1. Ch 12 實習

2. Introduction
We shall develop techniques to estimate and test three population parameters.
Population mean m
Population variance s2
Population proportion p
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3. Recall that when s is known we use the following
Inference About a Population Mean When the Population Standard Deviation Is Unknown
Recall that when s is known we use the following
statistic to estimate and test a population mean
When s is unknown, we use its point estimator s, and the z-statistic is replaced then by the t-statistic
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4. The t - Statistic t s The “degrees of freedom”,
(a function of the sample size)
determine how spread the
distribution is (compared to the
normal distribution)
The t distribution is mound-shaped,
and symmetrical around zero.
d.f. = v2
d.f. = v1
v1 < v2
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5. 自由度
統計學上的自由度（degree of freedom, df），是指當以樣本的統計量來估計總體的參數時， 樣本中獨立或能自由變化的資料的個數，稱為該統計量的自由度
Ex:
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6. How to calculus sample variance
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7. Testing m when s is unknown
Example 1
In order to determine the number of workers required to meet demand, the productivity of newly hired trainees is studied.
It is believed that trainees can process and distribute more than 450 packages per hour within one week of hiring.
Can we conclude that this belief is correct, based on productivity observation of 50 trainees
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8. Testing m when s is unknown
Example 1 – Solution
The problem objective is to describe the population of the number of packages processed in one hour.
The data are interval.
H0:m = H1:m > 450
The t statistic d.f. = n - 1 = 49
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9. Testing m when s is unknown
Solution continued (solving by hand)
The rejection region is t > ta,n – 1 ta,n - 1 = t.05,49 @ t.05,50 =
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10. Testing m when s is unknown
The test statistic is
Rejection region
1.676
1.89
Since 1.89 > we reject the null hypothesis in favor of the alternative.
There is sufficient evidence to infer that the mean productivity of trainees one week after being hired is greater than 450 packages at .05 significance level.
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11. Estimating m when s is unknown
Confidence interval estimator of m when s is unknown
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12. Estimating m when s is unknown
Example 2
An investor is trying to estimate the return on investment in companies that won quality awards last year.
A random sample of 83 such companies is selected, and the return on investment is calculated had he invested in them.
Construct a 95% confidence interval for the mean return.
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13. Estimating m when s is unknown
Solution (solving by hand)
The problem objective is to describe the population of annual returns from buying shares of quality award-winners.
The data are interval.
Solving by hand
From the data we determine
t.025,80
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14. Checking the required conditions
We need to check that the population is normally distributed, or at least not extremely nonnormal.
There are statistical methods to test for normality
From the sample histograms we see…
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15. A Histogram for Example 1
Packages
A Histogram for Example 2
Returns
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16. Summary of Test Statistics to be Used in a Hypothesis Test about a Population Mean
Yes
n > 30 ? No
s known ? No
Popul.
approx.
normal ?
Yes
Yes
Use s to
estimate s
s known ? No No
Yes
Use s to
estimate s
Increase n
to > 30
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17. Example 1
A federal agency responsible for enforcing laws governing weights and measures routinely inspects packages to determine whether the weight of the contents is at least as great as that advertised on the package. A random sample of 18 containers whose packaging states that the contents weigh 8 ounces was drawn. The contents were weighted and the results follows. Can we concluded at the 1% significance level that on average the containers are mislabeled? (Assume the random variable is normally distributed)
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18. Solution
H0:μ=8
H1:μ<8
There is enough evidence to conclude that the average container is mislabeled
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19. Example 2
How much money do winners go home with from the television quiz show Feopardy? To determine an answer, a random sample of winners was drawn and the amount of money each won was recorded and is listed here. Estimate with 95% confidence the mean winnings for all show’s players (Assume the random variable is normally distributed)
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20. Solution
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21. Example 3
A random sample of 10 college students was drawn from a large university. Their ages are 22, 17, 27, 20, 23, 19, 24, 18, 19, and 24 years. Assume the age is normal distributed.
a. Estimate the population mean with 90% confidence.
b. Test to determine if we can infer at the 5% significance level that the population mean is not equal to 20.
c. What is the required condition of the techniques used in the previous questions? What graphical device can you use to check to see if that required condition is satisfied?
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22. Solution a. Thus, LCL = 19.446, and UCL = 23.154.
b. H0: μ = 20 vs. H1: μ ≠ 20
Rejection region: | t | > t0.025,9 = 2.262
Test statistic: t = 1.285
Conclusion: Don't reject H0. We can't infer at the 5% significance level that the population mean is not equal to 20.
The condition is that ages in the population are normally distributed. A histogram of the data can be used to check if the normality assumption is satisfied.
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23. Inference About a Population Variance
Sometimes we are interested in making inference about the variability of processes.
Examples:
The consistency of a production process for quality control purposes.
Investors use variance as a measure of risk.
To draw inference about variability, the parameter of interest is s2.
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24. Inference About a Population Variance
The sample variance s2 is an unbiased, consistent and efficient point estimator for s2.
The statistic has a distribution called Chi-squared, if the population is normally distributed.
d.f. = 5
d.f. = 10
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25. Testing the Population Variance
Example 3 (operation management application)
A container-filling machine is believed to fill 1 liter containers so consistently, that the variance of the filling will be less than 1 cc (.001 liter).
To test this belief a random sample of 25 1-liter fills was taken, and the results recorded
Do these data support the belief that the variance is less than 1cc at 5% significance level?
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26. Testing the Population Variance
Solution
The problem objective is to describe the population of 1-liter fills from a filling machine.
The data are interval, and we are interested in the variability of the fills.
The complete test is: H0: s2 = 1
H1: s2 <1
We want to know whether the process is consistent
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27. Testing the Population Variance
Solving by hand
Note that (n - 1)s2 = S(xi - x)2 = Sxi2 – (Sxi)2/n
From the sample, we can calculate Sxi = 24,996.4, and Sxi2 = 24,992,821.3
Then (n - 1)s2 = 24,992,821.3-(24,996.4)2/25 =20.78
There is insufficient evidence
to reject the hypothesis that
the variance is less than 1.
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28. Testing the Population Variance
Rejection
region
20.8
Do not reject the null hypothesis
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29. Testing and Estimating a Population Variance
From the following probability statement P(c21-a/2 < c2 < c2a/2) = 1-a we have (by substituting c2 = [(n - 1)s2]/s2.)
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30. Example 4
With gasoline prices increasing, drivers are becoming more concerned with their cars’ gasoline consumption. For the past 5 years, a driver has tracked the gas mileage of his car and found that the variance from fill-up to fill-up was σ2=23 mpg2. Now that his car is 5 years old, he would like to know whether the variability of gas mileage has changed. He recorded the gas mileage from his last eight fill-ups; these are listed here. Conduct a test at a 10% significance level to infer whether the variability has changed.
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31. Solution
H0:σ2=23
H1:σ2≠23
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32. Example 5
During annual checkups physician routinely send their patients to medical laboratories to have various tests performed. One such test determines the cholesterol level in patients’ blood. However, not all tests are conducted in the same way. To acquire more information, a man was sent to 10 laboratories and in each had his cholesterol level measured. The results are listed here. Estimate with 95% confidence the variance of these measurements.
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33. Solution
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34. Example 6
Which of the following conditions is needed regarding the chi-squared test statistic for the test of variance?
a. The population random variable must be normal.
b. The test statistic must be a non-negative number.
c. The test statistic must have a chi-squared distribution with n - 1 degrees of freedom.
d. All of these choices are true.
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35. Inference About a Population Proportion
When the population consists of nominal data, the only inference we can make is about the proportion of occurrence of a certain value.
The parameter p was used before to calculate these probabilities under the binomial distribution.
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36. Inference About a Population Proportion
Statistic and sampling distribution
the statistic used when making inference about p is:
Under certain conditions, [np > 5 and n(1-p) > 5], is approximately normally distributed, with m = p and s2 = p(1 - p)/n.
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37. Testing and Estimating the Proportion
Test statistic for p
Interval estimator for p (1-a confidence level)
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38. Example 7
A dean of a business school wanted to know whether the graduates of her school used a statistical inference technique during their first year of employment after graduation. She surveyed 314 graduates and asked about the use of statistical technique. After tallying up the responses, she found that 204 used statistical inference within one year of graduation. Estimate with 90% confidence the proportion of all business school graduates who use their statistical education within a year of graduation.
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39. Solution
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40. Example 8
In some states the law requires drivers to turn on their headlights when driving in the rain. A highway patrol officer believes that less than one-quarter of all drivers follow this rule. As a test, he randomly samples 200 cars driving in the rain and counts the number whose headlights are turned on. H finds this number to be 41. Does the officer have enough evidence at the 10% significance level to support his belief?
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41. Solution There is enough evidence to support the officer’s belief
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42. Selecting the Sample Size to Estimate the Proportion
Recall: The confidence interval for the proportion is
Thus, to estimate the proportion to within W, we can write
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43. Selecting the Sample Size to Estimate the Proportion
The required sample size is
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44. Selecting the Sample Size
Two methods – in each case we choose a value for then
solve the equation for n.
Method 1 : no knowledge of even a rough value of This is a ‘worst case scenario’ so we substitute = .50
Method 2 : we have some idea about the value of This is a better scenario and we substitute in our estimated value.
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12.44

45. Example 9
As a manufacturer of golf clubs, a major corporation wants to estimate the proportion of golfers who are right-handed. How many golfers must be surveyed if they want to be within 0.02, with a 95% confidence?
a. Assume that there is no prior information that could be used as an estimate of .
b. Assume that the manufacturer has an estimate of found from a previous study, which suggests that 75% of golfers are right-handed.
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46. Solution a. b.
取n=1801
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