Two different formulas of an oxygenated motor fuel are being tested to study their road octane numbers. The variance of road octanenumber for formula 1 is o? = 1.5, and for formula 2 it is o,the mean octane numbers observed are = 89.5 fluid ounces and I2 = 93.1 fluid ounces. Assume normality.= 1.2. Two random samples of size n1 =15and n2 = 20 are tested, and(a) Test the hypotheses Ho : H1H2 versus H1 : µ1 < µ2 using a = 0.05. Round your answer to three decimal places (e.g. 98.765).%|Zov Ho.(b) Calculate a 95% two-sided confidence interval on the mean difference road octane number, x1 – X2. Round your answer to threedecimal places (e.g. 98.765).

In order to determine whether the mean octane number for formula 2 is significantly greater than…

Answered: Two different formulas of an oxygenated…

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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The given quantities below were calculated from a sample data set. Στο 338 I Lu=308 Ση2 = 4678 2 y = 4236 Determine the least squares regression line. Round values to four decimal places, if necessary. Σry = 3595 n = 30 Determine the correlation coefficient. Round the solution to four decimal places, if necessary.
Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 7.97.9 parts/million (ppm). A researcher believes that the current ozone level is at an excess level. The mean of 2424 samples is 8.18.1 ppm with a variance of 0.250.25. Does the data support the claim at the 0.10.1 level? Assume the population distribution is approximately normal. Step 3 of 5: Specify if the test is one-tailed or two-tailed.
Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 4.4 parts/million (ppm). A researcher believes that the current ozone level is not at the normal level. The mean of 1090 samples is 4.3 ppm. Assume the variance is known to be 1.44. Does the data support the claim at the 0.05 level? Step 1 of 5: Enter the hypotheses: Step 2 of 5: Enter the value of the z test statistic. Round your answer to two decimal places. Step 3 of 5: Specify if the test is one-tailed or two-tailed. Step 4 of 5: Enter the decision rule. Step 5 of 5: Enter the conclusion.
ol 106. Use the two scenarios presented in Figure 7.12 to answer the following question about the two sources of variability in ANOVA. 3.5 3.0 2.5 2.0 1.5 (i) HA: P₁ P2 ‡ P3 (iii) HA not all proportions are equal to this po 1.0 Figure 7.12: Within vs Between Variability in ANOVA onnisth Scenario A L (d) Recall that, O 2 H 3 2.05 2.10 2.15 2.20 2.25 2.30 2.35 (ii) Ho: M₁ M₂ = μ3 sans are (iv) HA: not all means are equal 720 qeaoluet alt sot mol 4 mobor 2924 (6) 2 3 4 gob za aim onbala (1) di batt (9) w-qodt fonqtotal bac o2 (6) (Troy ote12. (0) (a) Which scenario provides stronger evidence that the difference in factor level means represents true differences in the population means? Explain why in your own words. Scenario B 1 = stogol 05 O 1 oftaltete fest out of sulav gris I (b) Which scenario exhibits more variability within each factor level? Is this measured by SSError or SSFactor? Agulow (c) Which scenario exhibits more variability between each factor level? Is this measured by…
thill An experiment compares two different processes for producing steel plate. The measurements represent the thickness of the plate. The samples gave m = 8, X = 6.701, S1 = 0.108 n = 6, Y = 6.841, S2 = 0.155 %3D %3D Assume that the two populations are normal with equal variances, test Ho: H1 - H2 = 0 against H1 : µ1 - µ2#0 at a 0.05 significance level. 5.
An engineer wants to know if producing metal bars using a new experimental treatment rather than the conventional treatment makes a difference in the tensile strength of the bars (the ability to resist tearing when pulled lengthwise). At α=0.10, answer parts (a) through (e). Assume the population variances are equal and the samples are random. If convenient, use technology to solve the problem. Treatment Tensile strengths (newtons per square millimeter) Experimental 449 354 450 360 433 388 400 Conventional 370 376 374 424 378 450 438 404 352 376 (a) Identify the claim and state H0 and Ha. The claim is "The new treatment ▼ makes a difference does not make a difference in the tensile strength of the bars." What are H0 and Ha? The null hypothesis, H0, is ▼ mu 1 equals mu 2μ1=μ2 mu 1 less than or equals mu 2μ1≤μ2 mu 1 greater than or equals mu 2μ1≥μ2 . The alternative hypothesis, Ha,…
The length (in mm) of a spring subjected to a force FNewtons is of the form L = a + BF+ ɛ, where ɛ is assumed normally distributed with mean 0, variance o2. Given the sample statisticsn=D13, E F;= 138, E L; = 577, SEL = -43, SEF= 1735, estimate the expected length of the spring when a force of 11.2 Newtons is applied. (Give your answer correct to 2 decimal places.) Answer: Ti Check 江一 prime viden Type here to search 19 hp ins prt fu f12 f8 f9 f10 f7 f5 f6 『米 " JOI JDI & 6 7 8 5 96 %24 3
The average height X and weight Y of males in population have a bivariate normal distribution with means µX 1.80 m., µy 90.0 kgs and standard deviations Ox = 0.30 m., oy = 15.3 kgs respectively. The correlation coefficient between X and Y is p= 0.80.

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