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Exam 1 Outline 1) A ___________ is a measure that describes characteristics of a sample. 2) A ___________ is a measure that describes characteristics of a population. 3) The general practice of estimation theory consists of using ______________________ to estimate ____________________________ 4) The general requirement for a sample statistic to be an unbiased point estimator of its corresponding population parameter is that __________________________________ 5) The sample mean x is an unbiased point estimator of __________________________ because __________________________ 6) The sample median is a biased point estimator of the population median because ______ 7) If a sufficiently large number of independent simple random samples of equal size 16 are selected from the same population of numerical scores having mean 45 and standard deviation 6, (using the same sampling procedure), then the sampling distribution of the sample mean x will have mean x = _______ and standard deviation ________. x If the population proportion P is known to be 0.34, then the sampling distribution of the sample proportion p̂ will have mean p̂ = _____ and standard deviation p̂ =________ 8) If a sufficiently large number of independent simple random samples of equal size 9 are selected from the same normally distributed population of numerical scores having mean 36 and standard deviation 4, (using the same sampling procedure), then the shape of the sampling distribution of the sample mean x will be _______________________ and will have mean x = _______ and standard deviation ________. x If the population proportion P is known to be 0.34, then the sampling distribution of the sample proportion p̂ will have mean p̂ = _____ and standard deviation p̂ =________ 9) A population of scores has mean 102 and standard deviation 12. If nothing is known about the shape of the distribution of this population and a sufficiently large number of independent simple random samples of equal size 24 are selected from it, (using the same sampling procedure), what will be the shape of the sampling distribution of the sample mean x ? What will be the value of the mean of the sampling distribution x ? What will be the value of the standard error of the mean ? x 10) State the large-sample conditions for the sampling distribution of a sample proportion p̂ to be approximately the normal distribution. 11) Suppose that the proportion of left-handed people in the whole world is near 0.10. If a sufficiently large number of independent simple random samples of equal size 100 are selected from the population of people in the whole world, (using the same sampling procedure), what will be the shape of the sampling distribution of p̂ ? What will be the value of the mean of the sampling distribution p̂ ? What will be the value of the standard error of the proportion p̂ ? 12) Suppose that if we collect all samples of size 3 (with replacement) from a population of numerical scores, the resulting probability distribution of x is as follows: 1.0 1.5 x 1 P( x ) 641 16 2.0 2.5 3.0 3.5 4.0 3 16 15 32 3 16 1 16 1 64 a) Find the value of the population mean. b) Find the value of the population variance. 13) Suppose that if we collect all samples of size 2 (with replacement) from a population of numerical scores, the resulting probability distribution of the sample proportion of odd numbers p̂ is as follows: 1 p̂ 0 1 a) Find the value of the population proportion of odd numbers 2 P( p̂ ) 4 25 12 25 9 25 b) Find the standard error of p̂ c) Compute the value P(1 P) compare to the result in part b) n 14) A manufacturer of light bulbs wants to estimate the mean life of a large shipment of light bulbs to be delivered out of her factory very soon. She takes a random sample of 50 light bulbs and finds the mean life to be 1600 hours with standard deviation 12 hours. A) Give the point estimate value of the mean life of all light bulbs in the shipment. B) Describe the shape of the sampling distribution of the sample mean life of the light bulbs. Explain your reasoning with excruciating details. C) If the manufacturer selects a random sample of 24 light bulbs (instead of 50), describe the shape of the sampling distribution of the sample mean life of the light bulbs. Explain your reasoning with excruciating details. 15) A state insurance auditor needs to determine the proportion of claims that are paid by a health insurance company within two months of receiving the claim. She takes a random sample of 400 of such claims and finds that 90 were paid out within two months of receiving the claim. A) Give the point estimate value of the proportion of all claims paid out by this health insurance company within two months of receiving the claim. B) Describe the shape of the sampling distribution of the sample proportion of all claims paid out by this health insurance company within two months of receiving the claim. Explain your reasoning with excruciating details. C) If the auditor finds that only 14 (instead of 90) were paid out within two months of receiving the claim, describe the shape of the sampling distribution of the sample proportion of all claims paid out by this health insurance company within two months of receiving the claim. Explain your reasoning with excruciating details. 16) A) Find the correct value of if z = 2.09 2 B) Find the critical value z for a 91% confidence interval 2 C) Find the critical value z if z = z0.013 2 2 17) A radial tire manufacturer claims that the tires she produces last an average of 66,000 miles. A state inspector takes a random sample of 38 tires from this manufacturer and finds the mean life to be 65,000 miles with standard deviation 2,800 miles. A) The 93% confidence interval estimate of the mean life of all radial tires produced by this manufacturer is (64,177.86 ; 65,822.14). Interpret this interval with excruciating details. B) Construct a 98% confidence interval estimate of the mean life of all radial tires produced by this manufacturer. C) At the 93% confidence level, is the manufacturer’s claim correct? Explain with excruciating details. D) At the 98% confidence level, is the claim correct? Explain with excruciating details. E) Describe the shape of the sampling distribution of the sample mean life of the radial tires from this manufacturer. Explain with excruciating details. 18) A list of ages at retirement of randomly selected Victoria Secret’s Angels is shown here: 26, 28, 26, 29, 27, 28. A) Construct a 95% confidence interval estimate of the population mean age of retirement of Victoria Secret’s Angels. B) Under which condition(s) is the theory used for this confidence applicable? Explain with excruciating details. C) Many networks claim that the mean age of retirement of Victoria Secret’s Angels is 28. Use the 95% CI to explain with excruciating details whether this claim is true. 19) In a random sample of 984 U.S. adult consumers, 55% of the participants responded that they feel vulnerable to identity theft. A) Verify the large-sample condition(s). Explain with excruciating details. B) Construct a 96% CI estimate of the true proportion of all U.S. adult consumers who feel vulnerable to identity theft. C) Interpret the result with excruciating details. D) Many networks claim that the true proportion of all U.S. adult consumers who feel vulnerable to identity theft is at least 0.6. Use the 96% CI to explain with excruciating details whether this claim is true. 20) Many stores are interested in the proportion of adult consumers who purchase clothing online. Estimate the number of adult consumers that must be surveyed in order to be 99% confident that the sample percentage error is no more than 0.025. 21) It is believed that the percentage of men who leave it for the very last minute to purchase a Valentines gift for their sweethearts on Valentines day is near 75%. Estimate the number of men that must be surveyed in order to be 90% confident that the sample percentage error is no more than 0.045. 22) Suppose that the population variance of IQ scores for normal adults is 196. Estimate the number of adults that must sampled to estimate the mean IQ score for normal adults with a 99% CI and with a maximum error of 3 IQ points. Answers: 1) statistic 2) parameter 3) sample statistics; population parameters 4) the mean of the sampling distribution of the statistic is identically equal to the corresponding population parameter. In other words, using the correct probability distribution of the sample statistic, the expected value of the sample statistic is identically equal to the corresponding population parameter. 5) the population mean ; the mean of the sample distribution of x is identically equal to the population mean . In other words, E( x ) = . 6) the mean of the sample distribution of the sample median is not equal to the population median. In other words, E(sample median) (population median). 7) x = 45; (0.34)(1 0.34) 0.118 ; 8) also a normal distribution; x = 36; x 6 1.5 ; pˆ 0.34 ; pˆ 16 16 (0.34)(1 0.34) 0.158 ; 9) since the shape of the distribution x 4 1.333 ; pˆ 0.34 ; pˆ 9 9 of the original population is unknown and n = 24 < 30, then the Central Limit Theorem does not guarantee that the sampling distribution of x will be normally distributed (not even approximately). So, we cannot determine the shape of the sampling distribution of x ; x = 102 ; 12 2.449 x 24 10) n p̂ 15 and n(1 p̂ ) 15 ; 11) since the shape of the distribution of the original population is unknown and n p̂ = 100(0.10) = 10 < 15, then the Central Limit Theorem does not guarantee that the sampling distribution of p̂ will be normally distributed (not even approximately). So, we cannot determine the shape of the sampling distribution of p̂ ; 0.10 ; (0.10)(1 0.10) 0.030 pˆ pˆ 100 E( x ) = xP( x ) 1.0 1.5 2.0 2.5 3.0 3.5161 4.0641 = 2.391 2 3 2 1 2 1 2 b) 2 x 2 P( x ) 2 1.0 2 641 1.52 161 2.0 2 163 2.52 15 32 3.0 16 3.5 16 4.0 64 2.391 x 1 64 12)a) 1 16 3 16 15 32 3 16 x2 0.822 . Since x2 , then 2 n x2 (3)(0.822) 2.466 n 1259 1525 53 P E( p̂ ) = pˆ P( pˆ ) 0254 12 12 25 2 9 3 2 b) p̂ = 2 pˆ 2 P( pˆ ) P 2 0 2 254 12 2 12 25 1 25 5 13)a) p̂ 3 25 3 5 53 15 53 . The values are identically equal as expected. 2 n 14)a) 1600 hrs; b) the shape of the distribution of the original population is unknown. However, since n = 50 > 30, then the Central Limit Theorem guarantees that the sampling distribution of x will be approximately normally distributed (it will not be normally distributed, but it will be sufficiently close to a normal distribution); c) since the shape of the distribution of the original population is unknown and n = 24 < 30, then the Central Limit Theorem does not guarantee that the sampling distribution of x will be normally distributed (not even approximately). So, we cannot determine the shape of the sampling distribution of x ; 15)A) P pˆ x 90 0.225 ; B) examine the values n 400 of n p̂ and n(1 p̂ ). That is: n p̂ = 400(0.225) = 90 and n(1 p̂ ) = 400(1 0.225) = 310. Since n p̂ > 15 and n(1 p̂ ) > 15, the Central Limit Theorem guarantees that the sampling distribution of p̂ will be approximately normally distributed (it will not be normally distributed, but it will be sufficiently c) p̂ = P(1 P) 3 1 3 5 5 close to a normal distribution); C) In this case, pˆ x 14 0.035, so n p̂ = 14. Since the shape of n 400 the distribution of the original population is unknown and n p̂ = 14 < 15, then the Central Limit Theorem does not guarantee that the sampling distribution of p̂ will be normally distributed (not even approximately). So, we cannot determine the shape of the sampling distribution of p̂ 16)A) Since P(0 < z < 2.09) = 0.4817, then = 0.5 – 0.4817 = 0.0183. Thus, = 2(0.0183) = 0.0732 2 B) Notice that to 0.91 = 0.455 2 and the closest area for P(0 < z < z ) = 0.455 is 0.4554. This corresponds 2 z 2 = 1.70 ; C) z 2 = z0.013 means that 2 = 0.013. So, P(z > z 2 ) = 0.013 and P(0 < z < z 2 ) = 0.487 by subtracting 0.5 – 0.013 = 0.487. The area closest to 0.487 is 0.4871. Therefore, z 2 = 2.23 17)A) We are 93% confident that the interval (64,177.86 ; 65,822.14) does contain the true mean-life of all radial tires produced by this manufacturer. So, if we took all random samples of 38 tires (or a sufficiently large number of random samples of size 38, all from this same manufacturer and all using the same sampling method) and for each sample we constructed the corresponding confidence interval, then 93% of the resulting intervals will contain the true mean -life of all radial ( 2.33)( 2800) tires produced by this manufacturer. B) 65000 = 65000 1058.33 = (63941.67; 66058.33) 38 C) Since the claimed 66,000 miles average life is not in the 93% CI, (64,177.86 ; 65,822.14), then at the 93% confidence level, the sample data at hand did not provide sufficient evidence to affirm that the manufacturer’s claim is correct; D) Since the claimed 66,000 miles average life is in the 98% CI, (63941.67; 66058.33), then at the 98% confidence level, the sample data at hand did provide sufficient evidence to affirm that the manufacturer’s claim is correct; E) the shape of the distribution of the original population is unknown. However, since n = 38 > 30, then the Central Limit Theorem guarantees that the sampling distribution of x will be approximately normally distributed (it will not be normally distributed, but it will be sufficiently close to a normal distribution) x 2 164 2 2 x 4490 x 164 n 6 18)A) We must compute x n 6 27.333 and s 1.211 n 1 6 1 So, 95% CI = 27.333 ( 2.5706)(1.211) 6 = 27.333 1.271 = (26.062; 28.604) ; B) Since n = 6 < 30, it is assumed that the population of retirement ages of Victoria Secret’s Angels is normally distributed or not too deviated from a normal distribution shape. C) Since the claimed mean age 28 is in the 95% CI, (26.062; 28.604), then at the 95% confidence level, the sample data at hand did provide sufficient evidence to affirm that the networks’ claims are correct. 19)A) examine the values of n p̂ and n(1 p̂ ): n p̂ = 984(0.55) = 541.2; n(1 p̂ ) = 984(0.45) = 442.8 Since n p̂ > 15 and n(1 p̂ ) > 15, the Central Limit Theorem guarantees that the sampling distribution of p̂ will be approximately normally distributed (it will not be normally distributed, but it will be sufficiently close to a normal distribution); B) 0.55 2.05 ( 0.55)(1 0.55) 984 = (0.534; 0.566); C) We are 96% confident that the interval (0.534; 0.566) does contain the true population proportion of all adult consumers who feel vulnerable to identity theft. So, if we took all random samples of 984 U.S. adult consumers (or a sufficiently large number of random samples of size 984, all from the same population of U.S. adult consumers and all using the same sampling method) and for each sample we constructed the corresponding confidence interval, then 96% of the resulting intervals will contain the true population proportion of all adult consumers feeling vulnerable to identity theft. D) Since the claimed proportion 0.6 is not contained in the 96% CI, (0.534; 0.566), then at the 96% confidence level, the sample data at hand did not provide sufficient evidence to affirm that the networks’ claims are true. 20) n (2.575) 2 (0.5)(0.5) (0.025) 21) n 22) n (1.645) 2 (0.75)( 0.25) (0.045) 2 ( 2.575) 2 (196) (3) 2 2 = 2652.25. Use n = 2653 adult consumers. 250.558. Use n = 251 men. 144.4. Use n = 145 normal adults.