2. The inheritance for seed color for this cross is 9:3:3:1Expectedratio9/16|3/163/161/16ObservedExpected(E)DeviationDeviation(d)²di(0)682(0-E)E20117265Total =1120JPInterpolation:Interpretation:

expected ratio observed expected deviation 9/16 682 3/16 201 3/16 172 1/16 65…

Answered: 2. The inheritance for seed color for…

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 simple random sample without replacement was used to select 5 clusters from a population of 29 clusters. From each selected sample cluster simple random sampling without replacement was used to select samples of secondary units. For this data find an estimate of the population total and the population mean along with their estimated variances. clmemb clsize yval [1,] [2,] [3,] [4,] [5,] [6,] [7,] [8,] [9,] [10,] [11,] [12,] [13,] [14,] [15,] [16,] [17,] [18,] [19,] [20,] [21,] [22,] [23,] [24,] [25,] [26,] [27,] [28,] [29,] [30,] [31,] [32,] [33,] 1 20 46.07 20 45.54 20 43.07 1 1. 1 20 49.29 1 20 44.35 1 20 45.21 2 33 44.11 2 33 42.19 2 33 44.42 2 33 40.05 2 33 43.88 2 33 47.45 2 33 40.41 2 33 43.24 3 22 35.42 3 22 35.46 3 22 36.34 3 22 35.25 4 40 38.67 4 40 34.73 4 40 39.28 4 40 35.26 4 40 39.84 4 40 40.32 4 40 40.03 4 40 38.15 4 40 38.69 4 40 39.38 18 47.48 18 44.78 5 18 44.37 18 44.97 5 18 49.20
please state the approximate distribution of the sample mean
Please note the question is: What is the distribution of the t-test statistic if m, n -----> ∞ ?
Derive the PDF MGF mean and variance of Non Central Chi Square
When conducting a large sa (p – Po) Po(1- Po) where i for the denominator of this (A) The standard deviation (B) The standard deviation (C) The sample size is larg (D) The standard deviation (E) The standard deviation
What is the variance and standard deviation for the number of times the SAT is taken? x Frequency (f) fx f*x2= (f*x) x (x) (4) = (3)x(1) 5152727-(2547473)(2547473)/1518859=880035.3688/1518859=0.5794 (1) (2) (3)=(1)x(2) (4)=(3x1) Population Variance 1 721,769 721,769 721,769 2 601,325 1,202,650 2,405,300 3 166,736 500,208 1,500,624 4 22,299 89,196 356,784 5 6,730 33,650 168,250 Total 1,518,859 2,547,473 5,152,727 880035.3688 0.5794 n=1,518,859 ∑fx = 2,547,473 The population variance is 0.5794 The Standard Deviation? 0.76 Can you show me the steps to reach the standard deviation of .76 ?
Q.No.1. (Marks 08) Explain the following with suitable examples. Mutually exclusive events Sampling Correlation Regression
A researcher randomly selects 6 fathers who have adult sons and records the fathers' and sons' heights to obtain the data shown in the table below. Test the claim that sons are taller than their fathers at the α=0.10 level of significance. The normal probability plot and boxplot indicate that the differences are approximately normally distributed with no outliers so the use of a paired t-test is reasonable. Observation 1 2 3 4 5 6 Height of father (in inches) 70.1 74.5 69.5 65.1 69.6 65.9 Height of son (in inches) 69.5 75.9 71.7 69.4 71.5 70.7 What are the hypotheses for the t-test? Note that population 1 is fathers and population 2 is sons. A. H0: μ1=μ2 Ha: μ1<μ2 B. H0: μ1=μ2 Ha: μ1≠μ2 C. H0: μ1=μ2 Ha: μ1>μ2 D. H0: μ1≥μ2 Ha: μ1<μ2
stion 16: nts) A random sample of 200 psychology students was selected from a department's participant pool. A researchers determined each student's score on the NPI-16 and the number of his or her Facebook friends. The marginal distribution of NPI-16 scores was 4, with a standard deviation of 2. The marginal distribution "Facebook Friends" was 450, with a standard deviation of 200. The correlation between the two variables was .15. Assuming that NPI-16 is the predictor variable and Number of Facebook Friends is the outcome variable, compute the following: The lower limit of the 95% confidence interval around b. Note: Enter your answer to the second decimal place.
Question 15

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2. The inheritance for seed color for this cross is 9:3:3:1 Expected ratio Expected |(E) Deviation (d)² Observed Deviation di (0) 682 (0-E) 9/16 3/16 3/16 201 172 65 1/16 Total = 1120 P = JP