From the boxplot we find the
following things
1. First Quartile, Q1 = 8.97653
2. Median = 10.5594
3. Third Quartile = 2.29049
4. Whiskers to: 6.38556, 13.2207
5. One outlier is present. The value is 31.9955
6. The distribution is positively skewed
Since the data contain an outlier, we can not use moment measures of location and spread. Because, those measures depend upon all the observations. Hence, the effect of only a single observation can be severe to these measures. So, we depend on quantile measures of location and spread. Because, these do not depend upon all the observations, more precisely, on extreme observations and hence are more robust in nature. Here we will use median as a measure of location and quartile deviation as a measure of spread. Median = 10.5594 and Quartile deviation = 1.145245.
5.5.18 Generate a sample of 30 from an N(10, 2) distribution and a sample of 1...
mathematical statistic
5.4.11 Generate a sample of 1000 from an N (3, 2) distribution. (a) Calculate Fx for this sample. 280 Section 5.4: Data Collection (b) Plot a density histogram based on these data using the intervals of length 1 over the range (-5, 10). (c) Plot a density histogram based on these data using the intervals of length 0.1 over the range (-5,10) d) Comment on the difference in the look of the histograms in parts (b) and (c)....
Generate a population of size N= 10000 from an exponential distribution with mean θ= 10. a. Generate 1000 samples of sizen= 200 from the population and plot the density of the sample means. b. Generate a single sample of sizen= 200. Resample with sizen= 200 with replacement fromthis single sample 1000 times. Plot the density of the resample means. c. Comment on the two densities you have plotted
Using R programming language, supply the code for: Generate a random sample of size 10, 000 from gamma distribution with scale parameter equal to 1 and shape parameter equal to 2, and form it into a 1000 x 10 matrix. Use the apply() function on this matrix to compute the means of the 1000 rows. Note that the resulting vector comprises the mean of 1000 random samples of size 10 from the above distribution. Examine the distribution of the sample...
Generate 20 samples of size n = 30 from your population. To do this: i. Generating a sample of just size n = 30. Calculate the sample mean, X¯, for this sample. Record this value somewhere in your spreadsheet (you will need it later). ii. Repeat the previous step 19 more times, so that you end up with a spreadsheet with 20 columns, each column has 30 randomly generated values from your population, and you have calculated a sample mean...
Generate a random sample of 20,000 x values (based on a sample of size 30) from a Normal distribution with a mean of 26 and a standard deviation of 5. Be sure to use 30116 as your seed. Find the approximate mean and standard deviation of the sampling distribution of the sample means (x) based on your simulation. No credit will be awarded for responses that do not include R code and output.
How to do the following in R: Write a function to generate a random sample of size n from the Gamma(α,1) distribution by the acceptance-rejection method. Generate a random sample of size 1000 from the Gamma(3,1) distribution. (Hint: you may use g(x) ∼ Exp(λ = 1/α) as your proposal distribution, where λ is the rate parameter. Figure out the appropriate constant c).
, Samples In 30) drawn from a uniform distribution la Minitab was used to generate the samples. es 300, b 500) Variables 15 Observations Variable TypeFormValues Missing Sample 1 Quantitative Sample 2 Quantitative Numeric Sample 3 Quantitative Numeric Sample 4 Quantitative Sample 5 ive Sample 6 Quantitative Sample 7 Quantitative Observations Sample 8 Quantitative Numeric Sample 9 Quantitative Sample 10 Quantitative Sample 11 Quantitative Sample 12 Quantitative Sample 13 Quantitative Sample 14 Quantitative Sample 15 Quantitative Numeric Numeric Variable Numeric...
Suppose you generate 4 numbers from the distribution X = N(µ = 3, σ = 4). If you average these together, you get a new number. If you repeated this process over and over, the averages would form their own distribution, Y , which is, surprisingly, also a normal distribution. Write an expression for Y using the letter X, and then find the mean and standard deviation of Y . (See problem R1 also.)
Let X1,.. ,X be a random sample from an N(p,02) distribution, where both and o are unknown. You will use the following facts for this ques- tion: Fact 1: The N(u,) pdf is J(rp. σ)- exp Fact 2 If X,x, is a random sample from a distribution with pdf of the form I-8, f( 0,0) = for specified fo, then we call and 82 > 0 location-scale parameters and (6,-0)/ is a pivotal quantity for 8, where 6, and ô,...
R codeing simulation
For n = 20, simulate a random
sample of size n from N(µ, 2 2 ), where µ = 1. Note that we just
use µ = 1 to generate the random sample. In the problem below, µ is
an unknown parameter. Plot in different figures: (a) the likelihood
function of µ, (b) the log likelihood function of µ. Mark in both
plots the maximum likelihood estimate of µ from the generated
random sample
(2) For n-20,...