Body Mass Index: In a survey of adults with diabetes, the average body mass index (BMI) in a sample of 1928 women was 31.30 with a standard deviation of0.3. The BMI in a sample of 1559 men was 30.50, with a standard deviation of 0.6.Part 1 of 2(a) Construct a 99% confidence interval for the difference in the mean BMI between women and men with diabetes. Let H, denote the mean BMI forwomen with diabetes. Use the TI-84 calculator and round the answers to two decimal places.A 99% confidence interval for the difference in the mean BMI between women and men with diabetes isPart: 1 / 2Part 2 of 2(b) Does the confidence interval contradict the claim that the mean BMI is the same for both men and women with diabetes?The confidence interval (Choose one)this claim.

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Answered: Body Mass Index: In a survey of adults…

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 study of 96 bolts of carpet showed that their average length was 77.5 yards. The standard deviation of the population is 1.6 yards. Which of the following is the 90% confidence interval for the mean length per bolt of carpet?
#4data: 0.06, 0.11, 0.16, 0.15, 0.14, 0.08, 0.15. Use the sample data from #4 to construct a 98% confidence interval estimate for the standard deviation of the amount nitrogen-oxide emission for all cars.
The distribution of study times for first-year students follow a Normal distribution with a standard deviation σ = 48 minutes. Suppose a sample of 344 students found that the mean study time for these students was x¯= 170 minutes per week. Use this to find a 94.38% confidence interval to estimate the mean study time for all students. We are 94.38% confident that the mean study time for first-year students is between the following ( answer) minutes and ( answer) minutes. The answer needs to be the mean study time for first-year students is between what amount of minutes and the following minutes.
The standard deviation, S, for the annual medical cost of 24 households was found to equal $1689 find a 95% confidence interval for the standard deviation of the annual medical cost for all households. Assume medical costs are normally distributed.
to qualify for membership in a certain high-IQ society a person must score in the upper 2% of the population on an IQ test. the minimum IQ score to be in the upper 2% is a 140.5 if Iq scores are normanlly distributed with a standrd deviation of 10, what is the mean IQ score?
From a sample of 37 graduate students, the mean number of months of work experience prior to entering an MBA program was 36.63. The national standard deviation is known to be 19 months. What is a 90% confidence interval for the population mean? A 90% confidence interval for the population mean is 10 (Use ascending order. Round to two decimal places as needed.) ...
The mean number of English courses taken in a two-year time period by male and female college students is believed to be about the same. An experiment is conducted and data are collected from 29 males and 16 females. The males took an average of three English courses with a standard deviation of 0.8. The females took an average of four English courses with a standard deviation of 1.0. Are the means statistically the same? (Assume a 5% level of significance.)
A random sample of 10 children found that their average growth for the first year was 9.8 inches, with a standard deviation of 0.96 inches. Find the 85% confidence interval of the mean growth of all American children in the first year of life.What is the name of the confidence interval?
The manager of a fleet of automobiles is testing two brands of radial tires and assigns one tire of each brand at random to the two rear wheels of eight cars and runs the cars until the tires wear out. The data (in kilometers) follow. Find a 99% confidence interval on the difference in the mean life. Car Brand 1 Brand 2 1 36,925 34,318 2 45,300 42,280 3 36,256 35.548 4 32,100 31,950 5 37,210 38,015 6 48,360 47,800 7 38.200 37,810 8 33,500 33,215 Round your answer to 2 decimal places. Do not use commas.
In a school district, all sixth grade students take the same standardized test. The superintendant of the school district takes a random sample of 3030 scores from all of the students who took the test. She sees that the mean score is 169169 with a standard deviation of 7.06747.0674. The superintendant wants to know if the standard deviation has changed this year. Previously, the population standard deviation was 1313. Is there evidence that the standard deviation of test scores has decreased at the α=0.005α=0.005 level? Assume the population is normally distributed. Step 1 of 5: State the null and alternative hypotheses. Round to four decimal places when necessary. Step 2 of 5: Determine the critical value(s) of the test statistic. If the test is two-tailed, separate the values with a comma. Round your answer to three decimal places. Step 3 of 5: Determine the value of the test statistic. Round your answer to three decimal places. Step 4 of 5: Make the…
Scores on a standarized exam are known to follow a normal distribution with a standard deviation of 5. A researcher finds a random sample of 25 exam scores has a mean score of 72. Use this to find a 88% confidence interval for the mean exam score of all students taking the exam.
Suppose we wish to calculate a 90% confidence interval for the average amount spent on books by freshmen in their first year at a major university. The interval is to have a margin of error of $2. Assume that the amount spent on books by freshmen has a normal distribution with a standard deviation of $30. How many observations are required to achieve this margin of error?

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