Consider the hypothesis statement to the right using a=0.10 and the data to the right from two independent samples.a) Calculate the appropriate test statistic and interpret the resuit.b) Calculate the p-value and interpret the result.Click here to view page 1 of the standard normal table, Click here to view page 2 of the standard normal table.Hgi Hy-H2>0x, = 88o, = 24n, = 45x2 = 77O2 = 18n2 = 55a) The test statistic is 2.54.(Round to two decimal places as needed.)Determine the appropriate critical value(s).The critical value(s) is(are) 1.28.(Round to two decimal places as needed. Use a comma to separate answers as needed.)Since the test statisticfallsin the rejection region,rejectHn. There is sufficient evidence to conclude that the mean of population 1 is greater than the mean of population 2.b) The p-value is.(Round to three decimal places as needed.)

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Answered: Consider the hypothesis statement 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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Listed below are systolic blood pressure measurements (mm Hg) taken from the right and left arms of the same woman. Assume that the paired sample data is a simple random sample and that the differences have a distribution that is approximately normal. Use a 0.10 significance level to test for a difference between the measurements from the two arms. Identify the test statistic and p-value. Right arm Left arm 144 167 OA.T-3.07, p-value = 0.037 OB. T = -2.32, p-value = 0.081 OC. T = -1.93, p-value = 0.127 OD. T = -4.01, p-value = 0.016 133 161 132 179 138 145 136 144
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Please do not use/copy other's answer especially on chegg. Please answer this one using your own answer
The accompanying data table contains chest deceleration measurements (in g, whereg is the force of gravity) from samples of small, midsize, and large cars. Shown are the technology results for analysis of variance of this data table. Assume that a researcher plans to use a 0.05 significance level to test the claim that the different size categories have the same mean chest deceleration in the standard crash test. Complete parts (a) and (b) below. Click the icon to view the table of chest deceleration measurements A Click the icon to view the table of analysis of variance results. ..... a. What characteristic of the data specifically indicates that one-way analysis of variance should be used? O A. There are three samples of measurements. O B. The population means are approximately normal. O C. The measurements are categorized according to the one characteristic of size. O D. Nothing specifically indicates that one-way analysis of variance should be used. b. If the objective is to test…
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Listed below are systolic blood pressure measurements (mm Hg) taken from the right and left arms of the same woman. Assume that the paired sample data is a simple random sample and that the differences have a distribution that is approximately normal. Use a 0.05 significance level to test for a difference between the measurements from the two arms. What can be concluded? Right arm Left arm 150 142 120 131 167 160 179 156 In this example, Hd is the mean value of the differences d for the population of all pairs of data, where each individual difference d is defined as the measurement from the right arm minus the measurement from the left arm. What are the null and alternative hypotheses for the hypothesis test? O A. Ho: Hd = 0 H₁: Hd 0 Since the P-value is than the significance level, the null hypothesis. There sufficient evidence to support the claim of a difference in measurements between the two arms.
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