Generate a matrix containing 5000 samples from the Weibull distribution with a = 1 and B = 2, each of size n = 100. Compute the 500 means using the apply function and store it in x2matmn. Call this variable X2.
Q: Determine the variance of Z.
A: Helpful formulas (i)VaX=a2VX(ii)VC=0,where c is the constant(iii)VaX+ bY=a2VX+ b2VY
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A: The sample information for the data from the two populations is given as:nsDrama1415050Comedy1612020
Q: A Hollywood studio believes that a movie that is considered a drama will draw a larger crowd on…
A: The sample infomration for the data from the two populations is given as:nsDrama1415050Comedy1612020
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- You sample 1500 dung beetles in a population for 2 alleles A1 and A2 and get the following data: A1A1 825 individuals A1A2 525 individuals A2A2 135 Individuals Is this population in Hardy Weinberg Euilibrium, use a chi square test to determine?A random selection of volunteers at a research institute have been exposed to a typical cold virus. After they started to have cold symptoms, 10 of them were given multivitamin tablets formulated to fight cold symptoms. The remaining 10 volunteers were given placebo tablets. For each individual, the length of time taken to recover from the cold is recorded. At the end of the experiment the following data are obtained. Days to recover from a cold Treated with multivitamin 2.6, 5.9, 5.3, 5.8, 8.7, 6.2, 7.6, 6.4, 8.0, 4.0 Treated with placebo 3.0, 5.1, 4.7, 4.2, 4.7, 5.6, 5.9, 4.7, 7.1, 4.9 It is known that the population standard deviation of recovery time from a cold is 1.8 days when treated with multivitamin tablets, and the population standard deviation of recovery time from a cold is 1.5 days when treated with placebo tablets. It is also known that both populations are approximately normally distributed. The researchers claim that the mean recovery time, l, of the patients treated…, A manufacturer of sunflower oil uses machines to distribute liquid ingredients into bottles that move along a filling line. The machine that distributes is working properly when 8 ounces are distributed. Suppose that the average amount distributed in a particular sample of 35 bottles is 7.91 ounces. Assume that the variance of the process is known to be 0.03 ounces squared. Is there evidence that the machine should be stopped (when not 8 ounces are distributed), and production wait for repairs? The lost production from a shutdown is potentially so great that management feels that the level of confidence in the analysis should be 99%.
- Suppose that a study was conducted to determine if there is a significant difference in the average temperature of individuals before and after COVID-19 boosters were given. The temperatures of a sample of individuals were obtained before the boosters were given. The following day, their temperatures were again measured. The table below shows the data in degrees Celsius. (10 points) Temperature before Temperature after one day. Determine If there is a significant difference in the average temperature of Individuals who were given boosters at 5% level of significance. Assume that the differences in the temperatures follow the normal distribution.A chi-square test for goodness of fit is used to examine the distribution of individuals across five categories, and a chi-square test for independence is used to examine the distribution of individuals across six categories in a 2 X 4 matrix of categories. Which test has the larger value for df? Select one: a. The test for independence has the larger value. b. The goodness of fit has the larger value. c. The df value depends on the sizes of the samples that are used. d. Both tests have the same df value.A study was conducted to examine the difference between the average cholesterol levels in the blood of men and females. It can be assumed that cholesterol levels follow a normal distribution in both populations and that the distribution in the populations is similar (e. Assume that cholesterol levels follow a normal distribution in both populations with similar variances). An equal number of measurements were made on men and women. The results were entered into R and the R code and output can be seen below: t.test(kol$mael-kol$kyn,var.equal = T) %3D Two Sample t-test data: kol$mael by kol$kyn t = 5.7731, df = 18, p-value alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: 15.28775 32.78050 = 0.000018 sample estimates: mean in group Men mean in group women 219.2115 195.1774 1. Which statement is best based on the results of the study? Use a=0.05. a. The P-value is lower than a and therefore can be rejected HO b. The P-value is higher than a…
- 1.The least squares estimate of β0 is i. 2.3176 ii. 0.7176 iii. 0.8824 iv. 1.8990 2. The least squares estimate of β1 is i. 0.4773 ii. 0.2990 iii. 0.2061 iv. 0.9265 3. The linear correlation coefficient r is i. 0.4122 ii. 0.1700 iii. 0.2990 iv. 0.5683Suppose that a random man's weight from a certain population is M~ Normal (μ = 200, o²49) and a random woman's weight is W ~ Normal( 140, o² = 25). If M and W are independent, what is the variance of the difference between a man's weight and a woman's weight, i.e. Var (M-W)? (a) 340 (b) 24 (c) 140 (d) 200 (e) None of the aboveAssume a population with 5 elements: X1-0, X2-1, X3-2. X4-3,X5-4. Calculate mean and variance of X for random samples of size 3 taken with replacement Provide a numerical answer only.
- proveA study of fox rabies in a country gave the following information about different regions and the occurrence of rabies in each region. A random sample of n1 = 16 locations in region I gave the following information about the number of cases of fox rabies near that location. x1: Region I Data 1 9 9 9 7 8 8 1 3 3 3 2 5 1 4 6 A second random sample of n2 = 15 locations in region II gave the following information about the number of cases of fox rabies near that location. x2: Region II Data 1 1 5 1 6 8 5 4 4 4 2 2 5 6 9 What is the value of the sample test statistic? (Test the difference μ1 − μ2. Do not use rounded values. Round your final answer to three decimal places.)The probability that a trainee will remain with a company is 0.6. the probability that am employee earnsbmore than k10,000 per month is 0.5. the probability that an employee who is a trainee remained with the company or who earns more than k10,000 per month is 0.7. what is the probability that an employee earns more than k10,000 per month given that he is a trainee who stayed with the company
