A random sample of 120 SAT scores of students applying for merit scholarships showed anaverage of 1350 with a standard deviation of 240.If we want to provide a 95% confidence interval for the population mean SAT score, thedegrees of freedom for reading the t value isThe "t" value (3 decimals) for a 95% confidence interval estimation isThe margin of error (3 decimals) at 95% confidence isThe 95% confidence interval (3 decimals) for the SAT scores goes fromup to

The sample size is 120, the sample mean is 1350 and the sample standard deviation is 240.

Answered: A random sample of 120 SAT scores of…

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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Similar questions

For a random sample of 70 overweight men, the mean of the number of pounds that they were overweight was 28 . The standard deviation of the population is 3.8 pounds. Find the 99% confidence interval of the mean of these pounds. Round intermediate answers to at least three decimal places. Round your final answers to one decimal place.
Noise levels at various area urban hospitals were measured in decibels. The mean of the noise levels in 81 corridors was 62.04 decibels, and the standard deviation of the population was 6.34. Find the upper boundary of the 95% confidence interval of the true mean. Only the final answer should be rounded-off to TWO decimal places.
How large a sample should be selected to provide a 99% confidence interval with a margin of error of 15? Assume that the population standard deviation is 50.
A teacher gives a test to his class of 25 students. Since he wants to quickly obtain an estimate of the mean score on the exam he takes a random sample of 12 exams and determines that the average score is 82 with a standard deviation of 10. Construct a 95% confidence interval on the mean scores from this exam. Provide your formula used along with your result. Explain implications of what you have done. Interpret the final confidence interval you have obtained.
A survey taken several years ago found that the average time a person reading a newspaper was 10.8mins. The standard deviation was 3mins. To see whether the average time changed since the newspapers format was revised, the editor survered 36 individuals. The average time the 36 people spent reading the paper was 12.2mins. At $=0.02, is there a change in the average time an individual spends reading the paper ? Find the 98% confidence interval of the mean.
The confidence interval for a population mean is (7.32, 7.52). What are the sample mean and the margin of error? H = Margin of Error = Give your answers as decimals, to two places.
A sample of 28 adults attending a country fair determined that the mean age was 33 .given that the population standard deviation is known to be 12 determine the 99% confidence interval for the population mean age (round to two decimals)
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A normally distributed population is having standard deviation 2.4, a sample of 24 is taken randomly and the sample mean observed is 13.8 Find the 95% confidence interval. (round to 3 decimals)
We wish to estimate an average number of education years for a college student atWVC. We take a sample of 47 students and find that the mean is 5.3 years and standard deviation is 0.9years.Find the 95% confidence interval for the population mean.
You measure 40 turtles' weights, and find they have a mean weight of 63 ounces. Assume the population standard deviation is 7.5 ounces. Based on this, construct a 90% confidence interval for the true population mean turtle weight. Give your answers as decimals, to two places ounces
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