A 95% confidence interval for the mean body mass index (BMI) of young American women is 26.8 ± 0.6.One of your classmates interprets this interval in the following way: "We are 95% confident that future samples of young womenwill have mean BMI between 26.2 and 27.4."Is your classmate's interpretation correct? If not, what is the correct interpretation of the confidence interval?Correct. This interpretation correctly refers to the parameter (all young women).Incorrect. We are 95% confident that the interval from 26.2 and 27.4 captures the true mean BMI of all youngAmerican women.Incorrect. If we take many samples, the population mean BMI will be between 26.2 and 27.4 in about 95% of thosesamples.Incorrect. The mean BMI of young American women cannot be 28.Incorrect. We are 95% confident that the interval from 26.2 and 27.4 captures the BMI for every young Americanwoman.

Answered: Image /qna-images/answer/766afaf5-31f0-47b6-acfb-ac81e21efc68.jpg

Answered: A 95% confidence interval for the mean…

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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QUESTION 23 Ali constructed a 98% confidence interval for the mean amount spent per day in a shopping mall based on 100 customers. The confidence interval was found to be (AED 1150, AED 2150). Select all correct answers. The mean amount spent per day by the 100 customers selected in the sample falls between AED 1150 and AED 2150. We are 98% confident that the mean amount spent per day by customers falls between AED 1150 and AED 2150. The mean amount spent per day by all customers falls between AED 1150 and AED 2150. 98% of all customers spent between AED 1150 and AED 2150 per day.
If n=14, ¯xx¯(x-bar)=34, and s=4, construct a confidence interval at a 80% confidence level. Assume the data came from a normally distributed population.
Construct an 80 % confidence interval for the mean height of sons whose fathers are 70 inches tall. Round your answer to at least three decimal places. An 80 % confidence interval for the mean height of sons whose fathers are 70 inches tall is (this is part 2 of 2 of the math problem, please and thank you
QUESTION 4 Sara wants to determine the mean play time of media clips on her smartphone. A random sample of 36 media clips revealed a mean of 155 minutes and a standard deviation of 16 minutes. Find the upper confidence limit (UCL) of the 93% confidence interval estimate for the mean play time of media clips on Sara's smartphone. Round your answer to two decimal places.
Question 12 Astudent was asked to find a 90% confidence interval for the proportion of students who take notes using data from a random sample of size n = 86. Which of the following is a correct interpretation of the interval 0.13 p < 0.27? Check all that are correct. With 90% confidence, the proportion of all students who take notes is between 0.13 and0.27. U There is a 90% chance that the proportion of the population is between 0.13 and 0.27. O There is a 90% chance that the proportion of notetakers in a sample of 86 students wilL be between 0.13 and 0.27. O The proprtion of all students who take notes is between 0.13 and 0.27, 90% of the time. OWith 90% confidence, a randomly selected student takes notes in a proportion of their classes that is between 0.13 and 0.27.
If n=20, ¯xx¯(x-bar)=43, and s=10, construct a confidence interval at a 98% confidence level. Assume the data came from a normally distributed population.Give your answers to one decimal place.
You measure 30 textbooks' weights, and find they have a mean weight of 58 ounces. Assume the population standard deviation is 5.7 ounces. Based on this, construct a 99% confidence interval for the true population mean textbook weight. Give your answers as decimals, to two places I am 99% confident that the mean weight of textbooks is between ounces. and
Let's shake on it: A random sample of 12-ounce milkshakes from 11 fast-food restaurants had the following number of calo 642 504 473 580 450 375 700 608 476 613 472 Send date to Excel Assume the population standard deviation is o =87.
you conducted additional studies with these men and found the following: a. The RR for those who drank one drink per day was 1.50 (95% CI:1.32-1.67) in comparison to the abstainers. b. For those men who drank three or more drinks per day the RR compared to the abstainers was 4.20 (95% CI 2.68-5.85). Question: We can tell from the Confidence Intervals provided in #7 whether there are likely more people in the “one drink a day” group or the “3+ drinks per day” group. Which one likely has more people and how can we tell?
ANSWER PART B ONLY
Question 6 Is smoking during pregnancy associated with premature births? To investigate this question, researchers selected a random sample of 118 pregnant women who were smokers. The average pregnancy length for this sample of smokers was 255 days. From a large body of research, it is known that length of human pregnancy has a standard deviation of 16 days. The researchers assume that smoking does not affect the variability in pregnancy length. Find the 95% confidence interval to estimate the length of pregnancy for women who smoke. (Note: The critical z-value to use, Ze, is: 1.960) ) Your answer should be rounded to 3 decimal places.
Question 4 く > You must estimate the mean temperature (in degrees Fahrenheit) with the following sample temperatures: 32.6 -1.6 25.1 22.9 29.3 0.8 Find the 80% confidence interval. Enter your answer as an open-interval (i.e., parentheses) accurate to two decimal places (because the sample data are reported accurate to one decimal place). 80% C.I. =

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