Population Mean Estimates: Confidence Intervals and Sample Sizes

Question 1 0 / 1 point The population standard deviation for the height of college football players is 3.3 inches. If we want to estimate a 90% confidence interval for the population mean height of these players with a 0.7 margin of error, how many randomly selected players must be surveyed? (Round up your answer to nearest whole number) Answer: ___22 ___ ( 61) Hide question 1 feedback Z-Critical Value =NORM.S.INV(.95) = 1.645 n = �� 2 ∗ � 2 �� 2 n = 3.32 ∗ 1.6452.72 Question 2 1 / 1 point The population standard deviation for the height of college baseball players is 3.2 inches. If we want to estimate 90% confidence interval for the population mean height of these players with a 0.7 margin of error, how many randomly selected players must be surveyed? (Round up your answer to nearest whole number) Answer: ___ 57 _ __ Hide question 2 feedback

Z-Critical Value = NORM.S.INV(.95) = 1.645 n = �� 2 ∗ � 2 �� 2 n = 3.22 ∗ 1.6452.72 Question 3 1 / 1 point The FDA regulates that fresh Albacore tuna fish that is consumed is allowed to contain 0.82 ppm of mercury or less. A laboratory is estimating the amount of mercury in tuna fish for a new company and needs to have a margin of error within 0.03 ppm of mercury with 95% confidence. Assume the population standard deviation is 0.138 ppm of mercury. What sample size is needed? Round up to the nearest integer. Answer: ___ 82 ___ Hide question 3 feedback Z-Critical Value = NORM.S.INV(.975) = 1.96 n = �� 2 ∗ � 2 �� 2 n = .1382 ∗ 1.962.032 Question 4 1 / 1 point The population standard deviation for the height of college basketball players is 3.5 inches. If we want to estimate 97% confidence interval for the population mean height of these players with a 0.5 margin of error, how many randomly selected players must be surveyed? (Round up your answer to nearest whole number) Answer:

___ 231 ___ Hide question 4 feedback Z-Critical Value = NORM.S.INV(.985) = 2.17009 n = �� 2 ∗ � 2 �� 2 n = 3.52 ∗ 2.170092.52 Question 5 1 / 1 point The population standard deviation for the height of college basketball players is 3 inches. If we want to estimate a 99% confidence interval for the population mean height of these players with a 0.5 margin of error, how many randomly selected players must be surveyed? (Round up your answer to nearest whole number) Answer: ___ 239 ___ Hide question 5 feedback Z-Critical Value = NORM.SINV(.995) = 2.575 n = �� 2 ∗ � 2 �� 2 n = 32 ∗ 2.5752.52

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Question 6 1 / 1 point The population standard deviation for the height of college hockey players is 3.4 inches. If we want to estimate 90% confidence interval for the population mean height of these players with a 0.6 margin of error, how many randomly selected players must be surveyed? (Round up your answer to nearest whole number) Answer: ___ 87 _ __ Hide question 6 feedback Z-Critical Value = NORM.SINV(.95) = 1.645 n = �� 2 ∗ � 2 �� 2 n = 3.42 ∗ 1.6452.62 Question 7 1 / 1 point A random sample of college basketball players had an average height of 66.35 inches. Based on this sample, (65.6, 67.1) found to be a 94% confidence interval for the population mean height of college basketball players. Select the correct answer to interpret this interval. 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 8 1 / 1 point Suppose a marketing company wants to determine the current proportion of customers who click on ads on their smartphones. It was estimated that the current proportion of customers who click on ads on their smartphones is 0.42 based on a random sample of 100 customers. Compute a 92% confidence interval for the true proportion of customers who click on ads on their smartphones and fill in the blanks appropriately. ___ 0.334 _ __ (50 %) < p < ___ 0.506 _ __ (50 %) (round to 3 decimal places) Hide question 8 feedback Z-Critical Value = NORM.S.INV(.96) = 1.750686 LL = 0.42 - 1.750686 *.42 ∗ .58100 UL = 0.42 + 1.750686 *.42 ∗ .58100 Question 9 1 / 1 point

From a sample of 500 items, 30 were found to be defective. The point estimate of the population proportion defective will be: 30 0.06 0.94 0.60 Hide question 9 feedback 30/500 Question 10 1 / 1 point A survey asked people if they were aware that maintaining a healthy weight could reduce the risk of stroke. A 95% confidence interval was found using the survey results to be (0.54, 0.62). What is the correct interpretation of this interval? We are 95% confident that the interval 0.54 to 0.62 contains the population proportion of people that are aware that maintaining a healthy weight could reduce the risk of stroke. There is a 95% chance that the sample proportion of people that are aware that maintaining a healthy weight could reduce the risk of stroke is between 0.54 and 0.62. There is a 95% chance that the proportion of people that will have a stroke is between 54% and 62%. There is a 95% chance of having a stroke if you do not maintain a healthy weight. Question 11 1 / 1 point In a certain state, a survey of 600 workers showed that 35% belonged to a union. Find the 95% confidence interval of true proportion of workers who belong to a union. 210

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(0.0396, 0.0771) (0.29, 0.41) (34.9618, 35.0382) (0.3118, 0.3882) Hide question 11 feedback Z - Critical Value =NORM.S.INV(.975) = 1.96 LL = .35 - 1.96*.35 ∗ .65600 UL = .35 + 1.96*.35 ∗ .65600 Question 12 1 / 1 point You are told that a random sample of 150 people from Iowa has been given cholesterol tests, and 60 of these people had levels over the "safe" count of 200. Construct a 95% confidence interval for the population proportion of people in Iowa with cholesterol levels over 200. Place your LOWER limit, rounded to 3 decimal places, in the first blank___. For example, 0.678 would be a legitimate entry. Place your UPPER limit, rounded to 3 decimal places, in the second blank ___. For example, 0.789 would be a legitimate entry. 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 12 feedback Z-Critical Value =NORM.S.INV(.975) = 1.96 LL = .4 - 1.96*.4 ∗ .6150 UL = .4 +1.96*.4 ∗ .6150 Question 13 1 / 1 point Suppose you compute a confidence interval with a sample size of 25. What will happen to the confidence interval if the sample size increases to 50? Get larger Nothing Get smaller Hide question 13 feedback As the Sample Size increases, the Margin of Error decreases. Making the interval smaller. Question 14 1 / 1 point A sample of 9 production managers with over 15 years of experience has an average salary of $71,000 and a sample standard deviation of $18,000. Assuming that the salaries of production managers with over 15 years experience are normally distributed, you can be 95% confident that the mean salary for all production managers with at least 15 years of experience is between what two numbers. Place your LOWER limit, rounded to a whole number, in the first blank. Do not use a dollar sign, a comma, or any other stray mark. For example, 54321 would be a legitimate entry.___. Place your UPPER limit, rounded to a whole number, in the second

blank. Do not use a dollar sign, a comma, or any other stray mark. For example, 65432 would be a legitimate entry.___ ___ Answer for blank # 1: 57164 (50 %) Answer for blank # 2: 84836 (50 %) Hide question 14 feedback T-Critical Value = T.INV.2T(.05,8) = 2.306004 LL = 71000 - 2.306004*180009 UL = 71000 +2.306004*180009 Question 15 1 / 1 point If a sample has 25 observations and a 99% confidence estimate for�is needed, the appropriate value of the t-multiple required is___. Place your answer, rounded to 3 decimal places, in the blank. For example, 3.456 would be an appropriate entry. ___ Answer: 2.797 Hide question 15 feedback In Excel, =T.INV.2T(0.01,24) Question 16 1 / 1 point A sample of 40 country CD recordings of Willie Nelson has been examined. The average playing time of these recordings is 51.3 minutes, and the standard deviation is s = 5.8 minutes.

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Using an appropriate t-multiplier, construct a 95% confidence interval for the mean playing time of all Willie Nelson recordings. Place your LOWER limit, in minutes, rounded to 2 decimal places, in the first blank. For example, 56.78 would be a legitimate entry.___. Place your UPPER limit, in minutes, rounded to 2 decimal places, in the second blank. For example, 67.89 would be a legitimate entry.___ ___ Answer for blank # 1: 49.45 (50 %) Answer for blank # 2: 53.15 (50 %) Hide question 16 feedback T-Critical Value =T.INV.2T(.05,39) = 2.022691 LL = 51.3 - 2.022691*5.840 UL = 51.3 + 2.022691*5.840 Question 17 1 / 1 point It is known that the number of hours a student sleeps per night has a normal distribution. The sleeping time in hours of a random sample of 8 students is given below. See Attached Excel for Data. sleep hours data.xlsx

Compute a 92% confidence interval for the true mean time a student sleeps per night and fill in the blanks appropriately. We have 92 % confidence that the true mean time a student sleeps per night is between ___ 5.847 ___ (50 %) and ___ 7.728 ___ ( 50 %) hours. (round to 3 decimal places) Hide question 17 feedback T-Critical Value = T.INV.2T(.08,7) = 2.046011 �� =6.7875−2.046011 ∗ 1.2999318 �� =6.7875+2.046011 ∗ 1.2999318 Question 18 0 / 1 point Recorded here are the germination times (in days) for ten randomly chosen seeds of a new type of bean. See Attached Excel for Data. germination time data.xlsx

Assume that the population germination time is normally distributed. Find the 99% confidence interval for the mean germination time. (–3.250, 3.250) (13.063, 18.537) (12.550, 19.050) (12.347, 19.253) (13.396, 18.204) Hide question 18 feedback T-Critical Value = T.INV.2T(.01,9) = 3.249836 LL = 15.8 - 3.249836*3.35989410 UL = 15.8 + 3.249836*3.35989410 Question 19 1 / 1 point In a certain town, a random sample of executives have the following personal incomes (in thousands); Assume the population of incomes is 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

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35.862 < μ < 50.138 Hide question 19 feedback T- Critical Value = T.INV.2T(.05,13) = 2.160369 LL = 43 - 2.160369*13.626914 UL = 43 + 2.160369*13.626914 Question 20 1 / 1 point 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 distribution. Compute a 99% confidence interval for the population mean amount of beverage in 16-ounce beverage cans and fill in the blanks appropriately. 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 20 feedback T-Critical Value = T.INV.2T(.01,7) = 3.499483 �� =15.8−3.499483 ∗ .4407798 �� =15.8+3.499483 ∗ .4407798 Done

Homework 5 Attempt 2

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