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1) A school has 1500 students and they are going to vote as to whether they will completely convert the school completely off fossil fuels. How many students would you have to poll to be 95% confident of the outcome within +/- 3% of the vote? (25 points) 2) Download the file NOAAnew FOR FINAL to your computer from Canvas. We are going to use it for regression questions Upload it to Two Quantitative variables, choose delta temp as X and # disaster as Y (also other regression programs in statkey as needed, use the class 21 slides to help.) A) What is the model of number of disasters as a function of change in temperature? (10 points Full equation) B) What is a 95% confidence interval for the slope B (10 points) C) In this data, 1 data point at .61 degrees C was left out of the data, a. what is a 98% prediction interval for that point? (10 points) D) The assumptions going into the model are that: a. The mean value of the number of disasters at a given temperature rise is linear in temperature rise b. The actual number of disasters is approximately normally distributed around its mean with a constant variance. The true value that was left out was 16, Use the standard error of prediction (the stuff in the square root in the prediction interval) and the prediction from the regression line to calculate a t statistic. What is the Pvalue for the Null hypothesis that the point belongs to this line under these assumptions. (10 points) E) If I tell you that I picked this point on purpose as the 1 of 39 points furthest away from the line, does that change what you would say in E? How could you use Bonferonni reasoning to adjust the P value. (10 points) 3) For this exercise, load shanghaiearthquakes.csv into the appropriate statkey places for regression, x=time period 1, y =timeperiod 2 a. What are the slope and intercept for the relationship (10 points) b. Test if the slope =0 at the .1 level (all options use statkey) (15 points) 4)(30 points) : Below is a table of calls to a Poison center in Manhattan Kansas for exposures to disinfectants. A) Calculate the Row and column totals, and do usual Chisquare analysis to see if there is an association between year and age for the poisonings at the .05 level (15 points) Age 2018 2019 2020 0–5 76 68 81 6–19 18 17 29 20–59 27 28 45 ≥60 93 85 220 B) Are there any really unusual deviations from expected values.(5 points) C) Within each age group, 0-5, 6-59 (put two 6-19 and 20-59 together to get enough data), and 60 and up, run the Poisson difference tests we discussed to see if there are any interesting differences across the years. There will be 3 comparisons pre age group times 3 groups for 9 tests, Use FDR, not independent at the Q value of .1 to evaluate. (10 points) D) What about the approach in B means some of the P values are not independent? (5 points) 5) A method of estimating earth temperature based on local precipitation is to build models based on 100s of years of earth at several different constant green house gas level (So constant global temperature for 100s of years) , and compare the correlation of statistical relationships between predicted precipitation patterns in a region, and observed precipitation patterns for a region in a few year time period. (30 points) The following are independent p values of correlation of seasonal precipitation variation in northern California vs temperature in degrees C above 1950 base line for the years 2007 to 2012 Temperature rise Pvalue .05 .12 .1 .61 .15 .34 .2 .07 .25 .15 .3 .21 .35 .06 .4 .002 .45 .13 .5 .2 .55 .001 .6 .003 .65 .0001 .7 .1 .75 .3 .8 .07 .85 .001 .9 .15 .95 .36 1 .2 A) Use a False discovery rate of .05 to identify temperature rises in which the local precipitation rate over northern California is reasonably modeled. (20) B) Calculate the mean and median of these interesting temperatures. (10) 6) Below are the mean and standard deviation of Activity levels of 5 melanoma genes from 8 cell lines, Construct the analysis of variance table and test at the 0.01 level whether the means are the same (20 points) Gene number 196 200 247 4347 4384 mean -3.36 -.226 -1.69 2.03 1.92 sd 0.73 0.40 0.35 0.27 .47 n 8 8 8 8 8 After using the Anova above, determine at the 0.05 level, which pairs of mean activity levels differ significantly from 1 another at the 0.05 level. (20 points) 7) In a back yard vineyard in Napa valley with 10 grape vines in a row, if the weather works well (just right), rain in the spring and dry through summer, the yield for each vine is distributed roughly binomial with N=800, p=.71 . In a drought the yield is Binomial with N=800 and P=.6, while if the year is too wet, the yield of useful grapes per vine is N=700, P=.77. Under climate change the probability of a just right year is about .05, of a too wet year is .15, and a dry year is .8. On a just right year the wine can sell for 200 dollars/bottle, on a dry year the quality drops so it will sell for 100 dollars a bottle, on wet year it will sell for 25 dollars a bottle. (The yield for all 10 vines was more than 5600 grapes. (use the normal approximation to the binomial) Given this yield: What is the probability that you will be able to sell for 200 dollars a bottle? (50 points)