Sample Size and Confidence Intervals for Height and Proportions

Question 1 Select the correct answer for the blank: If everything else stays the same, the required sample size ____ reach the same margin of error. Answer: Increases Decreases Remains the same Question 2 The population standard deviation for the height of college basketball players is 3.5 inches. If we want to for the population mean height of these players with a 0.5 margin of error, how many randomly selected up your answer to nearest whole number) Answer: ___231 ___ Hide question 2 feedback Z-Critical Value = NORM.S.INV(.985) = 2.17009 n = n = Question 3 The population standard deviation for the height of college basketball players is 3 inches. If we want to e the population mean height of these players with a 0.6 margin of error, how many randomly selected pla your answer to nearest whole number) Answer: ___118 ___ Hide question 3 feedback Z-Critical Value = NORM.S.INV(.985) = 2.17009 n = n = Question 4

A random sample of college basketball players had an average height of 66.35 inches. Based on this sam confidence interval for the population mean height of college basketball players. Select the correct answ 94% of college basketball players have height between 65.6 and 67.1 inches. There is a 94% chance that the population mean height of college basketball players is between 65.6 and 67.1 inches. We are 94% confident that the population mean height of college basketball players is between 65.6 and 67.1 inches. We are 94% confident that the population mean height of college basketball players is 66.35 inches. Question 5 The population standard deviation for the height of college hockey players is 3.4 inches. If we want to es the population mean height of these players with a 0.6 margin of error, how many randomly selected pla your answer to nearest whole number) Answer: ___87 ___ Hide question 5 feedback Z-Critical Value = NORM.SINV(.95) = 1.645 n = n = Question 6 The population standard deviation for the height of college basketball players is 3 inches. If we want to e for the population mean height of these players with a 0.5 margin of error, how many randomly selected up your answer to nearest whole number) Answer: ___239 ___ Hide question 6 feedback Z-Critical Value = NORM.SINV(.995) = 2.575 n = n =

Question 7 The FDA regulates that fresh Albacore tuna fish that is consumed is allowed to contain 0.82 ppm of mer estimating the amount of mercury in tuna fish for a new company and needs to have a margin of error w confidence. Assume the population standard deviation is 0.138 ppm of mercury. What sample size is nee Answer: ___82 ___ Hide question 7 feedback Z-Critical Value = NORM.S.INV(.975) = 1.96 n = n = Question 8 A random sample of 150 people was selected and 12% of them were left handed. Find the 90% confiden handed people. (0.0764, 0.1636) (0.068, 0.172) (.12, .88) (–1.645, 1.645) (0.0436, 0.1164) Hide question 8 feedback Z-Critical Value = NORM.S.INV(.95) = 1.645 LL = .12 - 1.645* UL = .12 + 1.645* Question 9 From a sample of 500 items, 30 were found to be defective. The point estimate of the population proport 30 0.60 0.06 0.94 Hide question 9 feedback

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30/500 Question 10 Suppose a marketing company wants to determine the current proportion of customers who click on ads estimated that the current proportion of customers who click on ads on their smartphones is 0.42 based o Compute a 92% confidence interval for the true proportion of customers who click on ads on their smart appropriately. ___0.334 ___ (50 %) < p < ___0.506 ___ (50 %) (round to 3 decimal places) Hide question 10 feedback Z-Critical Value = NORM.S.INV(.96) = 1.750686 LL = 0.42 - 1.750686 * UL = 0.42 + 1.750686 * Question 11 In a random sample of 200 people, 135 said that they watched educational TV. Find the 95% confidence people who watched educational TV. Hide question 11 feedback Z - Critical Value =NORM.S.INV(.975) = 1.96 LL = .675 - 1.96* UL = .675 + 1.96* Question 12 A survey of 85 families showed that 36 owned at least one DVD player. Find the 99% confidence interva families who own at least one DVD player. Place your limits, rounded to 3 decimal places blank___ and the upper limit in the second blank___ When entering your answer do not use any labels or symbols other than the decimal point. Simply provi 0.123 would be a legitimate entry. Make sure you put a 0 before the decimal. ___

Answer for blank # 1: 0.285 (50 %) Answer for blank # 2: 0.562 (50 %) Hide question 12 feedback Z-Critical Value =NORM.S.INV(.995) = 2.575 LL = .4235 - 2.575* UL = .4235 + 2.575* Question 13 Senior management of a consulting services firm is concerned about a growing decline in the firm's wee firm expects each professional employee to spend at least 40 hours per week on work. In an effort to und management would like to estimate the standard deviation of the number of hours their employees spend typical week. Rather than reviewing the records of all the firm's full-time employees, the management ra from the available frame. The sample mean and sample standard deviations were 48.5 and 7.5 hours, res Construct a 88% confidence interval for the mean of the number of hours this firm's employees spend on week. Place your LOWER limit, in hours, rounded to 1 decimal place, in the first blank. For example, 6.7 wou Place your UPPER limit, in hours, rounded to 1 decimal place, in the second blank. For example, 12.3 w ___ Answer for blank # 1: 46.8 (50 %) Answer for blank # 2: 50.2 (50 %) Hide question 13 feedback T-Critical Value = T.INV.2T(.12,50) = 1.581805 LL = 48.5 - 1.581805 * UL = 48.5 +1.581805 * Question 14 You are told that a random sample of 150 people from Iowa has been given cholesterol tests, and 60 of t count of 200. Construct a 95% confidence interval for the population proportion of people in Iowa with Place your LOWER limit, rounded to 3 decimal places, in the first blank___. For example, 0.678 would UPPER limit, rounded to 3 decimal places, in the second blank ___. For example, 0.789 would be a legi Make sure you include the 0 before the decimal.

___ Answer for blank # 1: 0.322 (50 %) Answer for blank # 2: 0.478 (50 %) Hide question 14 feedback Z-Critical Value =NORM.S.INV(.975) = 1.96 LL = .4 - 1.96* UL = .4 +1.96* Question 15 After calculating the sample size needed to estimate a population proportion to within 0.05, you have be error (E) must be reduced to just 0.025. If the original calculation led to a sample size of 1000, the samp Place your answer, as a whole number in the blank. For example, 2345 would be a legitimate entry. ___ Answer: 4000 Hide question 15 feedback Use n = p*p* 1000= .5*.5 * 4000 = 3.16227 = Z Now solve for n, using Z n = .5*.5 * Question 16 The percent defective for parts produced by a manufacturing process is targeted at 4%. The process is m sizes n = 160 units. Suppose that today's sample contains 14 defectives. How many units would have to be sampled to be 95% confident that you can estimate the fraction of def information from today's sample--that is using the result that )? Place your answer, as a whole number, in the blank. For example 567 would be a legitimate entry.

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___ ___ Answer: 767 Hide question 16 feedback Z-Critical Value = NORM.S.INV(.975) = 1.96 n = n = Question 17 A randomly selected sample of college basketball players has the following heights in inches. See Attached Excel for Data. height data.xlsx Compute a 93% confidence interval for the population mean height of college basketball players based o appropriately. ___63.846 ___ (50 %) < μ < ___65.841 ___ (50 %) (round to 3 decimal places) Hide question 17 feedback T-Critical Value =T.INV.2T(.07,31) = 1.876701 Question 18 The amounts (in ounces) of randomly selected eight 16-ounce beverage cans are given below. See Attached Excel for Data. ounces data.xlsx Assume that the amount of beverage in a randomly selected 16-ounce beverage can has a normal distribu interval for the population mean amount of beverage in 16-ounce beverage cans and fill in the blanks ap A 99% confidence interval for the population mean amount of beverage in 16-ounce beverage cans is (

___15.255 ___ (50 %) , ___16.345 ___ (50 %) ) ounces. (round to 3 decimal places) Hide question 18 feedback T-Critical Value = T.INV.2T(.01,7) = 3.499483 Question 19 In a certain town, a random sample of executives have the following personal incomes (in thousands); A normally distributed. Find the 95% confidence interval for the mean income. See Attached Excel for Data. income data.xlsx 32.180 < μ < 55.543 35.132 < μ < 50.868 40.840 < μ < 45.160 39.419 < μ < 46.581 35.862 < μ < 50.138 Hide question 19 feedback T- Critical Value = T.INV.2T(.05,13) = 2.160369 LL = 43 - 2.160369* UL = 43 + 2.160369* Question 20 A tire manufacturer wants to estimate the average number of miles that may be driven in a tire of a certa Assume the population is normally distributed. A random sample of tires is chosen and are driven until t thousands of miles is recorded, find the 99% confidence interval using the sample data. See Attached Excel for Data miles data.xlsx (27.144, 33.356)

(27.256, 33.244) (26.746, 33.754) (–3.106, 3.106) (26.025, 34.475) Hide question 20 feedback T-Critical Value = T.INV.2T(.01,11) = 3.105807

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STATS WK VI

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