You are given the following summary statistics from a sample of 20 observations: (You may find it useful to reference the appropriatetable: chi-square table or Etable)cesMeanStandard DeviationSkewnessKurtosisa. Using the Jarque-Bera test, specify the null and alternative hypotheses to determine whether or not the data are normallydistributed.Ha: 50 and KOH: 50 and xOH: S andTest statsbo69.1411.381.142.65b-1. Calculate the value of the test statistic. (Round intermediate calculations to at least 4 decimal places and final answer to 3decimal places.)b-2. Find the p-value.p-value <0,01Onsabie 0108; HA: 50 and 00; 4: 50 and Kee0; M₁: 5>0 and x > 011.500#Mast

Given The data is as follows: Sample size, n=20 Mean=69.14 Standard deviation=11.38 Skewness=1.14…

Answered: Standard Deviation Skewness Kurtosis o.…

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 physical therapist wanted to know whether the mean step pulse of men was less than the mean step pulse of women. She randomly selected 54 men and 70 women to participate in the study. Each subject was required to step up and down a 6-inch platform. The pulse of each subject was then recorded. The following results were obtained. Two sample T for Men vs Women N Mean StDev SE Mean Men Women 98% CI for mu Men - mu Women (- 12.20, - 1.00) T-Test mu Men = mu Women (vs H2 O C. Ho: H1 = H2; Ha: H1 #H2 (b) Identify the P-value and state the researcher's conclusion if the level of significance was a = 0.01. What is the P-value? P-value =
Eemaining Time: 1 hour. 37 minutes, 54 seconds. Question Completion Status: QUESTION 10 A normally distributed random variable X with variance o = 400 and P(X> 130) = 0.6554. Then the mean of X is O 122 O 138 O 120 O 290 QUESTION 11 For the given data in stem-and- leaf display form, find the inter-quartile range. 3.2 | 4579 3.3 | 02558 3.4 | 45789 3.5 | 11678 3.6 | 0268 O 2.5 O 0.28 O 0.31
Number...4
use the appropr Given are five observations collected in a regression study on two variables. may Yi 2 6 9 13 20 9 19 7 26 21 Die or X to 2.44 X to 29.82 Develop the 95% confidence and prediction intervals when x = 10. (Round your answers to two decimal places.) confidence interval -1.04 prediction interval -11.02 hnology to answer this yu XX
Question 1c needed
Two samples are taken with the following sample means, standard deviations, and sample sizes. We have no reason to assume the variance are equal. T1 = 20 81 = 3 n₁ = 50 T₂ = 27 82 = 2 n₂ = 65 Estimate the difference in population means using a 88% confidence level in interval notation. Round answers to 2 decimal places. -7.770 -6.23 x
Answers: 2. t = .656, do not reject Ho: there is no evidence that this year's students are performing differently than students in previous years. 3. chi-square = 7.997, Do not reject H0; no evidence that the variance is significantly less than 175. chi-square = 11.250, Do not reject H0; no evidence that the variance is significantly greater than 175. 4. 3.973 to 13.5219
As a bonus assignment a former student checked if your professor gave a statisticallysignificant difference in grades between his male and female students. She based herstudy based on grades assigned in intermediate Econ courses (Econ 303, 305 and 317)and her sample included nm = 485 male students and nf = 264 female students. Theaverage grades received were xm = 84.6 and xf = 85.8 The population standad deviation were σ m = 12.0 and σ f = 11.4 8. From the same extra-credit study as in question 7 see above, this former student found that the proportion of female students in principle courses (Econ 203, 205) was ?̅? = 0.380, while the proportion of female students in intermediate courses (Econ 303, 305, 317) was ?̅? = 0.352. The principle courses sample size was np = 782, while the intermediate courses sample size was ni = 749. Test the hypothesis that female students are less in intermediate courses using a 90% confidence level and the p-value approach.
pounds, with a standard deviation of 15.81 pounds. At the 0.05 level of significance, test the claim that the weights of football players have a greater variance than the weights of baseball deviation of 40.09 pounds. A random sample of 25 baseball players had a mean weight of 202.9 6. A random sample of 48 football players had a mean weight of 246.6 pounds, with a standard ith a players. ie th sufficient evidence to conclude that Step 5: There
Use the AustStates data. Perform a repeated measures ANOVA using the 9 years as the paired variables and the states as the observations. Based on the Greenhouse-Geisser p-value of p=0.130, sphericity can be assumed for these data. The result of the analysis is F=9.563 on degrees of freedom, for a p-value of You the null hypothesis. 8 and 56, p<0.001, reject 8 and 56, p<0.001, fail to reject 1.0 and 7.3, p=0.016, reject 1.0 and 7.3, p=0.016, fail to reject
Answer the following questions: the test is to be used? The critical value is The test statistics/computed value is Based on the comparison of the critical value and the computed value, the decision is
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