chapter 8: estimating with confidence. 8.3 – estimating a population mean in the previous...
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Chapter 8: Estimating With Confidence
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8.3 – Estimating a Population Mean
In the previous examples, we made an unrealistic assumption that the population standard deviation was known and could be used to calculate confidence intervals.
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When the standard deviation of a statistic is estimated from the data
Standard Error:
s
n
When we know we can use the Z-table to make a confidence interval. But, when we don’t know it, then we have to use something else!
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Properties of the t-distribution:
• σ is unknown
• Degrees of Freedom = n – 1
• More variable than the normal distribution (it has fatter tails than the normal curve)
• Approaches the normal distribution when the degrees of freedom are large (sample size is large).
• Area is found to the right of the t-value
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Properties of the t-distribution:
• If n < 15, if population is approx normal, then so is the sample distribution. If the data are clearly non-Normal or if outliers are present, don’t use!
• If n > 15, sample distribution is normal, except if population has outliers or strong skewness
• If n 30, sample distribution is normal, even if population has outliers or strong skewness
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Example #1Determine the degrees of freedom and use the t-table to find probabilities for each of the following:
DF Picture Probability
P(t > 1.093) n = 11 10 1.093
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DF Picture Probability
P(t > 1.093) n = 11 10 1.0930.15
Example #1Determine the degrees of freedom and use the t-table to find probabilities for each of the following:
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Example #1Determine the degrees of freedom and use the t-table to find probabilities for each of the following:
DF Picture Probability
P(t < 1.093) n = 11 10
1.093
0.85
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Example #1Determine the degrees of freedom and use the t-table to find probabilities for each of the following:
DFPicture
Probability
P(t > 0.685) n = 24 230.685
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Example #1Determine the degrees of freedom and use the t-table to find probabilities for each of the following:
DFPicture
Probability
P(t > 0.685) n = 24 230.685
0.25
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Example #1Determine the degrees of freedom and use the t-table to find probabilities for each of the following:
DFPicture
Probability
P(t > 0.685) n = 24 23-0.685
0.25
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Example #1Determine the degrees of freedom and use the t-table to find probabilities for each of the following:
DFPicture
Probability
P(0.70<t<1.093) n = 11 10
1.0930.70
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.25 – .15 = 0.1
Example #1Determine the degrees of freedom and use the t-table to find probabilities for each of the following:
DFPicture
Probability
P(0.70<t<1.093) n = 11 10
1.0930.70
0.1
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Calculator Tip: Finding P(t)
2nd – Dist – tcdf( lower bound, upper bound, degrees of freedom)
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One-Sample t-interval:
1*ns
x tn
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Calculator Tip: One sample t-Interval
Stat – Tests – TInterval
Data: If given actual values
Stats: If given summary of values
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Conditions for a t-interval:
1. SRS
(population approx normal and n<15, or moderate size (15≤ n < 30) with moderate skewness or outliers, or large sample size n ≥ 30)
(should say)
10N n
(Population 10x sample size)
2. Normality
3. Independence
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Robustness:
The probability calculations remain fairly accurate when a condition for use of the procedure is violated
The t-distribution is robust for large n values, mostly because as n increases, the t-distribution approaches the Z-distribution. And by the CLT, it is approx normal.
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Example #2Practice finding t*
n Degrees of Freedom
Confidence Interval
t*
n = 10 99% CI
n = 20 90% CI
n = 40 95% CI
n = 30 99% CI
9
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Example #2Practice finding t*
n Degrees of Freedom
Confidence Interval
t*
n = 10 99% CI
n = 20 90% CI
n = 40 95% CI
n = 30 99% CI
9 3.250
19
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Example #2Practice finding t*
n Degrees of Freedom
Confidence Interval
t*
n = 10 99% CI
n = 20 90% CI
n = 40 95% CI
n = 30 99% CI
9 3.250
19 1.729
39
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Example #2Practice finding t*
n Degrees of Freedom
Confidence Interval
t*
n = 10 99% CI
n = 20 90% CI
n = 40 95% CI
n = 30 99% CI
9 3.250
19 1.729
39 2.042
29
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Example #2Practice finding t*
n Degrees of Freedom
Confidence Interval
t*
n = 10 99% CI
n = 20 90% CI
n = 40 95% CI
n = 30 99% CI
9 3.250
19 1.729
39 2.042
29 2.756
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Example #3As part of your work in an environmental awareness group, you want to estimate the mean waste generated by American adults. In a random sample of 20 American adults, you find that the mean waste generated per person per day is 4.3 pounds with a standard deviation of 1.2 pounds. Calculate a 99% confidence interval for and explain it’s meaning to someone who doesn’t know statistics.
P: The true mean waste generated per person per day.
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A: SRS:
Normality:
Independence:
N: One Sample t-interval
Says randomly selected
15<n<30. We must assume the population doesn’t have strong skewness. Proceeding with caution!
It is safe to assume that there are more than 200 Americans that create waste.
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I: 1*ns
x tn
df = 20 – 1 = 19
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I:
1.24.3 2.861
20
4.3 0.7677
3.5323, 5.0677
1*ns
x tn
df = 20 – 1 = 19
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C: I am 99% confident the true mean waste generated per person per day is between 3.5323 and 5.0677 pounds.
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Matched Pairs t-procedures:
Subjects are matched according to characteristics that affect the response, and then one member is randomly assigned to treatment 1 and the other to treatment 2. Recall that twin studies provide a natural pairing. Before and after studies are examples of matched pairs designs, but they require careful interpretation because random assignment is not used.
Apply the one-sample t procedures to the differences
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Confidence Intervals for Matched Pairs
n
Stx dnd*
1
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Example #4Archaeologists use the chemical composition of clay found in pottery artifacts to determine whether different sites were populated by the same ancient people. They collected five random samples from each of two sites in Great Britain and measured the percentage of aluminum oxide in each. Based on these data, do you think the same people used these two kiln sites? Use a 95% confidence interval for the difference in aluminum oxide content of pottery made at the sites and assume the population distribution is approximately normal. Can you say there is no difference between the sites?
New Forrest
20.8 18 18 15.8 18.3
Ashley Trails
19.1 14.8 16.7 18.3 17.7
Difference 1.7 3.2 1.3 -2.5 .6
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P: μn = New Forrest percentage of aluminum oxide
The true mean difference in aluminum oxide levels between the New Forrest and Ashley Trails.
μa = Ashley Trails percentage of aluminum oxide
μd = μn - μa = Difference in aluminum oxide levels
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A: SRS:
Normality:
Independence:
N: Matched Pairs t-interval
Says randomly selected
Says population is approx normal
It is safe to assume that there are more than 50 samples available
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I: df =
n
Stx dnd*
1
5 – 1 = 4
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I: df = 20 – 1 = 19
n
Stx dnd*
1
5
105469069.2776.286.
613866034.286.
4743.3 ,754.1
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C: I am 95% confident the true mean difference in aluminum oxide levels between the New Forrest and Ashley Trails is between –1.754 and 3.4743.
Can you say there is no difference between the sites?
Yes, zero is in the confidence interval, so it is safe to say there is no difference.
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Example #5The National Endowment for the Humanities sponsors summer institutes to improve the skills of high school language teachers. One institute hosted 20 Spanish teachers for four weeks. At the beginning of the period, the teachers took the Modern Language Association’s listening test of understanding of spoken Spanish. After four weeks of immersion in Spanish in and out of class, they took the listening test again. (The actual spoken Spanish in the two tests was different, so that simply taking the first test should not improve the score on the second test.) Below is the pretest and posttest scores. Give a 90% confidence interval for the mean increase in listening score due to attending the summer institute. Can you say the program was successful?
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Subject Pretest Posttest Subject Pretest Posttest
1 30 29 11 30 32
2 28 30 12 29 28
3 31 32 13 31 34
4 26 30 14 29 32
5 20 16 15 34 32
6 30 25 16 20 27
7 34 31 17 26 28
8 15 18 18 25 29
9 28 33 19 31 32
10 20 25 20 29 32
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P: μB = Pretest score
The true mean difference in test scores between the Pretest and Posttest
μd = μB - μA = Difference in test scores
μA = Posttest score
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A: SRS:
Normality:
We must assume the 20 teachers are randomly selected
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A: SRS:
Normality:
Independence:
N: Matched Pairs t-interval
We must assume the 20 teachers are randomly selected
15<n<30 and distribution is approximately normal, so safe to assume
It is safe to assume that there are more than 200 Spanish teachers
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I: df =
n
Stx dnd*
1
20 – 1 = 19
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I: df = 20 – 1 = 19
n
Stx dnd*
1
3.20321.45 1.729
20
1.45 1.2384
2.689, 0.2115
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C: I am 90% confident the true mean difference in test scores between the Pretest and Posttestis between –2.689 and –0.2115.
Can you say the program was successful?
Yes, zero is not in the confidence interval, so the pretest score is lower than the posttest score.