Anova: Single FactorSUMMARYGroupsGroup 1Group 2Group 3ANOVASource ofVariationCount151515SSBetween Groups 40.13333333Within Groups496.6666667Total536.8Sum727750761df24244Average48.46666667Variance10.2666666750 11.7142857113.495238150.73333333MSFP-valueF crit20.06666667 1.696912752 0.195570181 3.21994229311.82539683Required;Using a confidence level of 95% with a significance level of 0.05,i. State in words, the null and alternative hypothesis for this case (Groups 1,2&3).ii.Compute the degrees of freedom between groups (Source of Variation).

Given, sample size (n1) =15 sample size (n2) =15 sample size (n3) =15

Answered: SUMMARY Groups Group 1 Group 2 Group 3…

SUMMARY Groups Group 1 Group 2 Group 3 ANOVA Source of Variation Count 15 15 15 SS Sum 727 750 761 df Average Variance 48.46666667 10.26666667 50 11.71428571 13.4952381 50.73333333 MS F

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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13 a
Descriptive Statistics Mean 4.857802643 Standard Error 0.872236698 Median 3.2 Mode 1 Standard Deviation 4.777437152 Sample Variance 22.82390574 Kurtosis 2.10829171 Skewness 1.567625002 Range 18.45 Minimum 0.25 Maximum 18.7 Sum 145.7340793 Requared write conculution of this discriptive statistics
Standard Normal Distribution Table (page 1) Standard Normal Distribution Table (page 2) Area Area Standard Normal Distribution Standard Normal Distribution 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.0003 0.0005 0.0007 0.0003 0.0005 0.0007 0.0003 0.0005 0.0006 0.0003 0.0004 0.0006 0.0003 0.0004 0.0006 0.0003 0.0004 0.0006 0.0003 0.0004 0.0006 0.0003 0.0004 0.0005 0.0003 0.0004 0.0005 0.0007 0.0010 0.0002 0.0003 0.0005 0.0007 0.0010 0.5040 0.5438 0.5832 0.6217 0.6591 0.5080 0.5478 0.5871 0.6255 0.6628 0.5120 0.5517 0.5910 0.6293 0.6664 0.5199 0.5596 0.5987 0.6368 0.6736 0.5279 0.5675 0.6064 0.6443 0.6808 0.5319 0.5714 0.6103 0.5160 0.5557 0.5239 0.5636 0.6026 0.5359 0.5753 0.6141 -3.4 0.5000 0.5398 0.5793 0.0 -3.3 -3.2 0.1 0,2 0.5948 -3.1 -3.0 0.0010 0.0013 0.0009 0.0013 0.0009 0.0013 0.0009 0.0012 0.0008 0.0012 0.0008 0.0011 0.3 0.4 0.6179 0.6554 0.6480 0.6844 0.6517 0.6879 0.0008 0.0008 0.0011 0.6331 0.6406 0.6772 0.0011 0.6700…
ANOVA: Single Factor SUMMARY Groups Treatment 1 Treatment 2 Treatment 3 Treatment 4 ANOVA Source of Variation Between Groups Within Groups Total Count 16 19 17 15 SS 65.2023 409.9619 475.1642 Sum 87 111 66 53 Average Variance 5.4375 9.0625 5.8421 8.3626 3.8823 4.6103 3.5333 3.5524 df MS A B C F P-value Find the values that would go in boxes A, B, and C of this ANOVA table. Please identify each answer by typing the following below A = # , B = #, C = #
= According to a study conducted by an organization, the proportion of Americans who were afraid to fly in 2006 was 0 10. A random sample of 1,200 Americans results in 132 indicating that they are afraid to fly. Explain why this is not necessarily evidence that the proportion of Americans who are afraid to fly has increased. Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2) Select the correct choice below and, if necessary, fill in the answer box to complete your choice. O A. This is not necessarily evidence that the proportion of Americans who are afraid to fly has increased above 0.10 because the sample size n is more than 5% of the population OB. This is not necessarily evidence that the proportion of Americans who are afraid to fly has increased above 0.10 because the value of np(1-p) is less than 10. OC. This is not necessarily evidence that the proportion of Americans who are afraid to fly has…
In a survey, 38% of the respondents stated that they talk to their pets on the telephone. A veterinarian believed this result to be too high, so she randomly selected 170 pet owners and discovered that 61 of them spoke to their pet on the telephone. Does the veterinarian have a right to be skeptical? Use the a = 0.1 level of significance. Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2). Because npo (1- Po) = 40.1 > 10, the sample size is less than 5% of the population size, and the sample is given to be random, the requirements for testing the hypothesis are satisfied. (Round to one decimal place as needed.) What are the null and alternative hypotheses? Ho: | versus H,: (Type integers or decimals. Do not round.) 1 Standard Normal Distribution Table (page 1) Area Standard Normal Distribution 0.00 0.01 0.02 0.03 0.04 0,05 0.06 0,07 0.08 0,09 -34 -33 -32 -31 -30 0.0003 0.0005 0.0007 0.0010 0.0013 0.0003…
E Question Help Assume the readings on thermometers are normally distributed with a mean of 0°C and a standard deviation of 1.00°C. Find the probability that a randornly selected thermorneter reads between - 1.99 and -0.67 and draw a sketch of the region. Click to view nage 1 of the table Click to view page 2 of the table, Sketch the region. Choose the correct graph below. O A. OB. OC. -1.99 -0.67 -1.99 -0.67 -1.99 -0.67 The probability is. (Round to four decimal places as needed.) Click to select your answer(s). Save for Later O vA 10:57 esc DII & %23 backspace 3 6 7. 8 e r t y k
Treatment II III 2 N = 24 G = 72 EX2 = 292 3 M = 2 M = 3 T = 16 T= 24 T= 32 SS = 16 SS = 24 SS = 20 Use an ANOVA with a = .05 to determine whether there is a significant mean difference between the two treatments. (Round to two decimal places where needed.) Source df MS Feritical Between treatments Within treatments Total F Distribution Numerator Degrees of Freedom = 6 Denominator Degrees of Freedom = 16 0.0 1.0 3.0 4.0 5.0 6.0 7.0 9.0 11.0 120 Conclusion: Fail to reject the null hypothesis; there are no significant differences among the three treatments. Reject the null hypothesis; there are no significant differences among the three treatments. Reject the null hypothesis; there are significant differences among the three treatments. o Fail to reject the null hypothesis; there are significant differences among the three treatments.
Suppose a geyser has a mean time between eruptions of 80 minutes. If the interval of time between the eruptions is normally distributed with standard deviation 18 minutes, answer the following questions. Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2), (a) What is the probability that a randomly selected time interval between eruptions is longer than 89 minutes? The probability that a randomly selected time interval is longer than 89 minutes is approximately . (Round to four decimal places as needed.) Standard Normal Distribution Table (page 1) Area Standard Normal Distribution 0,03 0.00 0.01 0.02 0.04 0.05 0.06 0.07 0.08 0.09 -34 ー33 0.0003 0.0005 0.0007 0.0003 0.0005 0.0007 0.0003 0.0005 0.0006 0.0003 0.0004 0.0006 0.0003 0.0004 0.0006 0.0003 0.0004 0.0006 0.0003 0.0004 0.0005 0.0003 0.0003 0.0004 0.0005 0.0007 0.0010 0.0002 0.0003 0.0005 0.0007 0.0010 0.0004 0.0006 -32 2 -31 -30 0.0010 0.0013…
25
In the game of roulette, a wheel consists of 38 slots numbered 0, 00, 1, 2, 36. To play the game, a metal ball is spun around the wheel and is allowed to fall into one of the numbered slots. If the number of the slot the ball falls into matches the number you selected, you < win $35; otherwise you lose $1. Complete parts (a) through (g) below. Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2). (a) Construct a probability distribution for the random variable X, the winnings of each spin. P(x) X 35 - 1 (Type integers or decimals rounded to four decimal places as needed.) OOD
Compute the critical value z/2 that corresponds to a 88% level of confidence. Click here to view the standard norma distribution table (page 1D Click here to view the standard normal distribution table (page 2). Za/2 = (Round to two decimal places as needed.)