Confidence Intervals I

Confidence Intervals I Chapter 18 Estimating an unknown parameter Q: What proportion of M&Ms is orange? We take a sample of 400 M&Ms. Let’s say 120 of them are orange.
̂  120  0.3 400 How would the sample proportion
̂ vary if we took many samples of 400 M&Ms? N(0.3,0.02) 0.24 0.26 0.28 0.3 0.32 0.34 0.36 So, in 95% of all samples of size 400, the proportion of orange M&Ms will deviate from the true p by 0.04 (up or down…that’s 2 standard deviations above or below p). Statistical inference uses this fact about what would happen in MANY samples, to express our confidence in being able to capture the true proportion in our own ONE sample. Conclusion: Our sample of 400 M&Ms gave p=0.3. The resulting interval is 0.3 ± 0.04 (0.26, 0.34). We say that we are 95% confident that the unknown proportion p for all orange M&Ms is between 26% and 34%. Confidence Interval Is an attempt to estimate an unknown value from a sample. Basic Formula: Estimate ± critical value · standard error Works if conditions are met: o Random/representative sample o Independent o <10% of population o Success/fail (at least 10) np ≥ 10 nq ≥ 10 Interpret your Interval: I am ______% confident that the true proportion of __________ is between _______% and ________%. Explain/Interpret Confidence Level: I got my results using a method that should accurately estimate the true proportion of _________ from about _______ of 100 samples. ∗ - critical value - shows how many standard deviations away from the mean; corresponds with confidence level Example 1 Consider each situation described. Identify the population and the sample, explain what p and represent, and tell whether the methods of this chapter can be used to create a confidence interval. a) A consumer group hoping to assess customer experiences with auto dealers surveys 167 people who recently bought new cars; 3% of them expressed dissatisfaction with the salesperson. Population – all customers who recently bought new cars; Sample – 167 people surveyed about their experience;
proportion of all new car buyers who are dissatisfied with the salesperson;
̂ - proportion of new car buyers surveyed who are dissatisfied with the salesperson (3%). Conditions: npˆ = 167(0.03) = 5 and nqˆ = 162. Since only 5 people were dissatisfied, the sample is not large enough to use a confidence interval to estimate the proportion of dissatisfied car buyers. b) 240 potato plants in a field in Maine are randomly checked, and only 7 show signs of blight. How severe is the blight problem for the U.S. potato industry? Population – potato plants in the U.S.; Sample – 240 potato plants in a field in Maine;
– proportion of all potato plants in the U.S. that show signs of blight;
̂ – proportion of potato plants in the sample that show signs of blight (2.9%). Conditions: Independence assumption: It is not reasonable to think that signs of blight are independent. Blight is a contagious disease! Randomization condition: Although potato plants are randomly selected from the field in Maine, it doesn’t seem reasonable to assume that these potato plants are representative of all potato plants in the U.S. Success/Failure condition: npˆ = 7 and nqˆ = 233. There are only 7 (less than 10!) plants with signs of blight. The sample is not large enough. Three conditions are not met! Don’t use a confidence interval to attempt to estimate the percentage of potato plants in the U.S. that show signs of blight. Example 2 An insurance company checks police records on 582 accidents selected at random and notes that teenagers were at the wheel in 91 of them. a) Create a 95% confidence interval.
̂  0.156  = 0.844  = 582 0.156 ∗ 0.844  =  = 0.015 582 0.156 ± 1.96 ∗ 0.015 b) Explain what your interval means. (0.127,0.185) I am 95% confident that the true proportion of auto accidents involving teenagers is between 12.7% and 18.5%. c) Explain what “95% confidence” means. I got my results using a method that should accurately estimate the true proportion of accidents from about 95 of 100 samples. Example 3 In January 2007 Consumer Reports published their study of bacterial contamination of chicken sold in the United States. They purchased 525 broiler chickens from various kinds of food stores in 23 states and tested them for types of bacteria that cause food-borne illnesses. Laboratory results indicated that 83% of these chickens were infected with Campylobacter. a) Create a 95% confidence interval.
̂ = 0.83  = 0.17  = 525 0.83 ∗ 0.17  =  = 0.0164 525 0.83 ± 1.96 ∗ 0.0164 (0.7979,0.8621) b) Explain what your interval means. I am 95% confident that the true proportion of infected chickens is between 79.8% and 86.2%. c) Explain what “95% confidence” means. I got my results using a method that should accurately estimate the true proportion of infected chickens from about 95 of 100 samples. TI TIPS to calculate interval STAT TEST A: 1 – PropZInt (x, n, C-level) x = number of successes n = sample size C-level = confidence interval