Write and execute R code that accomplishes the following tasks: Part A: Use the rnorm() function to generate a sample of size N=1,000 from the normal distribution with meanu = 6 and standard deviation 5 = 0.5 . Record the sample as a vector v. Part B: Plot a histogram for the sample v.
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Given:
Sample size = 1000
Population mean = 6
Population standard deviation = 0.5
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- The Table 1 shows the LDL cholesterol level in a sample of ten heart patients. Table 1: Total LDL cholesterol level 132 162 133 145 148 139 147 160 150 153 Calculate: (a) measures of central tendency. (b) range and semi inter-quartile range. (c) variance and determine the shape of skewness.1. Was a dependent or independent t-test used to analyze the data? How do you know?For what types of data do we use chi-square tests?
- The weight of cereal boxes (y) follows a normal distributed with a variance of 0.1 squared ounces. If the average is 15.8 ounces, then P(y < 16) = If the average is 16.5 ounces, then P(y > 16.2)J In-shop othesis test for the difference between two population proportions 02562 2600368.qx3zqy7 Jump to level 1 The manager of a coffee shop is trying to determine if drive-through customers buy fewer items than customers who come into the shop. The manager randomly samples both populations and calculates the proportion of customers in each that buy three or more items. The results of the data collected are shown below. What are the population parameters? Y Pick Drive-through Successes 27 Successes What is the level of significance? Ex: 0.12 Observations 76 Observations 41 119 0.345 p-hat_1 0.355 p-hat_2 What is the null hypothesis Ho? Pick Confidence Level 0.1429 What is the alternative hypothesis Ha? Pick p-value Should Ho be rejected or does Ho fail to be rejected? Pick What conclusion can be drawn from the data? Pick evidence exists to support the claim that drive-through customers buy fewer items than in-shop customers. 2 p WThe durability of the hacksaw is measured in a quality control laboratory. A standard saw needs 2500 cuts. The cutting average of 28 saws selected randomly from a company was found to be 2600 and the variance as 17500. I wonder if the company's production is above the standard? (α = 0.01)
- Actuaries use various parameters when evaluating the cost of a life insurance policy. The variance of the life spans of a population is one of the parameters used for the evaluation. Each year, the actuaries at a particular insurance company randomly sample 30 people who died during the year (with the samples chosen independently from year to year) to see whether the variance of life spans has changed. The life span data from this year and from last year are summarized below. Current Last Year Year x1 = 75.8 | x,= 76.2 si= 47.61 s3= = 92.16 (The first row gives the sample means and the second row gives the sample variances.) Assume that life spans are approximately normally distributed for each of the populations of people who died this year and people who died last year. Can we conclude, at the 0.05 significance level, that the variance of the life span for the current year, of, differs from the variance of the life span for last year, o,? Perform a two-tailed test. Then complete the…Let X₁, X2,..., be a sequence of independent and identically distributed random variables, each with a mean value x and variance ok, both of which are assumed to be finite. Let N be a discrete random variable independent of X₁, X2,..., and assuming values in the set {0, 1, 2, ...}, with mean value N and variance ok (both are finite). Form the random-compound sum Y=1 Xk, with the convention Y = 0 whenever N = 0.Psi is a measure of compressive strength, or the ability of the material to carry loads and handle compression. The desired concrete psi rating used for sidewalks and residential driveways ranges from 2500psi to 3000psi obtained from mixing cement, stone, and sand in different ratios but with the same amount of water. The summary of the psi's of three (3) such concrete mixes made by three (3) different civil engineering students is given as follows: Mean vector: The variance-covariance matrix: Concrete mix Mean Concrete mix 1 Concrete mix 2 Concrete mix 3 Concrete mix 1 | Concrete mix 2 2700 18 12 2 3000 12 16 3 2400 Concrete mix 3 16 25 Let the random variable Y be the vector of the psi's of concrete mixes obtained by the students, i.e. Y1, denotes the psi of concrete mix made by student 1, Y2 is the psi of concrete mix made by student 2 and Y3 is the psi of concrete mix psi made by student 3. (a) Find (i) the multivariate probability distribution function (pdf) of Y. (ii) the…
- Total fat content (including saturated fat and trans fats) is a key characteristic of cheese. Assume that for a given type of cheese, total fat content (10-1 g/g of cheese) follows a normal distribution with a population mean of 2.85 (unit) and a population variance of 0.01 (unit2). Such an assumption is reasonable because fat content varies inside cheese and the “errors” around the mean total fat content (i.e. the value appearing on the etiquette) are distributed like a bell shape. a) What is the probability that the total fat content of this cheese exceeds 3 (unit)? b) What is the probability that the total fat content varies from 2.8 to 3 (unit)?Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 4.6 parts/million (ppm). A researcher believes that the current ozone level is at an excess level. The mean of 14 samples is 4.9 ppm with a variance of 1.2 Does the data support the claim at the 0.01 level? Assume the population distribution is approximately normal. Step 2 of 5 : Find the value of the test statistic. Round your answer to three decimal places.A manufacturing company produces bearings. One line of bearings is specified to be 1.64 centimeters (cm) in diameter. A major customer requires that the variance of the bearings be no more than .001 cm². The producer is required to test the bearings before they are shipped, and so the diameters of 16 bearings are measured with a precise instrument, resulting in the following values. Assume bearing diameters are normally distributed. Use the data and a = 0.01 to test to determine whether the population of these bearings is to be rejected because of too high a variance. 1.68 1.62 1.63 1.70 1.66 1.63 1.65 1.71 1.64 1.69 1.57 1.64 1.59 1.66 1.63 1.65 Appendix A Statistical Tables (Round your answer to 2 decimal places, e.g. 15.25.) The value of the test statistic is and we reject the null hypothesis ⇒