
Math 158-800 Assignment NAME Due 10/24/11 at 10:20 am. Soda
... Now consider a sample of size 26 (9 times smaller than 234). 11. Now what is the standard deviation of the sample proportion? __________ [nearest 0.001] 12. Using the Normal distribution, can one determine the probability that the percent of spam in a sample (of size 26) exceeds 35%? Explain. ...
... Now consider a sample of size 26 (9 times smaller than 234). 11. Now what is the standard deviation of the sample proportion? __________ [nearest 0.001] 12. Using the Normal distribution, can one determine the probability that the percent of spam in a sample (of size 26) exceeds 35%? Explain. ...
Lecture 11. Bayesian Regression
... Comment on the regression setup There are two subtle points regarding the Bayesian regression setup. First, a full Bayesian model includes a distribution for the independent variable X, p(X | ). Therefore, we have a joint likelihood p(X,Y | ,,) and joint prior p(,,). The fundamental assumpti ...
... Comment on the regression setup There are two subtle points regarding the Bayesian regression setup. First, a full Bayesian model includes a distribution for the independent variable X, p(X | ). Therefore, we have a joint likelihood p(X,Y | ,,) and joint prior p(,,). The fundamental assumpti ...
November 12
... – Step 3: Evaluate the threats on the basis of how likely they are to have an effect and plan to control for them. ...
... – Step 3: Evaluate the threats on the basis of how likely they are to have an effect and plan to control for them. ...
Two-proportion z
... Of 995 respondents, 37% reported they snored at least a few night a week. Split into two age categories, 26% of the 184 people under 30 snored, compared with 39% of 811 in the older group. Is this difference real (statistically significant) or due only to natural fluctuations. Use =0.05 Assumptions ...
... Of 995 respondents, 37% reported they snored at least a few night a week. Split into two age categories, 26% of the 184 people under 30 snored, compared with 39% of 811 in the older group. Is this difference real (statistically significant) or due only to natural fluctuations. Use =0.05 Assumptions ...
Chapter 4 Slides
... e) Find the probability at most 4 patients will survive. P(x≤4)=binomcdf(6,0.25,4) ...
... e) Find the probability at most 4 patients will survive. P(x≤4)=binomcdf(6,0.25,4) ...