chapter outline 6.1 confidence intervals for the mean (large samples) 6.2 confidence intervals for...
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Example: Point Estimate for Population μ Market researchers use the number of sentences per advertisement as a measure of readability for magazine advertisements. The following represents a random sample of the number of sentences found in 50 advertisements. Find a point estimate of the population mean, . (Source: Journal of Advertising Research) Larson/Farber 4th edTRANSCRIPT
Chapter Outline
• 6.1 Confidence Intervals for the Mean (Large Samples)
• 6.2 Confidence Intervals for the Mean (Small Samples)
• 6.3 Confidence Intervals for Population Proportions
Larson/Farber 4th ed 2
Example: Point Estimate for Population μ
Market researchers use the number of sentences per advertisement as a measure of readability for magazine advertisements. The following represents a random sample of the number of sentences found in 50 advertisements. Find a point estimate of the population mean, . (Source: Journal of Advertising Research)
Larson/Farber 4th ed 3
9 20 18 16 9 9 11 13 22 16 5 18 6 6 5 12 2517 23 7 10 9 10 10 5 11 18 18 9 9 17 13 11 714 6 11 12 11 6 12 14 11 9 18 12 12 17 11 20
Solution: Point Estimate for Population μ
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The sample mean of the data is
620 12.450
xxn
Your point estimate for the mean length of all magazine advertisements is 12.4 sentences.
Interval EstimateInterval estimate • An interval, or range of values, used to estimate a
population parameter.
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Point estimate
• 12.4
How confident do we want to be that the interval estimate contains the population mean μ?
( )
Interval estimate
Confidence Intervals for the Population Mean
A c-confidence interval for the population mean μ
•
• The probability that the confidence interval contains μ is c.
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where cx E x E E zn
Solution: Finding the Margin of Error
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5.01.9650
1.4
c csE z z
n n
You don’t know σ, but since n ≥ 30, you can use s in place of σ.
You are 95% confident that the margin of error for the population mean is about 1.4 sentences.
Example: Constructing a Confidence IntervalConstruct a 95% confidence interval for the
mean number of sentences in all magazine advertisements.
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Solution: Recall and E = 1.412.4x
12.4 1.411.0
x E
12.4 1.413.8
x E
11.0 < μ < 13.8
Left Endpoint: Right Endpoint:
Constructing Confidence Intervals for μ
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Finding a Confidence Interval for a Population Mean (n 30 or σ known with a normally distributed population)
In Words In Symbols
1. Find the sample statistics n and .
2. Specify , if known. Otherwise, if n 30, find the sample standard deviation s and use it as an estimate for .
xx n
2( )1
x xs n
x
Constructing Confidence Intervals for μ
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3. Find the critical value zc that corresponds to the given level of confidence.
4. Find the margin of error E.
5. Find the left and right endpoints and form the confidence interval.
Use the Standard Normal Table.
Left endpoint: Right endpoint: Interval:
cE zn
x Ex E
x E x E
In Words In Symbols
Interpreting the Results• μ is a fixed number. It is either in the
confidence interval or not.• Incorrect: “There is a 90% probability that
the actual mean is in the interval (22.3, 23.5).”
• Correct: “If a large number of samples is collected and a confidence interval is created for each sample, approximately 90% of these intervals will contain μ.
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Sample Size
• Given a c-confidence level and a margin of error E, the minimum sample size n needed to estimate the population mean is
• If is unknown, you can estimate it using s provided you have a preliminary sample with at least 30 members.
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2cz
nE
Example: Sample SizeYou want to estimate the mean number of sentences in a magazine advertisement. How many magazine advertisements must be included in the sample if you want to be 95% confident that the sample mean is within one sentence of the population mean? Assume the sample standard deviation is about 5.0.
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Solution: Sample Size
zc = 1.96 s = 5.0 E = 1
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221.96 5.0 96.04
1cz
nE
When necessary, round up to obtain a whole number.
You should include at least 97 magazine advertisements in your sample.
Section 6.2
Confidence Intervals for the Mean (Small Samples)
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The t-Distribution
• When the population standard deviation is unknown, the sample size is less than 30, and the random variable x is approximately normally distributed, it follows a t-distribution.
• .
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-xt sn
Confidence Intervals for the Population Meanσ unknown and n < 30
• The probability that the confidence interval contains μ is c.
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where csx E x E E tn
Example: Critical Values of tFind the critical value tc for a 95% confidence when the sample size is 15.
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Table 5: t-Distribution
tc = 2.145
Solution: d.f. = n – 1 = 15 – 1 = 14
Example: Constructing a Confidence IntervalYou randomly select 16 coffee shops and
measure the temperature of the coffee sold at each. The sample mean temperature is 162.0ºF with a sample standard deviation of 10.0ºF. Find the 95% confidence interval for the mean temperature. Assume the temperatures are approximately normally distributed.
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Solution:Use the t-distribution (n < 30, σ is unknown, temperatures are approximately distributed.)
•
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where csx E x E E tn
n = 16 x bar = 162 s = 10 want a 95% confidence interval
Solution: Constructing a Confidence Interval• n =16, x = 162.0 s = 10.0 c = 0.95
• df = n – 1 = 16 – 1 = 15
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Table 5: t-Distribution
tc = 2.131
Section 6.3
Confidence Intervals for Population Proportions
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Section 6.3 Objectives
• Find a point estimate for the population proportion• Construct a confidence interval for a population
proportion• Determine the minimum sample size required
when estimating a population proportion
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Point Estimate for Population pPopulation Proportion• The probability of success in a single trial of
a binomial experiment. • Denoted by pPoint Estimate for p• The proportion of successes in a sample. • Denoted by – – read as “p hat”
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number of successes in sampleˆ number in samplexp n
Example: Confidence Interval for pIn a survey of 1219 U.S. adults, 354 said that their favorite sport to watch is football. Construct a 95% confidence interval for the proportion of adults in the United States who say that their favorite sport to watch is football. 354/1219 = .2904
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So3543 ˆ 0.290402p
1 0.290402ˆ ˆ 0.7095981q p
Solution: Confidence Interval for p
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• Margin of error:
(0.290402) (0.709598)1.96ˆ ˆ 0.0251219cpqE z n
Solution: Confidence Interval for p
• Confidence interval:
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ˆ0.29 0.0250.265
p E
Left Endpoint: Right Endpoint:
0.265 < p < 0.315
ˆ0.29 0.0250.315
p E
Solution: Confidence Interval for p
• 0.265 < p < 0.315
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( )• 0.290.265 0.315
With 95% confidence, you can say that the proportion of adults who say football is their favorite sport is between 26.5% and 31.5%.
Point estimate
p̂p̂ E p̂ E