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## Chapter 6

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**Chapter 6**Confidence Intervals**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 • 6.4 Confidence Intervals for Variance and Standard Deviation**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**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**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**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.**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?**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 .**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.**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**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 .**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.**0.95**0.025 0.025 z z = 0 zc Solution: Finding the Margin of Error • First find the critical values -zc= -1.96 zc 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.)**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.**Confidence Intervals for the Population Mean**A c-confidence interval for the population mean μ • The probability that the confidence interval contains μis c.**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 .**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:**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**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.**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.**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 -zc= -1.645 zc= 1.645 zc= 1.645**Solution: Constructing a Confidence Interval σKnown**• Margin of error: • Confidence interval: Left Endpoint: Right Endpoint: 22.3 < μ < 23.5**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.**Using Ti-83 or 84 to find z confidence interval:**The same as: 22.3 < μ < 23.5**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.**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**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.**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.**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**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.**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**Section 6.2**Confidence Intervals for the Mean (Small Samples)**Section 6.2 Objectives**• Interpret the t-distribution and use a t-distribution table • Construct confidence intervals when n < 30, the population is normally distributed, and σ is unknown**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. • Critical values of t are denoted by tc.**In statistics, the t-distribution was first derived in 1876**by Helmertand Lüroth. In the English-language literature it takes its name from William Sealy Gosset's 1908 paper in Biometrika under the pseudonym "Student", published while he worked at the GuinnessBrewery in Dublin, Ireland. One version of the origin of the pseudonym is that Gosset's employer forbade members of its staff from publishing scientific papers, so he had to hide his identity. Another version is that Guinness did not want their competition to know that they were using the t-test to test the quality of raw material.The t-test and the associated theory became well-known through the work of Ronald A. Fisher, who called the distribution "Student's distribution". T-distribution’s mathematical formulas:**Properties of the t-Distribution**• The t-distribution is bell shaped and symmetric about the mean. • The t-distribution is a family of curves, each determined by a parameter called the degrees of freedom. The degrees of freedom are the number of free choices left after a sample statistic such as is calculated. When you use a t-distribution to estimate a population mean, the degrees of freedom are equal to one less than the sample size. • d.f. = n – 1 Degrees of freedom**d.f. = 2**d.f. = 5 Properties of the t-Distribution • The total area under a t-curve is 1 or 100%. • The mean, median, and mode of the t-distribution are equal to zero. • As the degrees of freedom increase, the t-distribution approaches the normal distribution. After 30 d.f., the t-distribution is very close to the standard normal z-distribution. The tails in the t-distribution are “thicker” than those in the standard normal distribution. t 0 Standard normal curve**Example: Critical Values of t**Find the critical value tc for a 95% confidence when the sample size is 15. Solution: d.f. = n – 1 = 15 – 1 = 14 Table 5: t-Distribution tc = 2.145**c = 0.95**t -tc = -2.145 tc = 2.145 Solution: Critical Values of t 95% of the area under the t-distribution curve with 14 degrees of freedom lies between t = +2.145.**Confidence Intervals for the Population Mean**A c-confidence interval for the population mean μ • The probability that the confidence interval contains μis c.**Confidence Intervals and t-Distributions**In Words In Symbols • Identify the sample statistics n, , and s. • Identify the degrees of freedom, the level of confidence c, and the critical value tc. • Find the margin of error E. d.f. = n – 1**Confidence Intervals and t-Distributions**In Words In Symbols Left endpoint: Right endpoint: Interval: Find the left and right endpoints and form the confidence interval.**Example: Constructing a Confidence Interval**You 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. Solution: Use the t-distribution (n < 30, σ is unknown, temperatures are approximately normally distributed.)**Solution: Constructing a Confidence Interval**• n =16, x = 162.0 s = 10.0 c = 0.95 • df = n – 1 = 16 – 1 = 15 • Critical Value Table 5: t-Distribution tc = 2.131**Solution: Constructing a Confidence Interval**• Margin of error: • Confidence interval: Left Endpoint: Right Endpoint: 156.7 < μ < 167.3**Solution: Constructing a Confidence Interval**• 156.7 < μ < 167.3 Point estimate 156.7 162.0 167.3 • ( ) With 95% confidence, you can say that the mean temperature of coffee sold is between 156.7ºF and 167.3ºF.**n =16, x = 162.0 s = 10.0 c = 0.95**df = n – 1 = 16 – 1 = 15 The same as: 156.7 < μ < 167.3**Use the normal distribution with**If is unknown, use s instead. Yes Yes No No Yes Use the normal distribution with No Use the t-distribution with and n – 1 degrees of freedom. Normal or t-Distribution? Is n 30? Is the population normally, or approximately normally, distributed? Cannot use the normal distribution or the t-distribution. Is known?