2- The test score is now normalised to have a mean of 0 and a standard deviation of 1.Using the same specification, the output is now:ntest |Coef.Std. Err.tP>|t|[95% Conf. Interval]female |-.0575288.0023799-24.170.000-.0621933-.0528643cons |.0288284.001684217.120.000.0255275.0321293cntryid |F(72, 519260) =2597.1110.000(73 categories)What is the expected difference in test score between boys and girls?

Answered: Image /qna-images/answer/f9aa4e6a-e58d-48c8-94fc-2ba8c8381220.jpg

Answered: 2- The test score is now normalised 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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of error, Claim: The standard deviation of pulse rates of adult males is more than 11 bpm. Fora random sample of 148 adult males, the pulse rates have a standard deviation of 11.6 bpm. Find the value of the test statistic. r7 *.* The value of the test statistic is (Round to two decimal places as needed.)
A normal distribution with a mean of 2mm and a standard deviation of 0.05mm a randomly generated number of 0.71 , need to calculate the standardised Z value, from my notes Z = (actual value - mean)/standard deviation. do i take the randomly generated number of 0.81 as the actual value ?
Gail averages 64 words per minute on a typing test with a standard deviation of 9.5 words per minute. Suppose Gail's words per minute on a typing test are normally distributed. Let X= the number of words per minute on a typing test. Then X∼N(64,9.5). If necessary, round to three decimal places. Provide your answer below: Suppose Gail types 89 words per minute in a typing test on Sunday. The z-score when x = 89 is . The mean is . This z-score tells you that x = 89 is standard deviations to the right of the mean.
An analyst for a transportation metro is asked to examine the time required (x) in minutes for a specific commute route. She records 200 randomly sampled times and finds: Σx = 11,000 and Sxx =1791. Find the average commute time, as well as standard deviation and coefficient of variation.
= Scores on a standardized exam are known to follow a normal distribution with standard deviation o 8.2. A researcher randomly selects 100 students and computes their average score. He reports that the mean score is 83, with a margin of error of 1.436. How confident are you that the mean score for all students taking the exam is in the interval (81.564, 84.436)?
The patient recovery time from a particular surgical procedure is normally distributed with a mean of 3.7 days and a standard deviation of 2.6 days.What is the z-score for a patient who takes 9 days to recover? (Round your answer to two decimal places.)
The lengths of pregnancies in a small rural village are normally distributed with a mean of 261 days and a standard deviation of 17 days. What percentage of pregnancies last fewer than 314 days? P(x < 314 days) = P(z<
IQ is normally distributed with a mean of 100 and a standard deviation of 15. Suppose one individual is randomly chosen. Let X = IQ of an individual. (a) Find the z-score for an IQ of 99, rounded to three decimal places. (b) Find the probability that the person has an IQ greater than 99. (c) Shade the area corresponding to this probability in the graph below. (Hint: The x- axis is the z-score. Use your z-score from part (a), rounded to one decimal place). Shade: Left of a value -1 -1.5 0 +). Click and drag the arrows to adjust the values. 1 2 (d) MENSA is an organization whose members have the top 2% of all IQs. Find the minimum IQ needed to qualify for the MENSA organization. (e) Sketch the graph, and write the probability statement. Edit Insert Formats B I U x₂ x² A => EM e & N 18 (f) The middle 50% of IQs fall between what two values? (g) Sketch the graph and write the probability statement. Edit Insert Formats BIUX₂ X² A E = = E EC P R N • Σ+ Σ Α Σ+ Σ Α
Find the area of the shaded region. The graph to the right depicts IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15. Click to view page 1 of th - × The area of the shaded r Standard Normal Table (Page 1) NEGATIVE z Scores Standard Normal (z) Distribution: Cumulative Area from the LEFT .00 .01 02 .03 .04 .05 .06 .07 .08 .09 -3.50 and lower .0001 -3.4 .0003 .0003 .0003 .0003 .0003 .0003 -3.3 .0005 .0005 .0005 .0004 .0004 .0004 .0003 0004 .0003 .0004 .0003 .0004 .0002 .0003 -3.2 .0007 .0007 .0006 0006 .0006 .0006 .0006 .0005 .0005 .0005 -3.1 .0010 .0009 .0009 .0009 .0008 .0008 .0008 .0008 0007 .0007 -3.0 .0013 .0013 0013 .0012 0012 .0011 .0011 .0011 0010 0010 -2.9 .0019 0018 0018 .0017 .0016 .0016 .0015 .0015 .0014 .0014 -2.8 .0026 .0025 .0024 .0023 .0023 .0022 .0021 .0021 .0020 .0019 -2.7 .0035 .0034 .0033 .0032 .0031 .0030 .0029 .0028 .0027 .0026 -2.6 .0047 .0045 0044 .0043 .0041 .0040 .0039 .0038 .0037 .0036 -2.5 .0062…
=90, perform a two-tailed t test witha-05 = 81.19 and Given lo % based on a sample of N= 4 scores with a mean of X s = 6.89.
A distribution of values is normal with a mean of 220 and a standard deviation of 8.Find the interval containing the middle-most 40% of scores: Incorrect Enter your answer using interval notation. Example: [2.1,5.6) Your numbers should be accurate to 1 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.
Scores on a standardized exam are known to follow a normal distribution with standard deviation σ = 21. A researcher randomly selects 125 students and computes their average score. He reports that the mean score is 73.8, with a margin of error of 2.920. How confident are you that the mean score for all students taking the exam is in the interval (70.88, 76.72)?

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