A random sample of 68 SAT scores of students applying for merit scholarships showed an average of 1400 with a standard deviation of 220. If we want to provide a 95% confidence interval for the population mean SAT score, the degrees of freedom for reading the t value is The "t" value (3 decimals) for a 95% confidence interval estimation is The margin of error (3 decimals) at 95% confidence is The 95% confidence interval (3 decimals) for the SAT scores goes from up to

A random sample of 68 SAT scores of students applying for merit scholarships showed an average of 1400 with a standard deviation of 220. If we want to provide a 95% confidence interval for the population mean SAT score, the degrees of freedom for reading the t value is The "t" value (3 decimals) for a 95% confidence interval estimation is The margin of error (3 decimals) at 95% confidence is The 95% confidence interval (3 decimals) for the SAT scores goes from up to

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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A random sample of 68 SAT scores of students applying for merit scholarships showed an average of 1400 with a standard deviation of 220. If we want to provide a 95% confidence interval for the population mean SAT score, the degrees of freedom for reading the t value is? The "t" value (3 decimals) for a 95% confidence interval estimation is? The margin of error (3 decimals) at 95% confidence is? The 95% confidence interval (3 decimals) for the SAT scores goes from up to?
The admissions officer at a small college compares the scores on the Scholastic Aptitude Test (SAT) for the school's in-state and out-of-state applicants. A random sample of 88 in-state applicants results in a SAT scoring mean of 1180 with a standard deviation of 41. A random sample of 17 out-of-state applicants results in a SAT scoring mean of 1107 with a standard deviation of 53. Using this data, find the 95% confidence interval for the true mean difference between the scoring mean for in-state applicants and out-of-state applicants. Assume that the population variances are not equal and that the two populations are normally distributed. Step 1 of 3: Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places. Step 2 of 3: Find the standard error of the sampling distribution to be used in constructing the confidence interval. Round your answer to the nearest whole number. Step 3 of 3: Construct the 95% confidence…
A sample of twenty newborn elephants had a sample mean of 225 pounds and a sample standard deviation of 15 pounds. However, it is known that the standard deviation for the population (sigma) is actually 18 pounds. Construct a 90% confidence interval for the weight of newborn elephants. Lower_______pounds Upper_______pounds Is it best to use a Z-interval or a T-interval?
A random sample of 20 grade 1 students in Pio Elementary School yielded a mean weight of 31.68 kg with sample standard deviation of 6.39 kg. What is the maximum error of 95% confidence interval for the mean weight of all grade 1 students in Pio Elementary School. Round off your answer to two decimal places
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A random sample of 84 students at a university showed an average age of 22 years and a sample standard deviation of 3 years. Find the margin of error for the 94% confidence interval. (Assume that the population of students is large relative to the sample size. Round your solution to 4 decimal places)
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A personnel manager has found that, historically, the scores on aptitude tests given to applicants for entry-level positions follow a normal distribution with standard deviation 20.37 points. A random sample of 10 test scores from the current group of applicants had a mean score of 69.89 points. Based on these sample results, a statistician found for the population mean a confidence interval extending from 57.329 to 82.451 points. Find the confidence level of this interval.