An SRS of 380 high school seniors gained an average of....
Homework Answers
from empirical rule:
a: =estimated mean -*/+z*std error =23.11 -3*48.85/sqrt(380)=15.592
b> =estimated mean +*/+z*std error =23.11 +3*48.85/sqrt(380)=30.628
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An SRS of 380 high school seniors gained an average of....
An SRS of 380 high...
An SRS of 380 high school seniors gained an average of i -23.51 points in their...
An SRS of 380 high school seniors gained an average of i -23.51 points in their second attempt at the SAT Mathematics exam. Assume that the change in score has a Normal distribution with standard deviation σ 48.51, we want to estimate the mean change in score u in the population of all high school seniors. (a) Using the 68-95-99.7 Rule or the z-table (Table A), give a 95% confidence interval (a, b) for based on this sample. Enter your...
An SRS of 380 high school seniors gained an average of ?¯=20.23 points in their second...
An SRS of 380 high school seniors gained an average of ?¯=20.23 points in their second attempt at the SAT Mathematics exam. Assume that the change in score has a Normal distribution with standard deviation σ=47.28. We want to estimate the mean change in score ?μ in the population of all high school seniors. (a) Using the 68 – 95 – 99.7 Rule or the ?-table (Table A), give a 95% confidence interval (?,?) for ? based on this sample....
An SRS of 450 high school seniors gained an average of x = 20 points in...
An SRS of 450 high school seniors gained an average of x = 20 points in their second attempt at the SAT Mathematics exam. Assume that the change in score has a Normal distribution with standard deviation ơ 49. (a) Find a 95% confidence interval for the mean change in score μ in the population of all high school seniors. (Enter your answers rounded to two decimal places.) lower bound of confidence interval: upper bound of confidence interval: (b) What...
An SRS of 500 high school seniors gained an average of x 25 points in their...
An SRS of 500 high school seniors gained an average of x 25 points in their second attempt at the SAT Mathematics exam. Assume that the change in score has a Normal distribution with standard deviation ơ-54. (a) Find a 90% confidence interval for the mean change in score μ in the population of all high school seniors. (Enter your answers rounded to two decimal places.) lower bound of confidence interval: upper bound of confidence interval: (b) What is the...
High school seniors' SAT scores are normally distributed with μ = 1050 and σ = 100
2.) High school seniors' SAT scores are normally distributed with μ = 1050 and σ = 100. If a student is selected at random, find the probability that her SAT score is: a.) above 1200 b.) below 890 c.) between 1000 and 1100 d.) What SAT score separates the smartest 4% of students? e). If 18 seniors are selected, find the probability that their mean SAT score is above 1150 3.) A survey of 200 college students revealed that 160 of them eat dessert...
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The level of nitrogen oxides (NOX) and nonmethane organic gas (NMOG) in the exhaust over the useful life (150,000 miles of driving) of cars of a particular model varies Normally with mean 90 mg/mi and standard deviation 4 mg/mi. A company has 25 cars of this model in its fleet. Using Table A, find the level L such that the probability that the average NOX +NMOG level for the fleet greater than L is only 0.02? (Enter your answer rounded...
The level of nitrogen oxides (NOX) and nonmethane organic gas (NMOG) in the exhaust over the...
The level of nitrogen oxides (NOX) and nonmethane organic gas (NMOG) in the exhaust over the useful life (150,000 miles of driving) of cars of a particular model varies Normally with mean 80 mg/mi and standard deviation 6 mg/mi. A company has 16 cars of this model in its fleet Using Table A, find the level L such that the probability that the average NOX + NMOG level i for the fleet greater than L is only 0.05? (Enter your...
(16.09) We have the survey data on the body mass index (BMI) of 640 young women....
(16.09) We have the survey data on the body mass index (BMI) of 640 young women. The mean BMI in the sample was x¯¯¯=26.6x¯=26.6. We treated these data as an SRS from a Normally distributed population with standard deviation σ=σ=7.8 . Find the margins of error for 99 % confidence based on SRSs of N young women. N margins of error (±±0.0001) 139 406 1558 10. (16.10) An SRS of 400 high school seniors gained an average of x¯¯¯x¯ =...
You want to estimate the mean SATM score for 250,000 high school seniors in California. Only abou...
You want to estimate the mean SATM score for 250,000 high school seniors in California. Only about 45% of California students take the SAT. These self-selected students are planning to attend college and are not representative of all California seniors. A simple random sample (SRS) of 500 California high school seniors is tested. The mean score of the sample is Y = 461 What could you say about the mean score, η = 508 in the population of all 250,000...
4. You want to estimate the mean SATM score for 250,000 high school seniors in California. Only a...
4. You want to estimate the mean SATM score for 250,000 high school seniors in California. Only about 45% of California students take the SAT. These self-selected students are planning to attend college and are not representative of all California seniors A simple random sample (SRS) of 500 California high school seniors is tested. The mean score of the sample is Y 461. What could you say about the mean score, n-508 in the population of all 250,000 seniors? Assume...
An SRS of 380 high school seniors gained an average of i -23.51 points in their second attempt at the SAT Mathematics exam. Assume that the change in score has a Normal distribution with standard deviation σ 48.51, we want to estimate the mean change in score u in the population of all high school seniors. (a) Using the 68-95-99.7 Rule or the z-table (Table A), give a 95% confidence interval (a, b) for based on this sample. Enter your...
An SRS of 450 high school seniors gained an average of x = 20 points in their second attempt at the SAT Mathematics exam. Assume that the change in score has a Normal distribution with standard deviation ơ 49. (a) Find a 95% confidence interval for the mean change in score μ in the population of all high school seniors. (Enter your answers rounded to two decimal places.) lower bound of confidence interval: upper bound of confidence interval: (b) What...
An SRS of 500 high school seniors gained an average of x 25 points in their second attempt at the SAT Mathematics exam. Assume that the change in score has a Normal distribution with standard deviation ơ-54. (a) Find a 90% confidence interval for the mean change in score μ in the population of all high school seniors. (Enter your answers rounded to two decimal places.) lower bound of confidence interval: upper bound of confidence interval: (b) What is the...
The level of nitrogen oxides (NOX) and nonmethane organic gas (NMOG) in the exhaust over the useful life (150,000 miles of driving) of cars of a particular model varies Normally with mean 90 mg/mi and standard deviation 4 mg/mi. A company has 25 cars of this model in its fleet. Using Table A, find the level L such that the probability that the average NOX +NMOG level for the fleet greater than L is only 0.02? (Enter your answer rounded...
The level of nitrogen oxides (NOX) and nonmethane organic gas (NMOG) in the exhaust over the useful life (150,000 miles of driving) of cars of a particular model varies Normally with mean 80 mg/mi and standard deviation 6 mg/mi. A company has 16 cars of this model in its fleet Using Table A, find the level L such that the probability that the average NOX + NMOG level i for the fleet greater than L is only 0.05? (Enter your...
You want to estimate the mean SATM score for 250,000 high school seniors in California. Only about 45% of California students take the SAT. These self-selected students are planning to attend college and are not representative of all California seniors. A simple random sample (SRS) of 500 California high school seniors is tested. The mean score of the sample is Y = 461 What could you say about the mean score, η = 508 in the population of all 250,000...
4. You want to estimate the mean SATM score for 250,000 high school seniors in California. Only about 45% of California students take the SAT. These self-selected students are planning to attend college and are not representative of all California seniors A simple random sample (SRS) of 500 California high school seniors is tested. The mean score of the sample is Y 461. What could you say about the mean score, n-508 in the population of all 250,000 seniors? Assume...
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