In a large design class, the scores on the "PDI" project were normally distributed with u = 100. Theprofessor examined a sample of n= 56 students who used extra resources instead of only one resource likethe other students in the class He found that the mean for students who used extra resources was M=104.125. He used a one-sample t test to analyze the results, and part of the SPSS output is shown below.%3DOne-Sample StatisticsStd ErrorMeanN.MeanStd Deviationpdi56104.125012.584351 68165One-Sample TestTest Value 10095% Confidence Interval of theDifferenceMeanDifferencedfSig (2-tailed)55LowerUpper7.4951pdi24530174.125007549A.What are the professor's null and altemative hypotheses?B.2. reject H,b. do not reject H,What is the professor's decision?C.resources. Identify this inform2tion and explain how you used it.You can use certain information from the table to answer Part B without referring to any other

Given: Output of the One samplt test

Answered: In a large design class, the scores on…

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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To calculate a z-score to compare against a critical value in the single sample z test, we divide the observed mean differences(sample mean-population mean) by what value?
An investigation of the relationship between traffic flow (measured in cars per day) and lead content (measured in micrograms per gram of dry weight) in the bark of trees near the highway yielded the following summary statistics: Sample Mean Sample Standard Deviation Lead Content (micrograms per gram dry weight) y¯=y¯= 680 sysy = 240 Traffic Flow (cars / day) x¯=x¯= 1750 sxsx = 800 The correlation between lead content and traffic flow was found to be r = 0.6 and a scatterplot showed the form to be linear. The least squares regression line for using X to predict Y was found to be y^=365+0.18xy^=365+0.18x. Which one of the following is a proper interpretation of the slope? Group of answer choices For each 1 car per day increase in traffic flow, we predict an corresponding increase of 0.18 in lead content of surrounding trees For each 0.6 car per day increase in traffic flow, we predict an corresponding increase of 365.18 in lead content of surrounding trees For each 1…
Given that the researcher wants to be sure the mean difference in the change in hemoglobin levels has a margin of error no more than 0.2 g/dL and the difference in the stand deviation between the mean hemoglobin levels for the two forms of iron supplements used to treat anemia is 1.2 g/dL, what is the desired number of recruits for the study, assuming the individuals will participate in a crossover trial, if the study’s attrition rate is expected to be .25? Group of answer choices n = 139 n = 554 n = 185 n = 55,320
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An investigation of the relationship between traffic flow (measured in cars per day) and lead content (measured in micrograms per gram of dry weight) in the bark of trees near the highway yielded the following summary statistics: Sample Mean Sample Standard Deviation Lead Content (micrograms per gram dry weight) y¯=y¯= 680 sysy = 240 Traffic Flow (cars / day) x¯=x¯= 1750 sxsx = 800 The correlation between lead content and traffic flow was found to be r = 0.6 and a scatterplot showed the form to be linear. Since Traffic flow is the X-variable, its mean and standard deviation have been labelled x¯ and sxx¯ and sx. Since Lead content is the Y-variable, its mean and standard deviation have been labelled y¯ and syy¯ and sy . Determine slope and y-intercept of the least squares regression line for using X to predict Y. Group of answer choices Slope y-intercept
Given that the researcher wants to be sure the mean difference in the change in hemoglobin levels has a margin of error no more than 0.2 g/dL and the difference in the stand deviation between the mean hemoglobin levels for the two forms of iron supplements used to treat anemia is 1.2 g/dL, what is the desired number of recruits for the study, assuming the individuals will participate in a crossover trial, if the study’s attrition rate is expected to be .25? Group of answer choices n= 139 n= 186 n= 554 n= 55,320
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