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LESSON 7

LESSON 7. LESSON 7: ESTIMATION AND CONFIDENCE INTERVALS

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LESSON 7

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  1. LESSON 7

  2. LESSON 7: ESTIMATION AND CONFIDENCE INTERVALS It is usually not feasible to inspect the entire population; means of small samples have more dispersion or scatter than means of large samples. The Central Limit Theorem states that if all samples of a particular size are selected from any population, the sampling distribution of the sample mean is approximately a normal distribution. This approximation improves with larger samples. Point estimates are sample measures of central tendency such as mean, median and mode. Sample measures of dispersion such as variance and standard deviation are also point estimates. Confidence intervals state the range within which a population parameter probably lies. The specified probability is called the level of confidence. A 95 percent confidence interval says that we have a 95 percent level of confidence that a similarly constructed interval will include the parameter being estimated. A random sample of 500 restaurants reveals a sample mean yearly income of waitresses of $ 35,743. The standard deviation of the population is 1092. What are the confidence limits at the 95% level of confidence?

  3. Confidence Interval - Mean Mean=35743 z=1.96 s =1092 n=500 Confidence Limits Lower Upper 35647 35839 (Distribution Plot using Minitab 16 ) We are 95% confident that the interval from 35647 to 35839 includes the true value of the mean. Solution: $35,743 - [(1.96)(1092)]/(500)^.5 < $35,743 < $35,743 + [(1.96)(1092)]/(500)^.5

  4. Confidence Interval -Mean Confidence Interval-Proportion Mean= 35743 Propor= 0.8 z= 1.96 z= 1.96 StdDev= 1092 n= 5000 n= 500 Confidence Limits Confidence limits Lower Upper Lower Upper 0.794 0.806 35647 35839 E= 0.006= margin of error .0.80 -E =.794 0.80 + E =.806 For the Confidence Interval for the Proportion, the Solution is: Sample proportion - margin of error < Population Prop. < Sample Prop. + margin of error 0.80 - 1.96 *[.8 * .2 / 5000]^.5 < 0.80 < 0.80 + 1.96*[.8 * .2 /5000]^.5 0.80 - 0.006 < 0.80 < 0.80 + 0.006 0.794 <0.800< 0.806

  5. A survey reported that 70% of those responding to a national survey of college freshmen were interested in taking summer courses. Using the sample size of 49, calculate a 95% confidence interval for the proportion of college students who are interested in taking summer courses. E=z*(p*q/n)^.5 = 1.96((.70)(.30)/49)^.5 = .124 = 12.4% Confidence interval .700 - .124 < .700 < .700 + .124 .576 < .700 < .824 A researcher wants to determine the proportion of pro-peace students at Ocean County College. He has no idea what the sample proportion will be. How large a sample is required to be 95% sure that the sample proportion is off by no more than 4%? Z= 1.96 p=0.5 = q since we do not know what the sample proportion will be. E=.04 n= 1.96^2 * 0.5 * 0.5 / 0.04^2 = 600.25 n=601 samples.

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