The deflection temperature under load for two different types of plastic pipe is beinginvestigated. Two random samples of 15 pipe specimens are tested, and the deflectiontemperatures observed are as follows (in oF):Туре 1: 206, 188, 205, 187, 194, 193, 207, 185, 189, 213, 192, 210, 194, 178, 205.Туре 2: 177, 197, 206, 201, 180, 176, 185, 200, 197, 192, 198, 188, 189, 203, 192.1. Determine the coefficient of dispersion of each type of plastic pipe.2. Interpret the results.Compare the SDUS of the 10th observation in each type. What conclusion can youmake?

Given the deflection temperatures of two random samples of 15 pipe specimens tested as Type 1…

Answered: The deflection temperature under load…

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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Suppose that the average salary of health systems professionals is normally distributed with µ=KES 40,000 and σ = KES 10,000. (20 Marks) What proportion of professionals will earn KES 24,800 or less? What proportion of professionals will earn KES 53,500 or more? What proportion of professionals will earn between KES 45,000 and KES 57,000? Calculate the 80th percentile. Calculate the 27th percentile
The deflection temperature under load for two different types of plastic pipe is being investigated. Two random samples of 15 pipe specimens are tested, and the deflection temperatures observed are as follows (in oF): Type 1: 206, 188, 205, 187, 194, 193, 207, 185, 189, 213, 192, 210, 194, 178, 205. Type 2: 177, 197, 206, 201, 180, 176, 185, 200, 197, 192, 198, 188, 189, 203, 192. 1. Determine the coefficient of dispersion of each type of plastic pipe. 2. Interpret the results. Compare the SDUS of the 10th observation in each type. What conclusion can you make?
3. At a land farm, a farmer desired to test the effect of a given fertilizer on soybean production. He chose 24 plots of land of which are equal surface area. He treated 12 plots with the fertilizer and the others were untreated. Otherwise the conditions were the same. The treated plots produced soybean with mean yield 5.1 bushels and a standard deviation of 0.36 bushels, while the untreated plots had mean yield 4.8 bushels and a standard deviation of 0.40 bushels. (a) Can we conclude that there is a significant improvement in soybean production because of the fertilizer if a significance level of 1% is used? (b) What is the P-value of the test?
A paint company has large outlet stores in two cities. Ten samples of the weekly sales (in tons) for the two cities are as follows. City A: n = 10, X10 = 148.5, S1 = = 4.3. City B: n = 10, Y 10 = 144.2, S2 = 3.8. At a = 0.05 level of significance, test if the weekly sales are different. The following ¨answers" have been proposed. (a) The null hypothesis is that the weekly sales are the same and the alternative is that the they are not the same. By using an F test, we reject the null hypothesis. (b) The null hypothesis is that the weekly sales are the same and the alternative is that the they are not the same. By using an F test, we do not reject the null hypothesis. (c) The null hypothesis is that the weekly sales are the same and the alternative is that the they are not the same. By using a i test, we do not reject the null hypothesis. (d) The null hypothesis is that the weekly sales are the same and the alternative is that the they are not the same. By using a i test, we reject the…
A sample of 150 matched pairs of banks was formed in a study comparing banks in Turkey and Germany. Each pair contained one bank from Turkey and one from Germany. The pairings were made in such a way that the two members were as similar as possible in regard to such factors as size and age. The ratio of total loans outstanding to total assets was calculated for each of the banks. For this ratio, the sample mean difference (Turkey - Germany) was 0.0412, and the sample standard deviation of the differences was 0.3039. Test, against a two-sided alternative, the null hypothesis that the two population means are equal. (Assume a = 0.01.) Click the icon to view the critical values of the Student's t distribution. Which of the following cor greater than tn-1, a/2 O B. Ho: Hx - Hy = 0 0< 1 - *rt :H O A. Ho: Hx -Hy =0 greater than - tn - 1, a 0# 서-저 : H O D. Ho: Hx-Hy = 0 OC. Ho: Hx-Hy #0 H: Hx-Hy =0 greater than t,-1. g 05 1 - Xt :'H less than -tn-1, a O F. Ho: Hx - Hy = 0 O E. Ho: Hx -Hy =0…
The manager of a fleet of automobiles is testing two brands of radial tires and assigns one tire of each brand at random of eight cars and runs the cars until the tires wear out. The data (in kilometers) follow. Is there any difference in mean life of these two brands of tires? Car Brand Brand 36,925 34.318 42,280 1 45.300 36,240 3 4 35.500 32,100 31,950 38,015 37.210 48,360 5 47,800 38,200 37,810 33.500 33,215 6. 7 8. 1. Is this samples independent or paired samples? Explain. 2. Write Ho and H1 3. Calculate Test statistics(T). 4. What is the critical value (T from tables)? 5. What is your conclusions if a = 0.05? 6. Estimate with 90% confidence the different in the means. Which brand would you prefer based on this calculation?
Two machines are used to fill plastic bottles with dishwashing detergent. The standard deviations of fill volume are known to be o, = 0.10 fluid ounces and o, = 0.15 fluid ounces for the two machines, respectively. Two random samples of n1 =12 bottles from machine 1 and n2 = 10 bottles from machine 2 are selected, and the sample mean fill volumes are I, = 30.87 fluid ounces and = 30.68 fluid ounces. Assume normality and use a = 0.05. Find the value of (a) zo and (b) P-value. а. none O b. zo = 3.24, P-value = 0.0036 z0 = 2.34, p-value = 0.0006 Oc. O d. zo = 3.42, p-value = 0.0003 %3D
Polyunsaturated fatty acids in the diet favorably affect several risk factors for cardiovascular disease. The principal dietary polyunsaturated fat is linoleic acid. Suppose we randomly selected twelve adults for our study. We took their blood pressure measurements (systolic blood pressure) at the start and recorded them for each adult. Then we placed required these adults to consume 23g/day of safflower oil which is high in linoleic acid for four weeks. At the end of the four weeks, we took another systolic blood pressure(SBP) measurement. The paired data is provided below: Subject 1 2 3 4 5 6 7 8 9 10 11 12 Baseline SBP 119.67 100.00 123.56 109.89 96.22 133.33 115.78 126.39 122.78 117.44 111.33 92.39 1-month SBP 117.33 98.78 123.83 107.67 95.67 128.89 113.22 121.56 126.33 110.39 107.00 93.06 Use the appropriate nonparametric test at alpha level of 0.05, to test whether the diet heavy in linoleic acid increases systolic blood pressure (SBP). Use Steps 0-5
Studies indicate that about 10 percent of polar bears weigh more than 1,000 pounds. A biologist studying the bearsthinks that percent might be too high. From a random sample of polar bears, the biologist found only 8 percent ofthe sample weighing over 1,000 pounds. Which of the following is the most appropriate method for the biologist’sstudy?(A) A one-sample -test for a sample proportion(B) A one-sample -test for a population proportion(C) A one-sample -test for a difference in population proportions(D) A two-sample -test for a difference in sample proportions(E) A two-sample -test for a difference in population proportions
(b) A rescarcher supccts that the fish in a particular elevated wetland have elevated mercury level. To confirm that, five striped bass in that wetland were caught and their tissues were tested for mercury. Then four striped bass in an unpolluted wetland were also caught and tested for the purpose of comparison. Table QI shows the fish mercury levels in mg/kg. Table QI shows the fish mercury levels in mg/kg. Sample from polluted wetland Sample from unpolluted wetland 0.580 0.382 0.711 0.276 0.571 0.570 0.666 0.366 0.598 Construct the 95% confidence interval for the difference in the population means. Then, test at the 5% level of significance, whether the researcher's suspicion is valid. Assume that the population variances are equal.
When does Two Related Samples occur?
Clothes dried outside smell better than those dried in a dryer because of a process called photolysis. Sunlight breaks down compounds that cause odor. A trained smeller is asked to smell a random sample of 20 towels. The smeller does not know it, but all of the 20 towels have been dried in the sun. The smeller is blindfolded and is not allowed to touch the towels but leans forward to smell the towel then rates the towel on a smell-scale of 1 to 10, where 1 is the an extremely unpleasant smell and 10 is a very good smell. A smell rating of 8 is typically attributed to a towel that is dried in the dryer. The mean smell rating for these 20 towels is approximately normal with a mean of 8.45 and the standard deviation is 0.887. A statistician performs a test of Ho μ = 8 versus H₁ μ> 8 where is the true mean smell rating for all towels dried in the sun. This test yields a P-value of 0.0176. What conclusion would you make at the significance level a = 0.05? Because the P-value of 0.0176 < α =…

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