HE AssignmentsSSuppose a simple random sample of size n=1000 is obta...(a) Describe the sampling distribution of p.A. Approximately normal, μ = 0.76 and o≈ 0.0135рOB. Approximately normal, μ = 0.76 and o≈ 0.0003рC. Approximately normal, μ = 0.76 and o~ 0.0002ррSuppose a simple random sample of size n = 1000 is obtained from a population whose size is N = 1,500,000 and whose population proportion with a specified characteristic is p = 0.76. Complparts (a) through (c) below.(b) What is the probability of obtaining x = 790 or more individuals with the characteristic?P(x ≥ 790) = 0 (Round to four decimal places as needed.)mylab.pearson.com(c) What is the probability of obtaining x = 720 or fewer individuals with the characteristic?P(x ≤ 720) = 1 (Round to four decimal places as needed.)■■■■■T■TextL√i>L(1)bHome | bartlebyMore●●+

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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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point(s) possible Assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of 1. Dra graph and find the probability of a bone density test score less than 0. .. Sketch the region. Choose the correct graph below. O A. O B. OD. Q ^ A A 0 0 The probability is Round to four decimal places as needed.) tatcrunch CH W O C. √₁ √₁ T (1) More
A population of values has a normal distribution with μ=185.7μ=185.7 and σ=57.5σ=57.5. You intend to draw a random sample of size n=57n=57.Find P25, which is the score separating the bottom 25% scores from the top 75% scores.P25 (for single values) = Find P25, which is the mean separating the bottom 25% means from the top 75% means.P25 (for sample means) =
Let x be a random variable that represents red blood cell count (RBC) in millions of cells per cubic millimeter of whole blood. Then x has a distribution that is approximately normal. For the population of healthy female adults, suppose the mean of the x distribution is about 4.78. Suppose that a female patient has taken six laboratory blood tests over the past several months and that the RBC count data sent to the patient's doctor are as follows. 4.9 4.2 4.5 4.1 4.4 4.3 (i) Use a calculator with sample mean and standard deviation keys to find x and s. (Round your answers to two decimal places.) x = s = (ii) Do the given data indicate that the population mean RBC count for this patient is lower than 4.78? Use ? = 0.05. (a) What is the level of significance?State the null and alternate hypotheses. H0: ? < 4.78; H1: ? = 4.78H0: ? = 4.78; H1: ? ≠ 4.78 H0: ? = 4.78; H1: ? < 4.78H0: ? = 4.78; H1: ? > 4.78H0: ? > 4.78; H1: ? = 4.78 (b) What sampling…
Assume that a randomly selected subject is given a bone density test Those test scores are nomally distributed with a mean of 0 and a standard deviation of 1 Draw a graph and find the probablity of a bone density test score greater than 0,34 Sketch the region. Choose the correct graph below. OA. OB. OD. 034 034 034 034 034 The probability isO (Round to four decimal places as needed) de Click to select your answer(s) Next emalhe tonsuffed ty robots (hey et thar smart
Briefly explain how increasing sample size influences each of the following. Assume that all other factors are held constant. Does tach increase, decrease, or stay the same? (M-u) (M-u)Vn The size of the Z-score Z. = M-u – OM (M-u) The size of Cohen's d d = (Mcr-u) (Mcr-u)vn The power P( 2Z> Mcritical-µu
Average Family Size The average family size was reported as 3.18 . A random sample of families in a particular school district resulted in the following family sizes: 5 4 5 4 4 3 6 4 3 3 5 5 6 3 3 2 4 5 2 2 2 3 2 Send datato Excel At =α0.10 , does the average family size differ from the national average?Assume the population is normally distributed. Use a TI-83 Plus/TI-84 Plus calculator
CHI-SQUARE TEST The Chi-square test will be used to compare phenotypic values derived from the actual crosses to the expected values from the theoretical crosses. This test is a simple statistical test to evaluate the amount of deviation from the expected equal frequency ratio in cach class The chi-square test can be used to test any ratio goodness of fit. The calculated chi-square value is obtained by the formula: X =E (O-E)/E. Where, O- observed number on the actual cross; observed value E- calculated or expected frequency (from the corresponding theoretical cross) E- summation symbol which means that you'll have to perform a calculation for every single data item in your data set. ОВJЕСTIVES The students should be able to: a. To compare phenotypic data derived from actual crosses to expected values using the Chi-square test b. To be able to interpret the results of chi-square values derived from computations of monohybrid crosses MATERIALS Observed data from the Monohybrid…
A person's body mass index (BMI) is computed by dividing the weight (kg) by the square of height (m). The accompanying table contains the BMI statistics for random samples of males and females. Assume that the two samples are independent simple random samples selected from normally distributedpopulations, and do not assume that the population standard deviations are equal. Let population 1 be females. Female BMI: n=69 x=29.14 s=7.41 Male BMI: n=79 x=28.18 s=5.24 The test statistic is The P-value is Construct a confidence interval appropriate for testing the claim in part (a). The ___% confidence interval estimate is ___<μ1−μ2<___.
A special bumper was installed on selected vehicles in a large fleet. The dollar cost of body repairs was recorded for all vehicles that were involved in accidents over a 1-year period. Those with the special bumper are the test group and the other vehicles are the control group, shown below. Each "repair incident" is defined as an invoice (which might include more than one separate type of damage). Statistic Test Group Control Group Mean Damage x¯1x¯1 = $ 1,101 x¯2x¯2 = $ 1,766 Sample Standard Deviation s1 = $ 696 s2 = $ 838 Repair Incidents n1 = 12 n2 = 9 (a) Construct a 90 percent confidence interval for the true difference of the means assuming equal variances. (Round answers to 3 decimal places. Negative values should be indicated by a minus sign.) (b) Repeat part (a), using the assumption of unequal variances with Welch's formula for d.f. (Round answers to 2 decimal places. Negative values should be indicated by a minus sign.) (d) Construct separate…
The annual rainfall in a certain region is modeled using the normal distribution shown below. The mean of the distribution is 36.5 cm and the standard deviation is 5.2 cm. In the figure, V is a number along the axis and is under the highest part of the curve. And, U and W are numbers along the axis that are each the same distance away from V. Use the empirical rule to choose the best value for the percentage of the area under the curve that is shaded, and find the values of U, V, and W. Percentage of total area shaded: (Choose one) ▼ 200 35 | 55 25 30 40 45 50 ( cm) Submit Continue |Privacy Center © 2022 McGraw Hill LLC. All Rights Reserved. Terms of Use DIl S0 FB F7 esc F3
Ch. 6.1 CI What sample size is needed to give the desired margin of error in estimating a population proportion with the indicated level of confidence? (16) A margin of error within +/- 1% with 99% confidence. (18) A margin of error within +/- 2% with 95% confidence. An initial small sample has p-hat=0.78
CI on Mean, Variance Unknown, Prediction and Tolerance Interval A CNC (computer numerical control) machine produces iron automobile crankshafts. Samples are measured and the inner diameter are found to be 1.01,0.97, 1.03, 1.04, 0.99, 0.98,0.99, 1.01, and 1.03 centimeters. For all computations, assume an approximately normal distribution. The sample mean and standard deviation for the given data are 1.0056, and s = 0.0246. Find a 99% confidence interval on the mean of diameter. Compute a 99% prediction interval on a measured diameter of a single crankshaft piece taken from the machine. Find the 99% tolerance limits that will contain most of the metal pieces produced by the CNC machine. a. Insufficient data to compute. Missing the sample size n b. CI on the mean: 0.9781 < µ< 1.0331. Prediction Interval: 0.9186

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