With 𝜎 unknown Problem 1: Given a sample mean of 2.8 with a sample size of 34, and standard deviation of .05; find the 95% confidence interval for the mean of the population. Problem 2: Given a sample proportion of 28% with a sample size of 34, find the 95% confidence interval for the proportion of the population. We took data and used the sample standard deviation along with the process for finding a confidence interval with 𝜎 known … ◦ However, we don’t really know 𝝈 in that case (we know s), so this is actually not the best way to construct a confidence interval in that situation! ◦ Instead, we can use a different distribution along with s in order to be more precise… Created for use in small samples and with standard deviation unknown Considers “degrees of freedom” ◦ If you have a average of 100 for a population of 10, the first 9 items are free to be whatever they want, but the 10th item must make the sum be 1,000, so it is not free. Therefore a population of 10 has 9 degrees of freedom We will always choose n – 1 degrees of freedom You will find the value of 𝒕𝜶/𝟐 in table A3 as part of the process of calculating E in the confidence interval for the mean. For this you need: ◦ Degrees of freedom: n – 1 ◦ Sum of the areas of the 2 tails (which is equal to 𝛼) Remember, 𝛼 = 1 − confidence level desired Find 𝑡𝛼 for a confidence level of 95% and a 2 sample size of 35 Find 𝑡𝛼 for a confidence level of 90% and a 2 sample size of 40 Find 𝑡𝛼 for a confidence level of 80% and a 2 sample size of 1001 Unless the original distribution is already normal, we must have a sample size of n > 30. If normal, there is no size requirement This method is only useful when 𝜎 is not known for the population Use z n > 30 and 𝜎 is known ◦ Data is normally distributed and 𝜎 is known Use t n > 30 and 𝜎 is not known ◦ Data is normally distributed and 𝜎 is not known If n is 30 or less and the data is not normally distributed, neither method applies. Just as it was in 7.3, the formula for the confidence interval is the sample statistic ± E, however, the way we calculate E changes. Confidence Interval: (𝑥 − 𝐸, 𝑥 + 𝐸) Formula for E: 𝐸=𝑡 ∙ 𝛼 2 𝑠 𝑛 The average incomes of the families of 36 randomly selected Bradley University students are listed below: 28 29 35 42 42 68 44 50 52 54 78 74 84 90 95 101 108 88 116 121 122 158 167 120 235 59 150 56 133 51 75 79 47 88 180 26 Determine a 95% confidence interval for the actual average income of a Bradley family P.365-367: #5-12 (simply state whether you would use t, z, or neither) #19-20 5. t 6.z 7. Neither 8.t 9. z 10. Neither 11. t 12.t 19. a.) 98.20ºF b.) 98.04ºF < m < 98.36ºF because the confidence interval limits do not contain 98.6ºF, the results suggest that the mean is less than 98.6ºF. 20. a.) 2.1 lb b.) 0.0 lb < m < 4.2 lb c.) There does appear to be a loss of weight, but the actual amount of lost weight is so small that the weight loss program does not appear to be practical. 46 subjects were treated with raw doses of garlic to see if it lowered their cholesterol. The change in was measured and resulted in the following statistics: ◦ 𝑛 = 46, 𝑥 = 0.4, 𝑠 = 21.0 Create a 95% confidence interval to estimate the actual average drop in cholesterol from doses of garlic. Complete and finish the worksheet. ****SKIP 2 and 5! What you do not finish is for HOMEWORK! Yes you need to show your work! Go back to the project you have been working on. Since you now know your confidence intervals were slightly off, update them using the new methods we learned today (do not erase any of the work you’ve done so far, and keep the answers you gave to the questions). Answer the following additional question in a 4-5 sentence paragraph: ◦ What changes did you notice in your results after now completing the process in the correct way? Was the change enough for you to feel it was worth redoing the calculations?