blugw6. Suppose that the following data are drawn from the population distribution in which μ = 7.0 and o = 2.5.Data8354 6 10 7 5 8(1) Compute the mean and the standard deviation of the sample distribution.(2) Compute the standard error of the mean (Notice that the population standard deviation is known).(3) Compute the 95% confidence interval.(4) Draw a curve (normal distribution) and indicate the 95% confidence interval.(5) Does the 95Cl include the population mean? What does the result mean

(6)Answer:- Given , Population mean μ= 7.0 and standard deviation σ = 2.5 Given sample data is,…

Answered: 6. Suppose that the following data are…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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24. What is the sample standard deviation from the Range Frequency Table from Excel?a. 15.00b. 16.61c. 15.65d. 14.38e. 14.7025. Why are these values different?a. Because Excel makes errorsb. Because when going from exact values to a range table, the values become approximatec. Because the spreadsheet calculates the values in different waysd. Because the variance is calculated differently
D A small amount of the trace element selenium, 50-200 micrograms (ug) per day, is considered essential to good health. Suppose that random samples of 50 adults were selected from two regions of the United States and that each person's daily selenium intake was recorded. The means and standard deviations of the selenium daily intakes for the two groups are shown in the table. Sample Mean (ug) Sample Standard Deviation (ug) Find a 95% confidence interval for the difference (₁-₂) in the mean selenium intakes for the two regions. (Round your answers to three decimal places.) 49 Region Phase One-Tw_docx 1 167.4 22.8 ug to Interpret this interval, O The probability that the difference in the average daily selenium intake for the populations in Region 1 and Region 2 is exactly in the middle of the Interval is 95%. A 2 140.5 17.3 O The probability that the difference in the average daily selenium intake for the populations in Region 1 and Region 2 is within the interval is 95%. We are 95%…
2. A pet association claims that the mean annual costs of food for dogs and cats are the same. The results for samples of the two types of pets are shown in the following table. On wished to test the claim. Assume both populations follow normal distributions Dogs Cats 21 - 16 T1= 239 n2 = 18 Iz= 203 S1 = 32 82 =28 with equal variances. Which of the following formula should be used to calculate the standardized test statistic? (a) Z = (21-32)- Do lu A n2 = (E1-2)-Do (b) Z V n1+2 (c) T (1-2)-Do (21-1)a7+(n2-1)s I1 11+11タ-3 V 1'n2 (d) T = d- Do sp/Vn (e) Z
4. QUESTION A precision instrument is guaranteed to be accurate within 2 units. A sample of four instrument readings on the same object yielded the measurements 353 351 351 355. 4.1 Construct a 95% confidence interval for the variance of the instrument's precision readings 4.2 It is believed that the variation in the instrument's precision readings is equal to 0.5. From your confidence interval, is there sufficient evidence to conclude that the variance differs from 0.5? Explain your answer
1. The strength of steel wire made by an existing process is normally distributed with a mean of 1550 and a standard deviation of 170. A batch of wire is made by a new process, and a random sample consisting of 25 measurements gives an average strength of 1316. Assume that the standard deviation does not change. Is there evidence at the 1% level of significance that the new process gives a larger mean strength than the old? To answer this question, the null hypothesis could be stated as A. μ= 1550 psi B. μ> 1550 psi C.µ + 1550 psi D. μ< 1550 psi
Consider the sample data 2, 4, and 6 with population standard deviation of 1.633. Compute the following using the formulas in Central Limit Theorem. N=3 n=2 sample mean sample standard deviation Choose... sample variance Choose... + Please answer all parts of the question.
10) A county is considering raising the speed limit on a road because they claim that the mean speed of vehicles is greater than 40 mi/hr. A random sample of 15 vehicles has a mean speed of 43 mil/hr and a standard deviation of 4.4 mi/hr. Assume that speeds follow a normal distribution. What statistical test should be used to test the claim? (pick one of the options) 1. t-test 2. 2.73 3. 15 4. 10.9 5. 14 6. z-test 7. 2.52 8. 2.64 Calculate the value of standardized test statistic. (pick one of the options) 1. t-test 2. 2.73 3. 15 4. 10.9 5. 14 6. z-test 7. 2.52 8. 2.64 What is the degree of the freedom for the distribution of the test statistics. (pick one of the options) 1. t-test 2. 2.73 3. 15 4. 10.9 5. 14 6. z-test 7. 2.52 8. 2.64
Find the variance and standard deviation
5. The standard deviation of a sample of the costs of 15 automobile repairs at a local garage was $120.82. Assuming that the variable is normally distributed, find the critical values, Xleft and right, that could be used to determine the 90% confidence interval of the true variance and standard deviation.
Suppose scores on exams in statistics are normally distributed with an unknown population mean and a population standard deviation of three points. A random sample of 36 scores is taken and gives a sample mean score of 68. Find a 90% confidence interval estimate for all the mean score on all exams or the true (population) mean of statistics exam scores. Use the format xx.xxx

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