At last, envision that the scientist doesn't have careful rest and response time information. That is, ceaseless information are not accessible for the two factors. All things being equal, the individual in question possibly knows whether members had moderate or speedy response times and satisfactory or insufficient measures of rest.

Enter the accompanying dataset into SPSS, at that point lead the proper chi square investigation to respond to the inquiries and decide whether rest time is identified with response time. Make certain to completely clarify the reasoning for your response to each address, including proof from the content and Learning Activity assets.

Moderate Reaction Time

Fast Reaction Time

Satisfactory Sleep

9

5

Insufficient Sleep

6

10

Review the four sizes of estimation (i.e., ostensible, ordinal, stretch, proportion). Clarify what size of estimation is utilized to gauge rest time in this model. How would you know?

Clarify what size of estimation is utilized to gauge response time. Clarify how you know.

State whether this situation requires a decency of fit test or a trial of autonomy. Clarify your answer.

Prior to processing the chi square, express the invalid theory and elective speculation in words (not recipes).

Distinguish the acquired ?2 utilizing SPSS and report it in your answer record.

Express the levels of opportunity and clarify how you determined it by hand.

Recognize the p esteem utilizing SPSS and report it in your answer archive.

Clarify whether you ought to hold or reject the invalid speculation and why.

Are the outcomes genuinely critical? How would you know?

Clarify what you can decide about the connection between rest time and response time.

Question 2 With 3 data points given by the column vectors 1 point xdat and ydat in MATLAB, the commands >> A = [ones(size(xdat)) xdat xdat.^2]; > >u = A \\ ydat produce the output u = 4.897 6.175 -57.32 What does this compute? Least squares approximation y(x) = 4.897 + 6.175x - 57.32x2 Least squares approximation y(x) = -57.32 + 6.175x + 4.897x2 Polynomial interpolation y(x) = 4.897 + 6.175x - 57.32x2 Polynomial interpolation y(x) = -57.32 + 6.175x + 4.897x2With 12 data points given by the column vectors xdat and ydat in MATLAB, the commands > > A = [ones(size(xdat)) xdat xdat.^2 xdat.^3 xdat. ^4]; > > u = A \\ ydat produce the output u = 0.08752 5.540 -47.59 4.148 -639.9 What does this compute? Least squares approximation y(x) = -639.9 + 4.148x - 47.59x2 + 5.540x3.08752x4 Least squares approximation y(x) = 0.08752 + 5.540x - 47.59x2 + 4.148x3 - 639.9x4 Polynomial interpolation y(x) = 0.08752 + 5.540x - 47.59x2 + 4.148x3 - 639.9x4 Polynomial interpolation y(x) = -639.9 + 4.148x - 47.59x2 + 5.540x3.08752x41. Given the mction at} = e' and nodes In t [1.1, 11 = [1.3, and 1': = [1.5, use them to construct interpolation polynomials of degree one [it = l] and two [rt = 2] to epproetimete 3' [I12]? and nd the actual error [or each approximation. Among the nodes In = Ill. 11 3:; = 11.3. and 12 = [1.5, how many different interpolation polymomiels of degree one {r1 = 1} can be mustructed for approximating INA}? [Do not use any node.) 2. Use the error formula. given in Theorem 13 in the textbook [or Theorem 4.1.3 in the lecture note} to nd an error bound for each approxiomtion obtained in Problem 1. Theorem 4.1.3 Suppose that e=ru