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.
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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.... An SRS of 380 high school seniors gained an average of -23.11 points in their second attempt at the SAT Mathematics exam. Assume that the change in score has a Normal distribution with standard deviation σ = 48.85. 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 z-table (Table A), give a 99.7%...
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...
(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 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...
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...
According to the 68-95-99.7 Rule, 99.7% of high school seniors had SAT scores between what and what?
1. A C level confidence interval is described as what? (3 points) Oan interval computed from a data sample by a method that has the probability C of producing an interval containing the true value of the parameter any interval within the margin of error O an interval with a margin of error C that is also correct C% of the time Oany interval that follows the 68-95-99.7 rules Oan interval computed from a data sample that guarantees that the...