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BCOR 1020 Business Statistics Lecture 17 – March 18, 2008
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Overview Chapter 8 – Sampling Distributions and Estimation –Confidence Intervals Mean ( ) with variance ( ) unknown –Sample size determination for the mean ( )
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Chapter 8 – Confidence Interval for a Mean ( ) with Unknown Suppose we want to find a confidence interval for when is unknown: Instead of using We will estimate with S and use We need to know how this statistic distributed to compute the confidence interval for when s is unknown?
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Chapter 8 – Confidence Interval for a Mean ( ) with Unknown has a Student’s t distribution with (n – 1) degrees of freedom, denoted T ~ t(n–1). Assumes the population from which we are sampling is Normal. t distributions are symmetric and shaped like the standard normal distribution. The t distribution is dependent on the size of the sample. (As n increases, the t distribution approaches the standard normal.)
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Chapter 8 – Confidence Interval for a Mean ( ) with Unknown Degrees of Freedom: Degrees of Freedom (d.f.) is a parameter based on the sample size that is used to determine the value of the t statistic. Degrees of freedom (denoted by ) tell how many independent observations are used to calculate our estimate of , S. = n - 1 For a given confidence level, t is always larger than z, so a confidence interval based on t is always wider than if z were used.
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Chapter 8 – Confidence Interval for a Mean ( ) with Unknown Comparison of z and t: For very small samples, t-values differ substantially from the normal. As degrees of freedom increase, the t-values approach the normal z-values. For example, for n = 31, the degrees of freedom are: What would the t-value be for a 90% confidence interval? = 31 – 1 = 30
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Chapter 8 – Confidence Interval for a Mean ( ) with Unknown Comparison of z and t: For = 30, the corresponding t-value is 1.697. Compare this to the 90% z-value, 1.645.
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Clickers Use the table below to find the t-value for a 95% confidence interval from a sample of size n = 6. A = 3.365B = 2.015 C = 1.960D = 2.571
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Chapter 8 – Confidence Interval for a Mean ( ) with Unknown Use the Student’s t distribution instead of the normal distribution when the population is normal but the standard deviation is unknown and the sample size is small. Constructing the Confidence Interval with the Student’s t Distribution: x + tx + t snsn The confidence interval for (unknown ) is Solving the inequality inside the probability statement P(–t < < t ) = 1 – leads to the interval
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Chapter 8 – Confidence Interval for a Mean ( ) with Unknown Graphically…
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Chapter 8 – Confidence Interval for a Mean ( ) with Unknown Example: GMAT Scores Here are GMAT scores from 20 applicants to an MBA program: x = 510 s = 73.77 Construct a 90% confidence interval for the mean GMAT score of all MBA applicants. Since is unknown, use the Student’s t for the confidence interval with = 20 – 1 = 19 d.f. First find t 0.90 from Appendix D.
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Chapter 8 – Confidence Interval for a Mean ( ) with Unknown Example: GMAT Scores For = 20 – 1 = 19 d.f., t 0.90 = 1.729 (from Appendix D).
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Chapter 8 – Confidence Interval for a Mean ( ) with Unknown Example: GMAT Scores The 90% confidence interval is: x - t snsn x + t snsn < < 510 – 1.729 73.77 20 < < 510 + 1.729 73.77 20 510 – 28.52 < < 510 + 28.52 We are 90% certain that the true mean GMAT score is within the interval 481.48 < < 538.52.
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Chapter 8 – Confidence Interval for a Mean ( ) with Unknown Using Appendix D: Beyond = 50, Appendix D shows in steps of 5 or 10. If the table does not give the exact degrees of freedom, use the t-value for the next lower. This is a conservative procedure since it causes the interval to be slightly wider. Using MegaStat: MegaStat gives you a choice of z or t and does all calculations for you.
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Clickers A sample of 20 final exams from BCOR1020 has an average of and a standard deviation of S = 8. Assuming the population of final exam scores is normally distributed, find the appropriate t distribution critical value to determine the 95% confidence interval for the mean score. (t-dist. Table on overhead) A) t = 1.725 B) t = 1.729 C) t = 2.086 D) t = 2.093
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Clickers A sample of 20 final exams from BCOR1020 has an average of and a standard deviation of S = 8. Assuming the population of final exam scores is normally distributed, find the 95% confidence interval for the mean score. (t-dist. Table on overhead) A) B) C) D)
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Chapter 8 – Confidence Interval for a Mean ( ) with Unknown Confidence Interval Width: Confidence interval width reflects - the sample size, - the confidence level and - the standard deviation. To obtain a narrower interval and more precision - increase the sample size or - lower the confidence level (e.g., from 90% to 80% confidence)
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Chapter 8 – Sample Size Determination for C. I. for a Mean ( ) To estimate a population mean with a precision of + E (allowable error), you would need a sample of what size? Sample Size to Estimate (assuming is known): This formula is derived by comparing the intervals + E and +. Solving the following equation for n: Gives us the formula: Always round up!
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Chapter 8 – Sample Size Determination for C. I. for a Mean ( ) Method 1: Take a Preliminary Sample: Take a small preliminary sample and use the sample s in place of in the sample size formula. Method 2: Assume Uniform Population: Estimate rough upper and lower limits a and b and set = [(b-a)/12]½. Method 3: Assume Normal Population: Estimate rough upper and lower limits a and b and set = (b-a)/4. This assumes normality with most of the data with + 2 so the range is 4 . Method 4: Poisson Arrivals: In the special case when = is a Poisson arrival rate, then = = How to Estimate an unknown ?
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Chapter 8 – Sample Size Determination for C. I. for a Mean ( ) Example: GMAT Scores Recall from our last lecture that the population standard deviation for GMAT scores is = 86.8. How large should our sample be if we would like to have a 95% confidence interval for average GMAT scores that is no wider than + 20 points? We know = 86.8,E = 20,z = 1.96 for 95% C.I. So, Since we always round up, we will draw a sample of size n = 73.
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Clickers Suppose we want to find a confidence interval for the mean score on BCOR1020 final exams. Assuming the population of final exam has a standard deviation of = 8, how large should our sample be if we would like a 95% confidence interval no wider than + 2 points? A) n = 30 B) n = 62 C) n = 100 D) n = 246
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