1. Section 8.3 Estimating Population Means (Small Samples) HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2008 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved.

2. σ unknown, and a small sample (n < 30) AND population is normally distributed Large sample: Use z, the normal distribution. Small sample: Use the Student t-distribution. z t This is where we were in the previous lesson. TODAY (added content by D.R.S.)

3. HAWKES LEARNING SYSTEMS math courseware specialists Criteria for estimating the population mean for small samples: All possible samples of a given size have an equal probability of being chosen. The size of the sample is less than 30 (n < 30). The distribution of the population is approximately normal. The population’s standard deviation is unknown. Confidence Intervals 8.3 Estimating Population Means (Small Samples) When all of the above conditions are met, then the distribution used to calculate the margin of error for the population mean is the Student t-distribution.

4. HAWKES LEARNING SYSTEMS math courseware specialists Margin of Error, E, for Small Samples: Confidence Intervals 8.3 Estimating Population Means (Small Samples) When calculating the margin of error, round to one more decimal place than the original data, or the same number of places as the standard deviation. t  /2  the critical t-value  1 – c s  the sample standard deviation n  the sample size

5. Assuming the population is approximately normal and all samples have an equal probability of being chosen, find the margin of error for a sample size of 10 given that s = 15.5 and the level of confidence is 95%. Find the margin of error: HAWKES LEARNING SYSTEMS math courseware specialists n  10, s  15.5, c  0.95,  1 – 0.95  0.05 t 0.05/2  t 0.025  Solution: Confidence Intervals 8.3 Estimating Population Means (Small Samples) 2.262 Printed table lookup: C=95% or α=.05 or α/2=0.025 d.f. = n – 1 = 9

6. A student records the repair cost for 20 randomly selected computers. A sample mean of $216.53 and a standard deviation of $15.86 are subsequently computed. Determine the 98% confidence interval for the mean repair cost for all computers. Assume the criteria for this section are met. Construct a confidence interval: HAWKES LEARNING SYSTEMS math courseware specialists n  20,  216.53, s  15.86, c  0.98,  1 – 0.98  0.02 t 0.02/2  t 0.01  Solution: Confidence Intervals 8.3 Estimating Population Means (Small Samples) 2.539 $207.53 <  < $225.53 ($207.53, $225.53) 216.53 – 9.00 <  < 216.53+ 9.00 Printed table lookup: C=98% or α=.02 or α/2=0.01 d.f. = n – 1 = 19

7. Using TI-84 to find critical value of t Similar to invNorm (area to left ) = critical z But t involves degrees of freedom, too invT(area to left, df) = critical t value invT is only available on TI-84 TI-83/Plus owners: – Use printed tables if asked for critical value – If asked only for the interval, you do have TInterval (added content by D.R.S.)

8. Using TI-84 to find critical value of t From the two examples they demonstrated with primitive formulas, we use invT C = 95%, α = 0.05, α/2 = 0.025, n = 10, df = 9 C = 98%, α = 0.02, α/2 = 0.01, n = 20, df = 19 Their table values were 2.262 and 2.539 (added content by D.R.S.)

9. A student records the repair cost for 20 randomly selected computers. A sample mean of $216.53 and a standard deviation of $15.86 are subsequently computed. Determine the 98% confidence interval for the mean repair cost for all computers. Assume the criteria for this section are met. Construct a confidence interval: HAWKES LEARNING SYSTEMS math courseware specialists Solution: Confidence Intervals 8.3 Estimating Population Means (Small Samples) Inputs Outputs Note that these results differ from theirs by a penny. This is because the calculator’s T critical value is more precise. (added content by D.R.S.) Using TI-84 STAT, TESTS, TInterval