≠ μo
If the primary concern is deciding whether the population mean is less than some specified value, then the alternative takes the form:
Ha : μ < μo
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3. Defining the null and alternative hypotheses
If the primary concern is deciding whether the population mean is greater than some specified value, then the alternative takes the form:
Ha : μ > μo
A hypothesis test is called two-tailed if the alternative hypothesis takes the form “different from” and is called one-tailed if the alternative hypothesis is of the form “greater than” or “less than”.
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4. Making a decision
How do we decide whether or not to reject the null hypothesis in favor of the alternative hypothesis?
Take a random sample from the population
If the sample data are consistent with the null hypothesis, then do not reject the null hypothesis
If the sample data are inconsistent with the null hypothesis, then reject the null hypothesis and conclude that the alternative hypothesis is true.
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5. Hypothesis Testing Procedure
Set up the null and alternative hypotheses.
Define the test statistic
Define the rejection region based on the significance level
Calculate the value of the test statistic and determine whether or not to reject the null hypothesis
State your conclusion in terms of the original problem
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6. Define the test statistic
For tests regarding the mean of a population when the population standard deviation, σ, is known, the test statistic is given by 𝑥 − 𝜇 𝑜 𝜎/ 𝑛 where 𝜇 𝑜 is the value specified by the null hypothesis. Note that the test statistic will have a standard normal distribution if the null hypothesis is true.
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7. Define the rejection region
The rejection region is the set of values for the test statistic that leads to rejection of the null hypothesis. The rejection region will depend on (1) the requested significance level and (2) the form of the alternative hypothesis (one-tailed or two-tailed).
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8. Type I and Type II Errors
Null is true
Null is false
Decision:
Do not reject the null
Correct decision
Incorrect decision
Type II Error
Reject the null
Type I Error
We want to guard against making a type I error. That is, we want to insure that the probability of making a type I error is small.
The significance level (α) of a hypothesis test is the probability of making a type I error.
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9. Defining the Critical Values
Suppose that a hypothesis test is to be performed at a specified significance level. Then, the critical value(s) must be chosen so that if the null hypothesis is true, the probability that the test statistic will fall into the rejection region is equal to α. (See illustrations)
P 513 Weiss
𝑧 0.10
𝑧 0.05
𝑧 0.025
𝑧 0.01
𝑧 0.005
1.28
1.645
1.96
2.33
2.575
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10. Drawing conclusions
Suppose that a hypothesis test is conducted using a small significance level. Then we can draw one of the two following conclusions:
If the null hypothesis is rejected, we conclude that the alternative hypothesis is true. We can also say that the test results are “statistically significant” at the α level.
If the null hypothesis is not rejected, we conclude that the data do not provide sufficient evidence to support the alternative hypothesis. We can also say that the test results are “not statistically significant” at the α level.
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11. Example
A company that produces snack foods uses a machine to package 454-gram bags of pretzels. We will assume that the net weights are normally distributed and that the population standard deviation of all such weights is 7.8 grams. A random sample of 25 bags of pretzels has an average net weight of At the 5% significance level, do the data provide sufficient evidence to conclude that the packaging machine is not working properly?
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12. Example
The RR Bowker Company of New york collects information on the retail prices of books. In 1993, the mean retail price of all history books was $ Based on a sample of 40 history books taken this year, the mean price for the sample is $ The population standard deviation is known to be $ At the 1% significance level, do the data provide sufficient evidence to conclude that this year’s mean retail price of all history books has increased over the 1993 mean of $40.69?
p. 515
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13. Example
According to the Food and Nutrition Board of the National Academy of Sciences, the recommended daily allowance of calcium for adults is 800mg. A random sample of 18 people with incomes below the poverty level yields an average daily calcium intake of 747.4mg. At the 5% significance level, do the data provide sufficient evidence to conclude that the mean calcium intake of people with incomes below the poverty level is less than the recommended daily allowance of 800mg? Assume σ=188mg.
Also do this example in Excel and discuss p-values
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14. P-values
A p-value is the probability of obtaining a value more extreme as defined by the alternative hypothesis) than the value of the test statistic.
The p-value can be used as follow to arrive at a conclusion for a hypothesis test:
Reject the null hypothesis if p-value < α
Fail to reject the null hypothesis if p-value ≥ α
Show illustration
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15. Hypothesis test for the mean of a normal population (σ unknown)
For tests regarding the mean of a normal population when the population standard deviation, σ, is NOT known, the test statistic is given by 𝑡= 𝑥 − 𝜇 𝑜 𝑠/ 𝑛 For a specified significance level, α, the critical value for the test is given by 𝑡 𝛼/2 with n-1 degrees of freedom for 2-tailed tests and 𝑡 𝛼 with n-1 degrees of freedom for 1-tailed tests Note that this test is referred to as the one-sample t-test.
Show critical regions (p.550)
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16. Example
According to the US Energy Information Administration, the mean energy expenditure of all US households in 1993 was $ That same year, 36 randomly selected households living in single family detached homes reported the energy expenditures shown in the attached Excel table. At the 5% significance level, do the data provide sufficient evidence to conclude that in 1993, households living in single family detached homes spent more for energy, on average, than the national average of $1282?
Also do this example in Excel and discuss p-values
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17. Independent Samples (σ’s unknown)
Suppose that X is a normally distributed random variable that is measured on each of two populations where the mean for the first population is 𝜇 1 and 𝜎 1 is not known and the mean of the second population is 𝜇 2 and and 𝜎 2 is unknown. Then, for independent samples of sizes 𝑛 1 and 𝑛 2 from the two populations:
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18. Test Statistic
The test statistic is given by 𝑡= 𝑥 1 − 𝑥 2 𝑠 1 2 𝑛 1 + 𝑠 2 2 𝑛 2 t follows has approximately a t-distribution with degrees of freedom as defined by equation 9.7 in the book.
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19. Example
Example 9.5 p 313
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20. Independent Samples (σ’s unknown, equal)
Suppose that X is a random variable that is measured on each of two populations where the mean for the first population is 𝜇 1 and the mean of the second population is 𝜇 2 . Assume that 𝜎 1 = 𝜎 2 = σ, which is unknown. Then, for independent samples of sizes 𝑛 1 and 𝑛 2 from the two populations, the pooled standard deviation is given by 𝑠 𝑝 2 = 𝑛 1 −1 𝑠 1 2 +( 𝑛 2 −1) 𝑠 2 2 𝑛 1 + 𝑛 2 −2
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21. Test Statistic
The test statistic is given by 𝑡= 𝑥 1 − 𝑥 2 𝑠 𝑝 2 𝑛 1 + 𝑠 𝑝 2 𝑛 2 t follows a t-distribution with ( 𝑛 1 + 𝑛 2 −2) degrees of freedom.
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22. When can we assume equal variances?
Assume equal variances based on past experience
Formal test to show equal variances
If you can assume equal variances, you now have a test statistic with an exact distribution and you have a more powerful test (e.g., it is more likely that we will be able to reject the null hypothesis when in fact it is false)
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23. Example
Suppose that you’re a financial analyst for Charles Schwab. You want to see if there is a difference in dividend yield between stocks listed on the NYSE and NASDAQ. Assuming that the variances are equal, conduct a hypothesis test at the 5% significance level.
NYSE
NASDAQ n 21 25
Sample mean
3.27
2.53
Sample std dev
1.30
1.16
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24. Example
Test Statistic: Critical Value: (interpolate) Decision: Reject the null hypothesis Conclusion: Conclude that the data provide significant evidence that there is a difference between the dividend yield from the NYSE and the NASDAQ
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25. Comparing Two or More Population Means
Two different options:
Independent samples – the sample selected from one population has no effect or bearing on the sample selected from the other population
Dependent samples – there is some sort of pairing or matching between the observations from the two populations (e.g., before and after measurements)
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26. Paired Samples
Suppose that we want to decide whether a newly developed gasoline additive increases gas mileage.
Paired samples remove a source of extraneous variation which results in smaller sampling error and a more powerful test
Pair consists of a car driven with the additive and then without the additive
Analyze the difference between gas mileages for each car with and without the additive
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27. Mileage without additive
Paired Samples
Car
Mileage with additive
Mileage without additive
Paired difference 1
25.7
24.9
0.8 2
20.0
18.8
1.2 3
28.4
27.7
0.7 4
13.7
13.0 5
17.8
1.0 6
12.5
11.3 7
27.8
0.6 8
8.1
8.2
-0.1 9
23.1
0.0 10
10.4
9.9
0.5
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28. Test Statistic
The test statistic is the same as the test statistic for the 1-sample t-test, except that it is based on the sample differences 𝑡= 𝑑 − 𝐷 0 𝑠 𝑑 / 𝑛 t follows has a t-distribution with n-1 degrees of freedom
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