ENGR 320 Probability and Statistics 3Distributions and Conf

ENGR 320 - Probability and Statistics 3-Distributions and Confidence Intervals Open Book, Open Notes Exam Fall 2015 November 24, 2015 Exam 1. Ninety percent of bricks from Acme Inc. meet local building codes. Eight-five percent of bricks from Unbreakable Inc. meet the same codes. If 400 bricks are purchased fronm each company, for evaluation, what is the probability that more than 725 bricks meet code? (50 points) 2. The top speeds of ten cheetahs at a research facility was compared on a day with an average morning temperature of 75 degrees Farenheit. The tests were done again two days later when the average morning temperature was 55 degrees Farenheit. Find a 95% confidence interval for the difference in average top cheetah speeds between the warm and cool day. (50 points) Cheetah Warm Day Cool Day Numbr Top Speed Top Speed Number (MPH) 64.5 70.3 69.9 (MPH) 63.6 68.9 68.2 67.6 69.7 63.5 676 60.8 62.561.5 67.4 66.0 65.9 68.9 67.9 68.3 67.1 868 0 10 68.967

Solution

1.

A- Bricks from Acme Inc. meet local building code

B- Bricks from unbreakable Inc. meet the same code

P(A) = 0.90 and P(B) = 0.85

Number of bricks from Acme Inc. meet local building code out of 400 bricks = 0.90 * 400 = 360

Number of bricks from unbreakable Inc. meet the same code out of 400 bricks = 0.85*400 = 340

Total number of bricks having same code = 360+340 = 700

the probability that more than 725 bricks meet code = 700/725 = 0.96.

2. To calculate the confidence interval of the difference of mean of the given sample is calculated using R- software. here we apply the t.test for testing and find the confidence interval at the 95 % confidence interval we calculate the difference of mean. and resulat is given below;

> x1<-c(64.5,70.3,69.9,69.7,63.5,62.5,68.3,67.1,66.8,68.9)
> x2<-c(63.6,68.9,68.2,67.6,60.8,61.5,67.4,66.0,65.9,67.9)
> x<-c(x1,x2)
> y<-c(rep(0,10),rep(1,10))
> t.test(x~y,alternative = \"two.sided\",conf.level = 0.95)

        Welch Two Sample t-test

data: x by y
t = 1.0815, df = 17.989, p-value = 0.2938
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-1.291473 4.031473
sample estimates:
mean in group 0 mean in group 1
          67.15           65.78

Here p- value is much greater than the 0.05 its means that there is no difference between the average cheetah speed during the wrm and cold days.