Wright blog

  Report on drug and stress level by using R. Provide a full summary report on the result of ANOVA testing and what does it mean. More specifically, report the following:  Df, Sum, Sq Mean, Sq, F value, Pr(>F)  df: degrees of freedom 2 sum sq: sum of squares 82.11 sq mean: mean squares 41.06 fvalue: F statistic value 21.36 PR(>F) p value or significance    4.08x ^-5

  Define the relationship model between the  predictor  and the  response  variable: 1.2 Calculate the coefficients? its a scatter plot and coefficient are: (Intercept) x  19.205597 3.269107 An attempt was made and its too late to ask for help.

  A. Consider a population consisting of the following values, which represents the number of ice cream purchases during the academic year for each of the five housemates. 8, 14, 16, 10, 11 a. Compute the mean of this population.  11.8 b. Select a random sample of size 2 out of the five members.  8 and 14 c. Compute the mean and standard deviation of your sample.  11 d. Compare the Mean and Standard deviation of your sample to the entire population of this set (8,14, 16, 10, 11). Mean of sample is less then mean of population. standard deviation is smaller as well.  Suppose that the sample size  n  = 100 and the population proportion  p  = 0.95. Does the sample proportion  p  have approximately a normal distribution? yes Since p = .95, q = .05.  p * n = .95 * 100 = 95 q * n = .05 * 100 = 5 What is the smallest value of  n  for which the sampling distribution of  p  is approximately normal?   6 beca...

  1) Null = 70 and the alternate is != 70. At .05 level of significance we cannot reject the null since T -1.8  does not lie in the region z > 19.6 or z < -1.96 P value is .0719 which means there is a 7.19% chance that the value is true. If the standard deviation was specified at 1.75 we would reject the null. If the sample mean is 69 lbs and the standard deviation is 3.5 we would reject the null as t -2 lies in the rejection region.     2) If x ̅ = 85, σ = standard deviation = 8, and n=64, set up 95% confidence interval estimate of the population mean μ. a 95% confidence interval the population mean is 83.04 < u < 86.96   I think I did this right. without seeing another version of the program I don't know if I got the right output.