9. In an experiment conducted online at the University of Mississippi, study participants are asked toreact to a stimulus. In one experiment, the participant must press a key on seeing a blue screen.Reaction time (in seconds) to press the key is measured. The same person is then asked to press akey on seeing a red screen, again with reaction time measured. The results from six randomlysampled study participants are shown below. Construct a 99% confidence interval about thepopulation mean difference and interpret your results.Participant36.Blue0.5820.4810.8410.2670.6850.450Red0.4080.4070.5420.4020.4560.533

9. Obtain the 99% confidence interval for the differences between the means. The 99% confidence…

Answered: 9. In an experiment conducted online at…

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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. An educator was trying to determine the effect of remote learning on middle school aged children. They randomly selected 250 students who spent a full year with in person learning and 190 of them showed similar or improved scores on a set of metrics. They also randomly selected 350 student who spent the year remotely learning and 252 of them showed similar or improved scores.a. Construct a 95% confidence interval for the difference of proportions? b. Based on the confidence interval can you conclude there is adifference between the two groups?
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An article reported the results of a peer tutoring program to help children learn to read. In the experiment, the children were randomly divided into two groups: the experimental group received peer tutoring along with regular instruction, and the control group received regular instruction with no peer tutoring. There were n1 = n2 = 30 children in each group. The Gates-MacGintie Reading Test was given to both groups before instruction began. For the experimental group, the mean score on the vocabulary portion of the test was x1 = 344.5, with sample standard deviation s1 = 49.7. For the control group, the mean score on the same test was x2 = 353.8, with sample standard deviation s2 = 50.1. Use a 5% level of significance to test the hypothesis that there was no difference in the vocabulary scores of the two groups before the instruction began. (a) What is the level of significance?State the null and alternate hypotheses. H0: ?1 = ?2; H1: ?1 ≠ ?2H0: ?1 = ?2; H1: ?1 < ?2 H0: ?1 =…
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