Two statistically independent random have Variances of Variables X and new ran- 54 dom T है। 2 and o² =25. are 9 variables Two defined by U= 3x +4Y V = 5X-2Y
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- To carry out the lack or lack of fit test, it is necessary to partition the sum of squares due to random error in two: a. One due to the fitted model and one due to the pure error b. One due to lack of fit and one due to the fitted model c. One due to the total variation of the experiment and another due to the estimated model d. One due to lack of fit and one due to pure error Explain clearly each option pleaseA study of blood samples of 64 men shows a mean count of 5million RBC per cubic ml., with a variance of 0.06. A tabulation of64 women from the same population group show a mean countof 4.5 million, with a variance of 0.13. What is the probability offinding a woman with a count of less than 5 million?A research study was conducted about gender differences in 'sexting'. The researcher believed that the proportion of girls involved in 'sexting' is different from the proportion of boys involved. The data collected in the spring of among a random sample of middle and high school students in a large school district in is summarized next: out of 380 females 43 sent 'sexts' and out of 410 males 64 sent 'sexts'. Is the proportion of girls sending sexts different from the proportion of boys 'sexting'? Choose the appropriate procedure. Procedure: Assumptions: (select everything that applies) Population standard deviations are known Normal populations Sample sizes are both greater than 30 Population standard deviation are unknown but assumed equal Paired samples Population standard deviation are unknown Simple random samples The number of positive and negative responses are both greater than 10 for both samples Independent samples
- Don't know how to solveResearches are interested in whether or not the average fuel economy for compact cars is thesame as sedan cars. A sample of 36 compact cars and 25 sedan cars is taken.The sample of compacts returns a mean of ̄X = 38 miles per gallon and sample variance of s^2 = 25.The sample of sedans returns a sample mean ̄Y = 36 with a sample variance of s^2 = 49.Let μ 1 be the population mean of compacts and μ 2 for sedans. a. Create a 95% confidence interval for μ 1 .b. Now create a 95% confidence interval for μ 1 μ 2 .c. Say the interval from part b. contains 0. What can you say about the difference in the averagefuel economy of compact cars versus economy carsIf I was calculating the variance of this list of numbers (attached), do I need to include the "no home games" as a 0 and base the calculation on 14 entries? Or do I just leave them out completely and base the calculation on 12 entries?
- Consider data on every game played by the Brooklyn Nets in 2014 (82 games) that includes the variables margin, - the Net's margin of victory (number of points the Nets scored minus the number of points their opponent scored) for game i, and • home; - a dummy variable equal to 1 when the Nets are the home team (game i was played in their home arena) and equal to 0 when they are the away team (game i was played in the opponent's arena). I use the least-squares method to estimate the following regression model margin = a + ßhome; + ei Below is the Stata output corresponding to the estimated regression line: regress margin home if team===== "Brooklyn Nets" . Source Model Residual Total margin home _cons SS 1459.95122 15252.0488 16712 df 1459.95122 1 80 190.65061 None of the above 81 206.320988 Coef. Std. Err. 8.439024 3.049595 -5.219512 2.156389 MS t Number of obs F(1, 80) Prob > F R-squared O The Nets lost more games than they won in 2014 P>|t| 2.77 0.007 -2.42 0.018 Adj R-squared = Root…The aim of a study is to test the ratio of variances in the weight of two groups of rats being under a certain drugs A ( group 1) and B ( group 2). A random sample of 17 rats was selected from group 1and 11 rats from group 2, Then their weight were recorded. The results are shown in the following table 2 J غير مجاب عليه بعد الدرجة من 0 2.0 sd Median Mean n 3.0 3.5 17 group 1 2.0 2.20 11 group 2 علم هذا السؤال 1.05 1.15 F19,16,0.025 =0.386 F15,12,0.05 =0.404 F16,10,0.025=0.335 F19.16,0.05 =0.451 F15,12,0.1=0.496 F1610,0.05=0.401 F16,19,0.025 =0.371 F1215,0.05 =0.382 F10,16,0.025 =0.286 F16,19,0.05 =0.437 F1215,0.1=0.475 F10,16,0.05=0.354 Construct a 95% confidence interval for the ratio of the variances. [ Write only the final result in the box below ] 直 awhy are large x2 values sometimes obtained when the null hypothesis true?
- Each passenger on a plane brings one piece of luggage. The weight of a piece of luggage is normally distributed with mean 50 pounds and variance 150 pounds^2. The weight is independent across passengers. The total luggage weight on the plane cannot exceed 1000 pounds. What is the maximum number of passengers allowed such that with at least 0.9 probability the luggage weight is below the limit.7When a least squares line is fit to the 11 observations in the service time data, we obtain SSE= 210.8204. Calculate s? and s. (Round s2 to 4 decimal places. Round s to 5 decimal places.) s^2 This is a numeric cell, so please enter numbers only.
