Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Basic Premise • A sample will represent a population • But a single mean, proportion, or standard deviation derived from the sample might not be accurate. • The confidence interval says if, instead of a sample, we survey the entire population (called a _________), there is a specific probability (called the ___________________) that the population’s mean, proportion, or standard deviation will fall within the confidence interval. Concepts and Definitions (Match them!!) Point estimate Interval estimate Degree of confidence Degrees of freedom Margin of error Critical value Standard normal (Z) distribution Student (T) distribution Chi-squared distribution Distribution for standard deviation The sample size minus 1 A single estimate of the population calculated from the sample The boundary between what's in and what's our of the interval estimate Distribution for small samples The difference between the mean and the interval boundaries The probability that the interval estimate contains the population parameter A range that probably contains the population parameter Distribution for large samples Question to Ponder • If an mean interval estimate of (34 to 47) has a 95% degree of confidence, it would be ___________ for the population mean to be 32. A. Impossible B. Unusual C. Likely D. Weird Questions to Ponder • The _________ the sample size, the ________ the interval. A. larger, larger B. smaller, smaller C. larger, smaller D. smaller, larger • The _________ the degree of confidence, the ________ the interval. A. larger, larger B. smaller, smaller C. larger, smaller D. smaller, larger Skills and Procedures • Finding confidence interval for – Mean – Proportion – Standard deviation (and variance) • Finding sample size needed to achieve a margin of error and degree of confidence. – Mean – Proportion Mean X s n Degree of confidence 550 15 25 99% 4.5 0.25 55 95% $12,000 $850 30 90% Point estimate Interval estimate Proportion Sample size Successes 64 24 120 25 10 p q DoC 99% 0.85 22 95% 99% 0.1 90% Point Interval estimate Sample size For Mean Degree of confidence Margin of error Standard deviation 95% 5 10 90% 1.2 2.0 99% 100 350 95% 1500 2700 n Sample size for Proportion Degree of confidence Margin of error p 99% 10% 0.65 95% 3% Unknown 90% 5% 0.05 95% 7.5% Unknown n Standard Deviation Degree of Confidence s n 99% 18 45 95% 0.45 32 90% 145 75 90% 2.78 81 95% 1500 23 95% 15.4 12 Point Interval estimate Case study #1 • An insurance company that is considering opening an office in town would like to know the average age of its adults. Our class sampled their parents and found the mean age is 52, the standard deviation is 7. • Identify the point estimate:____________________ • Decide on a degree of confidence (________) and calculate the confidence interval: _________________ • How many people would we have to sample if we want to reduce the margin of error by one-half? Case Study #2 • The insurance company would also like to know the percent of cars that have antilock brakes. Of the 120 people we polled, 96 said they had ABS • Calculate the point estimate:____________________ • Decide on a degree of confidence (________) and calculate the confidence interval: _________________ Case study #3 • The insurance company would like to know the percent of residents that own hybrid cars, and they would like a margin of error of 7.5% • Select a degree of confidence (_________) and calculate the sample size needed to meet the company’s needs. ________ • The insurance company would like to know the percent of residents that have been in accidents. They would like a margin of error of 7.5%. A study done in a nearby town suggests 8% • Select a degree of confidence (_______) and calculate the sample size needed to meet the company’s needs: _______ Case study #4 • An insurance company would also like to know the average accident repair bill. A sample of 41 accident records showed an average bill of $1250. The standard deviation was $525. • Identify the point estimate:____________________ • Decide on a degree of confidence (________) and calculate the confidence interval: _________________ Case Study #5 • An insurance company would also like to know the percent of households that have teenage drivers. Of the 75 people we asked at the local supermarket, 16 said they had teenage drivers • Calculate the point estimate:____________________ • Decide on a degree of confidence (________) and calculate the confidence interval: _________________ Answers • Mean – (541.6, 558.4) – (4.43, 4.57) – (11,745, 12255) • Proportion – – – – p = 0.375, q = 0.625, point = 0.375, (0.219, 0.531) x = 102, q = 0.15, point = 0.85, (0.786, 0.914) p = 0.88, q = 0.12, point = 0.88, (0.70, 1.06) x = 1, q = 0.9, point = 0.1, (-0.07, 0.27) More answers • Mean sample size – – – – 16 8 82 13 • Proportion Sample Size – – – – 151 1068 52 169 More Answers • Standard Deviation – – – – – – Point = 18, Interval = (14.6, 26.2) Point = 0.45, interval = (0.366, 0.611) Point = 145, interval = (131.1, 173.4) Point = 2.78, interval = (2.463, 3.200) Point = 1500, interval = (1160.1, 2123.1) Point = 15.4, interval = (10.91, 26.15) More answers • Case study #1 – Point = 52 – Assume n = 25 and Degree of Confidence = 95%, then Confidence interval = (49.1, 54.8) • Case Study #2 – Point estimate = 0.8, – If degree of confidence is 90%, then confidence interval = (0.740, 0.860) – If degree of confidence is 95%, then confidence interval = (0.728, 0.872) – If degree of confidence is 99%, then confidence interval = (0.706, 0.894) More answers • Case Study #3 (hybrid cars) – If degree of confidence is 90%, n = 121 – If degree of confidence is 95%, n = 171 – If degree of confidence is 99%, n = 295 • Case study #3 (accidents) – If degree of confidence is 90%, n = 36 – If degree of confidence is 95%, n = 51 – If degree of confidence is 99%, n = 88 More answers • Case study #4 – – – – Point estimate = 1250 If DoC = 90%, interval estimate = (1115.10, 1384.90) If DoC = 95%, interval estimate = (1089.30, 1410.70) If DoC = 99%, interval estimate = (1038.80, 1461.20) • Case study #5 – – – – Point = 0.21 If DoC = 90%, interval estimate = (0.136, 0.291) If DoC = 95%, interval estimate = (0.121, 0.306) If DoC = 99%, interval estimate = (0.091, 0.335)