Recent content by MWR

Unfortunately, I am unable to view your picture. Also, I am not able to upload a .docx file either.

Have I completed this 2 x 3 factorial ANOVA chart correctly? A sample of ¬N = 36 is recruited to participate in a study about speech errors. The researchers believe that there is an interaction between whether the speaker is distracted (distracted or not distracted) and the difficulty of the...

Thank you, Jameson, for your help. I greatly appreciate it. Regarding the second part of the question, would we do another t-test to show the differences between pre-training and post-training?

So this would be my answer to number 1 (I did it by hand so the numbers are slightly different). Pre-training scores: 6, 5, 10, 17, 2, 12, 15, 6, 9 = 82/9 = 9.11 mean (6 – 9.11)^2 = 9.67 (5-9.11)^2 = 16.89 (10-9.11)^2 = 0.79 (17-9.11)^2 = 62.25 (2-9.11)^2 = 50.55 (12-9.11)^2 = 8.35 (15-9.11)^2...

Makes perfect sense, yes. So how do we tie this into answering number 1? Do we provide our answer relative to the p-value being greater than 0.05 indicating that, yes, this sample, or department, engages in more interpersonal behavior than the rest of the organization?

Also, I keep getting 4.91 as the SD with a variance of 24.11. - - - Updated - - - Now I get 0.16. Thus not making it significant. In that case, we reject the null hypothesis, correct?

So with that I get 1.54. And a two-sided probability of 0.2?

I believe it is asking if the pre-training sample average is different from the rest of the organization, which is 7. (9.11-7)/(1.74 x 2.65) = (2.11)/(4.61) = 0.46. I am a little confused as to how to use this information to determine the answer.

So the appropriate test is the t-test, or z-test. Here is what I have: x numbers represent pre-training. y numbers represent post-training. Sx = 82 Mx = 9.11 Sx^2 = 24.11 Sy = 66 My = 7.33 Sy^2 = 20.75 After calculating the square root of Sx^2 and Sy^2, I obtain t = 0.16, which is...

How do I go about analyzing this scenario? Consider the following: You’ve been hired by a company who is interested in finding a way to decrease incidents of offensive interpersonal behavior at work. The company has identified one department that seems to have a big problem with offensive...

Could someone double-check my answers to these derivative problems? -- f(x) = e^(x^3 + 2) = 3x^2 e^(x^3 + 2)-- f(x) = (e^-4x)/(x^3 + 7) = e^-4x [(-4x^3 - 3x^2 - 28)/(x^3 + 7)^2]-- f(x) = x^3 ln x = 3x^2 ln(x) + x^2-- f(x) = (ln^3 sqrt(x^2))/(x^3) = (ln)3-- f(x) = (^7 sqrt(x+8))/(5-6x)^8) =...

This makes perfect sense. Thanks so much for the clarification. :-)

For DT/DN, I get -1.7t + 45.4. So, does this mean I multiply that by zero, since that's the derivative of S(T)? Wouldn't I get zero again?

Thanks, Mark. Could you help me finish out my problem? I'm not exactly sure how. S(T) = (-0.03T^2+1.6T-13.65)^-1 N(T) = -0.85T^2 +45.4T - 547 So the derivative is zero for S(T) when T = 26.67. How do you then find DS/DN, S(T) with respect to N(T)? Do we use the chain rule? Exactly how...

Hi, I am working on a rate of change problem and appear stumped with my calculations. S(T) = (-0.03T^2 + 1.6T - 13.65)^-1 S'(T) = (-1)((-0.03T^2 + 1.6T - 13.65)^-2 (-0.03 (2T) + 1.6)) S'(T) = - (-0.06T + 1.6) / (-0.03T^2 + 1.6T - 13.65)^2 S'(T) = (0.06T-1.6) / (-0.03T^2+1.6T-13.65)^2 = 0...