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Section 6.1. Confidence Intervals for the Mean (Large Samples). Section 6.1 Objectives. Find a point estimate and a margin of error Construct and interpret confidence intervals for the population mean Determine the minimum sample size required when estimating μ.
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Section 6.1 Confidence Intervals for the Mean (Large Samples) Larson/Farber 4th ed
Section 6.1 Objectives • Find a point estimate and a margin of error • Construct and interpret confidence intervals for the population mean • Determine the minimum sample size required when estimating μ Larson/Farber 4th ed
Point Estimate for Population μ Point Estimate • A single value estimate for a population parameter • Most unbiased point estimate of the population mean μ is the sample mean Larson/Farber 4th ed
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) 9 20 18 16 9 9 11 13 22 16 5 18 6 6 5 12 25 17 23 7 10 9 10 10 5 11 18 18 9 9 17 13 11 7 14 6 11 12 11 6 12 14 11 9 18 12 12 17 11 20 Larson/Farber 4th ed
Solution: Point Estimate for Population μ The sample mean of the data is Your point estimate for the mean length of all magazine advertisements is 12.4 sentences. Larson/Farber 4th ed
Point estimate 12.4 • Interval Estimate Interval estimate • An interval, or range of values, used to estimate a population parameter. ( ) Interval estimate How confident do we want to be that the interval estimate contains the population mean μ? Larson/Farber 4th ed
c ½(1 – c) ½(1 – c) z -zc z = 0 zc Critical values Level of Confidence Level of confidence c • The probability that the interval estimate contains the population parameter. c is the area under the standard normal curve between the critical values. Use the Standard Normal Table to find the corresponding z-scores. The remaining area in the tails is 1 – c . Larson/Farber 4th ed
c = 0.90 ½(1 – c) = 0.05 ½(1 – c) = 0.05 z z = 0 zc Level of Confidence • If the level of confidence is 90%, this means that we are 90% confident that the interval contains the population mean μ. zc zc= 1.645 -zc= -1.645 The corresponding z-scores are +1.645. Larson/Farber 4th ed
Sampling Error Sampling error • The difference between the point estimate and the actual population parameter value. • For μ: • the sampling error is the difference – μ • μ is generally unknown • varies from sample to sample Larson/Farber 4th ed
Margin of Error Margin of error • The greatest possible distance between the point estimate and the value of the parameter it is estimating for a given level of confidence, c. • Denoted by E. • Sometimes called the maximum error of estimate or error tolerance. When n 30, the sample standard deviation, s, can be used for . Larson/Farber 4th ed
Example: Finding the Margin of Error Use the magazine advertisement data and a 95% confidence level to find the margin of error for the mean number of sentences in all magazine advertisements. Assume the sample standard deviation is about 5.0. Larson/Farber 4th ed
0.95 0.025 0.025 z z = 0 zc Solution: Finding the Margin of Error • First find the critical values zc zc= 1.96 -zc= -1.96 95% of the area under the standard normal curve falls within 1.96 standard deviations of the mean. (You can approximate the distribution of the sample means with a normal curve by the Central Limit Theorem, because n ≥ 30.) Larson/Farber 4th ed
Solution: Finding the Margin of Error 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. Larson/Farber 4th ed
Confidence Intervals for the Population Mean A c-confidence interval for the population mean μ • The probability that the confidence interval contains μis c. Larson/Farber 4th ed
Constructing Confidence Intervals for μ Finding a Confidence Interval for a Population Mean (n 30 or σknown with a normally distributed population) In Words In Symbols • Find the sample statistics n and. • Specify , if known. Otherwise, if n 30, find the sample standard deviation s and use it as an estimate for . Larson/Farber 4th ed
Constructing Confidence Intervals for μ In Words In Symbols Use the Standard Normal Table. • Find the critical value zc that corresponds to the given level of confidence. • Find the margin of error E. • Find the left and right endpoints and form the confidence interval. Left endpoint: Right endpoint:Interval: Larson/Farber 4th ed
Example: Constructing a Confidence Interval Construct a 95% confidence interval for the mean number of sentences in all magazine advertisements. Solution: Recall and E = 1.4 Left Endpoint: Right Endpoint: 11.0 < μ < 13.8 Larson/Farber 4th ed
Solution: Constructing a Confidence Interval 11.0 < μ < 13.8 11.0 13.8 12.4 • ( ) With 95% confidence, you can say that the population mean number of sentences is between 11.0 and 13.8. Larson/Farber 4th ed
Example: Constructing a Confidence Interval σKnown A college admissions director wishes to estimate the mean age of all students currently enrolled. In a random sample of 20 students, the mean age is found to be 22.9 years. From past studies, the standard deviation is known to be 1.5 years, and the population is normally distributed. Construct a 90% confidence interval of the population mean age. Larson/Farber 4th ed
c = 0.90 ½(1 – c) = 0.05 ½(1 – c) = 0.05 z z = 0 zc Solution: Constructing a Confidence Interval σKnown • First find the critical values zc= 1.645 zc -zc= -1.645 zc= 1.645 Larson/Farber 4th ed
Solution: Constructing a Confidence Interval σKnown • Margin of error: • Confidence interval: Left Endpoint: Right Endpoint: 22.3 < μ < 23.5 Larson/Farber 4th ed
Solution: Constructing a Confidence Interval σKnown 22.3 < μ < 23.5 Point estimate 22.3 22.9 23.5 • ( ) With 90% confidence, you can say that the mean age of all the students is between 22.3 and 23.5 years. Larson/Farber 4th ed
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. Larson/Farber 4th ed
Interpreting the Results The horizontal segments represent 90% confidence intervals for different samples of the same size. In the long run, 9 of every 10 such intervals will contain μ. μ Larson/Farber 4th ed
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. Larson/Farber 4th ed
Example: Sample Size You 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. Larson/Farber 4th ed
0.95 0.025 0.025 z z = 0 zc Solution: Sample Size • First find the critical values zc= 1.96 zc -zc= -1.96 zc= 1.96 Larson/Farber 4th ed
Solution: Sample Size zc = 1.96 s = 5.0 E = 1 When necessary, round up to obtain a whole number. You should include at least 97 magazine advertisements in your sample. Larson/Farber 4th ed
Section 6.1 Summary • Found a point estimate and a margin of error • Constructed and interpreted confidence intervals for the population mean • Determined the minimum sample size required when estimating μ Larson/Farber 4th ed