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5-6 6.3-7
8-10 | Tests Chapter 1-3
4 5-6 6.3-7 8-10 Final Review
Elementary Statistics
(STA2023)
Chapter Test 6.3 - 7 * μ = population mean x̄ = Mean
Here are summary statistics for randomly selected weights of newborn girls: n = 226, x̄ =26.2 hg, s = 7.2 hg. Construct a confidence interval estimate of the mean. Use a 98% confidence level. Are these results very different from the confidence interval: 24.8 hg < μ < 28.4 hg with only 18 sample values, x̄ = 26.6 hg, and s = 2.9 hg? What is the confidence interval for the population mean mu? 25.1 hg < μ < 27.3 hg (Round to one decimal place as needed.) 226 – 1 = 225 1 - .98 / 2 = .01 TI84 Plus Steps  TI84 Plus 2nd VARS invT: .01 / 225 = 2.343 26.2 – (2.343 x 7.2 / √226) = 25.0778509 (round to one decimal place (25.1) 26.2 + (2.343 x 7.2 / √226) = 27.3221491 (round to one decimal place (27.3) * √ = square root σX = standard deviation μ = population mean An IQ test is designed so that the mean is 100 and the standard deviation is 15 for the population of normal adults. Find the sample size necessary to estimate the mean IQ score of statistics students such that it can be said with 99% confidence that the sample mean is within 6 IQ points of the true mean. Assume that σX = 15 and determine the required sample size using technology. Then determine if this is a reasonable sample size for a real-world calculation. The required sample size is 42. (Round up to the nearest integer.) TI84 Plus Steps TI84 Plus 2nd VARS 1.0 - .99 / 2 = 0.005 invNorm: .05 / 0 / 1 = 2.58 (round to 2 decimals) (2.58 x 15 / 6)2 = 41.6025 σX = standard deviation μ = population mean Salaries of 34 college graduates who took a statistics course in college have a mean, x of $68,800 . Assuming a standard deviation, σX of $17928, construct a 95% confidence interval for estimating the population mu . $62,774 < μ < $74,826 (Round to the nearest integer as needed.) Explanation: 1 - .95 / 2 = .025 TI84 Plus Steps TI84 Plus 2nd VARS invNorm: .025 / 0 / 1 = 1.96 (round to 2 decimals) 1.960 x 17928 / √34 = 6026.268 (round to 3 decimals) 68800 – 6026.268 = 62773.752 (round up to 62,774) 68800 + 6026.268 = 74826.268 (round off to 74,826) * √ = square root
σX =
standard deviation
μ = population mean Assume that females have pulse rates that are normally distributed with a mean of μ = 73.0 beats per minute and a standard deviation of sigma equals 12.5 beats per minute. Complete parts (a) through (c) below. If 1 adult female is randomly selected, find the probability that her pulse rate is less than 80 beats per minute. The probability is 0.7122 (Round to four decimal places as needed.) Explanation: 80 – 73 / 12.5 = .56 TI84 Plus Steps TI84 Plus Normalcdf: -9999 / .56 / 0 / 1 = 0.7122 b. If 25 adult females are randomly selected, find the probability that they have pulse rates with a mean less than 80 beats per minute. The probability is 0.9974. (Round to four decimal places as needed.) Explanation: 25 / 2 = 12.5 12.5 / √25 = 2.5 (80 – 73) / 2.5 = 2.8 TI84 Plus Steps TI84 Plus Press 2ND | VARS Normalcdf: -9999 / 2.8 / 0 / 1 = .9974 * √ = square root
In a survey of 3042 adults aged 57 through 85
years,
it was found that 85.1% of them used at least one prescription medication. Complete parts (a) through (c) below. How many of the 3042 subjects used at least one prescription medication? 2589 (Round to the nearest integer as needed.) 3042 · 0.851 = 2589 Construct a 90% confidence interval estimate of the percentage of adults aged 57 through 85 years who use at least one prescription medication. 84.0% < p < 86.2% (Round to one decimal place as needed.) Za / 2 100% - 90% / 2 = 1 - 0.90 / 2 = 0.05 Ti84 plus 2nd | VARS invNorm: .5 / 0 / 1 = 1.64 (rounded to 2) 1.64√.851 x (1 - .851) / 3042 = 1.64√.851 x .149 / 3042 = .0106 .851 - 0.0106 = 0.8404 = 84.0% (multiply 0.8404 · 100) .851 + 0.0106 = 0.8616 = 86.2% (multiply 0.8616 · 100) * √ = square root
A magazine provided results from a poll of 1,500
adults who were asked to identify their favorite pie.
Among the 1500 respondents, 14% chose chocolate pie, and the margin of error was given as plus or minus 5 percentage points. What values do p̂, q̂, n, E, and p represent? If the confidence level is 99 % what is the value of a? The value of p̂ is the sample proportion. (p̂ is a p-hat) The value of q̂ is found from evaluating 1 - p̂ (q̂ is a q-hat) The value of n is the sample size. The value of E is the margin of error. The value of p is the population proportion. If the confidence level is 99%, what is the value of alpha? a = 0.01 (Type an integer or a decimal. Do not round.) 100% - 99% = 1% (0.01)
μ = population mean
Sx = sample standard deviation x̄ = Mean Refer to the accompanying data display that results from a sample of airport data speeds in Mbps. Complete parts (a) through (c) below. TInterval (13.046, 22.15) x̄ = 17.598 Sx = 16.01712719 n = 50 a. Express the confidence interval in the format that uses the "less than" symbol. Given that the original listed data use one decimal place, round the confidence interval limits accordingly. 13.05 Mbps < μ < 22.15 Mbps (Round to two decimal places as needed.) (Given) 13.046 (rounded to 13.5) b. Identify the best point estimate of μ and the margin of error. The point estimate of μ is 17.60 Mbps. (Round to two decimal places as needed.) x = 17.598 rounded to 17.60 c. The margin of error is E = 4.55 Mbps. (Round to two decimal places as needed.) 17.598 – 13.046 = 4.552 μ = population mean Sx = sample standard deviation x̄ = Mean Refer to the accompanying data display that results from a sample of airport data speeds in Mbps. Complete parts (a) through (b) below TInterval (13.046, 22.15) x̄ = 17.598 Sx = 16.01712719 n = 50 What is the number of degrees of freedom that should be used for finding the critical value ta/2? df = 49 (Type a whole number.) 50 – 1 = 49 b. Find the critical value ta/2 corresponding to a 95% confidence level. ta/2 = 2.01 (Round to two decimal places as needed.) TI84 Plus 2nd VARS invT: .025 / Df: 49 = 2.00957 (round to 2.01) Which of the following statistics are unbiased estimators of population parameters? Choose the correct answer below. Select all that apply. A. Sample variance used to estimate a population variance. B. Sample standard deviation used to estimate a population standard deviation. C. Sample mean used to estimate a population mean. D. Sample range used to estimate a population range. E. Sample proportion used to estimate a population proportion. F. Sample median used to estimate a population median. μ = population mean Which of the following would be a correct interpretation of a 99% confidence interval such as 4.1 < μ < 5.6? Choose the correct answer below. It means that 99% of sample means fall between 4.1 and 5.6. We are 99% confident that the interval from 4.1 to 5.6 actually does contain the true value of p. It means that 99% of all data values are between 4.1 and 5.6. There is a 99% chance that g will fall between 4.1 and 5.6. p̂ is a p-hat. Express the confidence interval (0.069, 0.135) in the form of p̂ - E < p < p̂ + E. The value of q̂ 0.069 < p < 0.135 (Type integers or decimals.) μ = population mean * √ = square root In a test of the effectiveness of garlic for lowering cholesterol, 42 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes (before - after) in their levels of LDL cholesterol (in mg/dL) have a mean of 4.8 and a standard deviation of 19.6 Construct a 90 % confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment. What does the confidence interval suggest about the effectiveness of garlic in reducing LDL cholesterol? What is the confidence interval estimate of the population mean mu? -0.29 mg/dL < μ < 9.89 mg/dL (Round to two decimal places as needed.) ta/2 1.0 - .09 / 2 = .05 42 – 1 = 41 TI84 Plus 2nd VARS invT: .05 | 41 = 1.683 (round to 3 decimals) 1.683(19.6 / √42) = 5.090 (round to 3 decimals) 4.8 – 5.090 = -0.29 4.8 + 5.090 = 9.89 * √ = square root A genetic experiment with peas resulted in one sample of offspring that consisted of 429 green peas and 160 yellow peas. a. Construct a 90% confidence interval to estimate of the percentage of yellow peas. b. It was expected that 25% of the offspring peas would be yellow. Given that the percentage of offspring yellow peas is not 25%, do the results contradict expectations? A. Construct a 90% confidence interval. Express the percentages in decimal form. 0.2416 < p < 0.3016 (Round to three decimal places as needed.) Given that the percentage of offspring yellow peas is not 25%, do the results contradict expectations? No, the confidence interval includes 0.25, so the true percentage could easily equal 25% Yes, the confidence interval does not include 0.25, so the true percentage could not equal 25% 160 + 429 = 589 160 ÷ 589 = 0.2716 (round to 4) 1 - 0.90 ÷ 2 = .05 Ti84 plus 2nd | VARS invNorm: .05 | 0 | 1 = 1.644 (rounded to 3) q = 1 - .2716 = .7284 1.64√.2716 (0.7284 ÷ 589) = .030 (rounded to 4) Lower: .2716 - 0.0300 = 0.2416 Upper: .2716 + 0.0300 = 0.3016 * √ = square root Listed below are the amounts of net worth (in millions of dollars) of the ten wealthiest celebrities in a country. Construct a 90% confidence interval. What does the result tell us about the population of all celebrities? Do the data appear to be from a normally distributed population as required? 242 210 176 160 156 154 150 150 150 150 What is the confidence interval estimate of the population mean mu? $151.5 million < μ < $188.1 million (Round to one decimal place as needed.) Explanation: 1 - .90 / 2 = .05 n = 10 – 1 = 9 invT area: .05 df: 9 TI84 Plus 2nd VARS invT: .05 | 9 | = -1.833112923 (round and switch to positive) 1.833 Now we find the standard deviation: This format and SAMPLE Option. 242, 210, 176, 160, 156, 154, 150, 150, 150, 150 https://www.calculator.net/standard-deviation-calculator.html or Ti-84 plus: stat | edit | enter numbers. stat | calc | 1-var | List: L1 | FreqList: blank | calculate. sample standard deviation (smaller one) Sx = 31.57 1.833 · 31.57 / √10 = 169.8 – 18.3 = 151.5 169.8 + 18.3 = 188.1 * √ = square root Use the sample data and confidence level given below to complete parts (a) through (d). A research institute poll asked respondents if they felt vulnerable to identity theft. In the poll, n = 925 and x = 562 who said "yes." Use a 95 % confidence level. Find the best point estimate of the population proportion p. Find the best point estimate of the population proportion p. (Round to three decimal places as needed.) 0.608 562 / 925 = 0.608 Identify the value of the margin of error E. .031 (Round to three decimal places as needed.) 1 - 0.95 / 2 = 0.025 TI84 Plus 2nd VARS invNorm: .025 | 0 | 1 = -1.96 1.96 √.608 (1-.608) / 925 = 0.03146 Construct the confidence interval. 0.577 < p < 0.639 (Round to three decimal places as needed.) .608 - 0.031 = 0.577 .608 + 0.031 = 0.639 Write a statement that correctly interprets the confidence interval. Choose the correct answer below. A. One has 95% confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion. B. 95% of sample proportions will fall between the lower bound and the upper bound. C. There is a 95% chance that the true value of the population proportion will fall between the lower bound and the upper bound. D. One has 95% confidence that the sample proportion is equal to the population proportion. The __________ is the best point estimate of the population mean. sample mean A __________ ___________ is a single value used to approximate a population parameter. point estimate The brand manager for a brand of toothpaste must plan a campaign designed to increase brand recognition. He wants to first determine the percentage of adults who have heard of the brand. How many adults must he survey in order to be 90% confident that his estimate is within five percentage points of the true population percentage? Complete parts (a) and (b) below. a. Assume that nothing is known about the percentage of adults who have heard of the brand. N = 271 (Round up to the nearest integer.) b. Assume that a recent survey suggests that about 85% of adults have heard of the brand. n = 138 (Round up to the nearest integer.) Explanation: 1 - .9 / 2 = .1 TI84 Plus 2nd VARS invNorm: .1 | 0 | 1 = 1.6449 (round to 4 decimals) 0.50 is the estimated proportion (when noting is known, 0.5 is assumed) E = .05 a. 1.64492 · 0.5(1 - .5) / 0.052 = 270.569601 (round to 271) b. 1.64492 · 0.85(1 - .85) / 0.062 = 137.9904 (round to 138) Here are summary statistics for randomly selected weights of newborn girls: n = 226, x̄ =26.2 hg, s = 7.2 hg. Construct a confidence interval estimate of the mean. Use a 98% confidence level. Are these results very different from the confidence interval: 24.8 hg < μ < 28.4 hg with only 18 sample values, x̄ = 26.6 hg, and s = 2.9 hg? What is the confidence interval for the population mean mu? 25.1 hg < μ < 27.3 hg (Round to one decimal place as needed.) 226 – 1 = 225 1 - .98 / 2 = .01 TI84 Plus 2nd VARS invT: .01 | 225 = 2.343 26.2 - (2.343 x 7.2 / √226) = 25.0778509 26.2 + (2.343 x 7.2 / √226) = 27.3221491 Construct a confidence interval estimate of the mean. Use a 98% confidence level. Are these results very different from the confidence interval: 24.8 hg < μ <28.4 hg with only 18 sample values, x̄ = 26.6 hg, and s = 2.9 hg? What is the confidence interval for the population mean mu? 25.1 hg < μ < 27.3 hg (Round to one decimal place as needed.) 226 – 1 = 225 1 - .98 / 2 = .01 TI84 Plus 2nd VARS invT: .01 | 225 = 2.343 26.2 – (2.343 x 7.2 / √226) = 25.0778509 26.2 + (2.343 x 7.2 / √226) = 27.3221491 σ = standard deviation An IQ test is designed so that the mean is 100 and the standard deviation is 15 for the population of normal adults. Find the sample size necessary to estimate the mean IQ score of statistics students such that it can be said with 99% confidence that the sample mean is within 6 IQ points of the true mean. Assume that σ = 15 and determine the required sample size using technology. Then determine if this is a reasonable sample size for a real-world calculation. The required sample size is 42. (Round up to the nearest integer.) TI84 Plus 2nd VARS 2.0 - .99 | 2 = .005 invNorm: .05 | 0 | 1 = 2.58 (round to 2 decimals) (2.58 x 15 / 6)2 = 41.6025 μ = population mean Salaries of 34 college graduates who took a statistics course in college have a mean, x̄ of $68,800. Assuming a standard deviation, σ of $17928, construct a 95% confidence interval for estimating the population μ. $62774 < u < $74826 (Round to the nearest integer as needed.) 1 - .95 / 2 = .025 TI84 Plus 2nd VARS invNorm: .025 | 0 | 1 = 1.96 (round to 2 decimals) 1.960 x 17928 / √34 = 6026.268 (round to 3 decimals) 68800 – 6026.268 = 62773.752 68800 + 6026.268 = 74826.268 Assume that females have pulse rates that are normally distributed with a mean of μ = 75.0 beats per minute and a standard deviation of sigma equals 12.5 beats per minute. Complete parts (a) and (b) below. a. If 1 adult female is randomly selected, find the probability that her pulse rate is between 68 beats per minute and 82 beats per minute. The probability is 0.4545 (Round to four decimal places as needed.) 68 – 75 / 12.5 = -.56 82 – 75 / 12.5 = .56 TI84 PLUS Press 2ND | VARS Normalcdf: -9999 | -.56 | 0 | 1 = .9974 = 0.2877 (4 decimal places) Normalcdf: -9999 |.56 | 0 | 1 = .9974 = 0.7122 (4 decimal places) 0.7122 - 0.2877 = 0.4245 b. If 4 adult females are randomly selected, find the probability that they have pulse rates with a mean between 68 beats per minute and 82 beats per minute. The probability is 0.7372 (Round to four decimal places as needed.) 12.5 / √4 = 6.25 68 – 75 / 6.25 = -1.12 82 – 75 / 6.25 = 1.12 TI84 PLUS Press 2ND | VARS Normalcdf: -9999 | -1.12 | 0 | 1 = 0.9974 = 0.1314 (4 decimal places) Normalcdf: -9999 | 1.12 | 0 | 1 = 0.9974 = 0.8686 (4 decimal places) 0.8686 - 0.1314 = 0.7372 A magazine provided results from a poll of 2,000 adults who were asked to identify their favorite pie. Among the 1500 respondents, 13% chose chocolate pie, and the margin of error was given as plus or minus 5 percentage points. What values do p̂, q̂, n, E, and p represent? If the confidence level is 99 % what is the value of a? The value of p̂ is the sample proportion. (p̂ is a p-hat) The value of q̂ is found from evaluating 1 - p̂ (q̂ is a q-hat) The value of n is the sample size. The value of E is the margin of error. The value of p is the population proportion. If the confidence level is 99%, what is the value of alpha? a = 0.01 (Type an integer or a decimal. Do not round.) 100% - 99% = 1% (0.01) |
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