Math 4, Unit 8, Central Limit Thm/Confidence Intervals Name: ______Confidence Intervals Date: ______
Confidence Interval & Confidence Level: We all “know” that normal body temperature is 98.6ºF. Some researchers, Wasserman, Mackowiak, and Levine, at the University of Maryland, suspected that 98.6ºF might be a little high. Once a mean of a sample is found, it will represent the best estimate for the population’s mean; it does not, however, give us any indication of just how good the best estimate is. Statisticians developed the confidence interval. A confidence interval is a range (i.e. interval) 98.6 of values used to estimate the true value of a population parameter. A confidence interval is 98.6 98.0 associated with a confidence level, such as 0.95 (or 95%). The confidence level gives us the 97.3 success rate of the procedure used to construct the confidence 97.2 interval. The confidence level is often 98.6 expressed as the probability (or area) 1 – , 98.8 98.3 where is the complement of the 97.6 confidence level. For example, for a 0.95 1 - 98.4 (95%) confidence level, = 0.05. A visual 98.0 is given to the right. 98.6 98.5 98.2 Margin or Error: 2 2 98.6 98.0 When we collect a sample of data, such as the set of 97.0 50 body temperatures to the right, we can calculate Level of 97.3 confidence a za 99.6 the sample mean, x, and that sample mean is 2 2 1 – 98.2 typically different from the population mean, . 99.0 The difference between the sample mean and the 90% 5% 1.645 97.0 population mean is an error. We recently learned 98.7 s 98.7 that is the standard deviation of sample means. 95% 2.5% 1.96 98.0 n 98.4 s 98.8 Using and za notation, we now use the 98% 1% 2.33 97.4 n 2 99.4 following equation to express the margin of error E: 99% 0.5% 2.575 97.8 s 98.4 E= za g 97.6 2 n 98.9 98.2 za is simply the z-score associated with 2 98.0 the confidence level. For example, if the 98.4 confidence level is 95%, there is 2.5% at 97.7 95% 98.6 the lower end of the normal curve and 98.0 2.5% at the upper end. Since z-scores deal 98.4 with area to the left of a point on the 98.4 normal distribution, we are looking for the 2.5% 2.5% 98.8 z-score that yields a 0.975. That score is 99.5 98.6 1.96. The confidence interval is simply x ± E. 98.6 EXAMPLE 1: For the body temperatures to the right, n = ______and x = ______(nearest 98.6 98.0 th 100 ). Assume that the sample is a simple random sample and that is 97.5 somehow known to be 0.62ºF. Using a 95% confidence interval, find a) the 98.6 margin of error and b) the confidence interval for . 98.6 EXAMPLE 2: A sample of 100 observations is collected and yields x = 75 and = 8. Find a 95% confidence interval for the true population average.
EXAMPLE 3: Kennesaw State University claims the average starting salary of its graduates is $38,500. A sample of 100 KSU students is sampled and yields an average starting salary of $36,800 with a standard deviation of $9,369. Using a 95% confidence level what can you say about KSU's claim?
EXERCISES:
In exercises 1 – 4, find the critical value za that corresponds to the given confidence level. 2 1. 98% 2. 95% 3. 96% 4. 99.5%
In exercises 5 – 8, use the given confidence interval and sample data to find (a) the margin of error E and (b) a confidence interval for estimating the population mean . 5. Use a 99% confidence level with example 3.
6. Starting salaries of college graduates who have taken a statistics course: 95% confidence; n = 28, x = $45,678, the population is normally distributed, and is known to be $9900. 7. Ages of drivers occupying the passing lane whose speeds are less than the speed limit and with their left signal flashing: 99% confidence interval; n = 50, x = 80.5 years, and = 4.6 years
8. During TV commercials breaks, the time between uses of the remote control by males: 90% con- fidence; n = 25, x = 5.24 seconds, the population is normally distributed, and = 2.50 seconds.
9. In order to monitor the ecological health of the Florida Everglades, various measurements are recorded at different times. The bottom temperatures are recorded at the Garfield Bight station and the mean of 30.4ºC is obtained for 61 temperatures on 61 different days. Assuming that = 1.7ºC, find a 95% confidence interval estimate of the population mean of all such temperatures. What aspect of this problem is not realistic?
10. When people smoke, the nicotine they absorb is converted to cotinine, which can be measured. A sample of 40 smokers has a mean cotinine level of 172.5. Assuming that is known to be 119.5, find a 90% confidence interval estimate of the mean cotinine level of all smokers.
11. The two intervals (114.4, 115.6) and (114.1, 115.9) are confidence intervals for = true average resonance frequency (in hertz) for all tennis rackets of a certain type. (a) What is the value of the sample mean resonance frequency? (b) The confidence level for one of the intervals is 90% and for the other it is 99%. Which is which and how can you tell? 12. An economist wants to estimate the first year’s mean income of college graduates who have had the profound wisdom to take a statistics course. How many such incomes must be found if we want to be 95% confident that the sample mean is within $500 of the true population mean? Assume that a previous study of such incomes has shown a = $6250.
13. In order to keep from recalibrating their machinery, the U.S. Mint wants to 5.63 5.73 5.60 be 95% sure that their quarters have a mean of 5.65 grams. They collect the 5.68 5.59 5.60 weights of 30 quarters (to the right). If quarter’s weights standard deviation 5.62 5.63 5.57 is 0.068 grams, will they have to recalibrate? 5.60 5.66 5.71 5.53 5.67 5.62 5.58 5.60 5.72 5.60 5.74 5.57 5.58 5.57 5.70 5.59 5.62 5.60 5.66 5.73 5.49
14. SAT scores are normally distributed with a mean of 1000 and standard 940 850 deviation of 120. Mr. Markley’s calculus students earned the scores 1140 1170 shown to the right. Find the 99% confidence interval estimate of the 1040 1330 mean. What conclusion can be drawn concerning the SAT’s mean score 1120 1220 and the calculus class’s mean score? 990 990 1070 1120 1080 1000 1190 1100