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- A. Clearly state an appropriate null hypothesis and an alternative hypothesis. B. What proportion of deaths occurred while the windows were set at a vertical orientation? C. What statistical test would you use to test the null hypothesis? D. Carry out the statistical test from part (c). Is there evidence that window angle affects the mortality rates of birds?ne students took the SAT test twice without doing any sort of preparatory course between the two tests. Their scores are listed below. Use the data to test the claim that in the population of all such students, the differences between the scores have mean 0. Student A B C D E F G H I First score 480 510 530 540 550 560 600 620 660 Second score 460 500 530 520 580 590 560 640 690 a. Define the parameter A. mu Subscript d Baseline equals Mean of the first score B. mu Subscript d Baseline equals Mean of the second score C. mu equals Mean score D. mu Subscript d Baseline equals Mean of (First score - Second score) b. State the null and alternative hypotheses A. Upper H 0 : mu Subscript d Baseline equals 0 Upper H 1 : mu Subscript d Baseline greater than 0 B. Upper H 0 : mu Subscript d Baseline…Refer to the accompanying data display that results from a sample of airport data speeds in Mbps. Complete parts (a) through (c) below. TInterval (13.046,22.15) x=17.598 Sx=16.01712719 n=50 Identify the best point estimate of μ and the margin of error. The point estimate of μ is nothing Mbps.
- A manufacturer of high-resolution video terminals must control the tension on the mesh of fine wires that lies behind the surface of the viewing screen. Too much tension will tear the mesh, and too little will allow wrinkles. The tension is measured by an electrical device with output readings in millivolts (mV). Some variation is inherent in the production process. Here are the tension readings from a random sample of 20 screens from a single day’s production: 269.5 297.0 269.6 283.3 304.8 280.4 233.5 257.4 317.5 327.4 264.7 307.7 310.0 343.3 328.1 342.6 338.8 340.1 374.6 336.1 From this data, we compute ? = 306.32, ? = 36.21 Compute a 90% confidence interval for the mean tension μ of all the screens produced on this day.Which of the following is(are) TRUE? I. A necessary assumption for the parametric test to be valid is that the data on the vitamin content of oranges and lemons should be both normally distributed. II. The parameter of interest is the difference between the mean vitamin content (in milligrams per 10 grams) of oranges and lemons. A. I only B. II only C. Both I and II D. Neither II nor IIRegression analysis was applied between sales data (y) and advertising data (x) and the following information was obtained. SSR = 800SST = 1250Sample Size (n) = 12 The standard error of the estimate is
- Need help on part D! How does lateral acceleration—side forces experienced in turns that are largely under driver control—affect nausea as perceived by bus passengers? An article reported data on x = motion sickness dose (calculated in accordance with a British standard for evaluating similar motion at sea) and y = reported nausea (%). Relevant summary quantities are n = 17, xi = 223.9 , yi = 196 , xi2 = 3056.65 , xiyi = 2759.6 , yi2 = 2976 . Values of dose in the sample ranged from 6.0 to 17.6. (a) Assuming that the simple linear regression model is valid for relating these two variables (this is supported by the raw data), calculate and interpret an estimate of the slope parameter that conveys information about the precision and reliability of estimation. (Calculate a 95% CI. Round your answers to three decimal places.) , (b) Does it appear that there is a…The managing partner of an advertising agency believes that his company's sales are related to the industry sales. He uses Microsoft Excel’s Data Analysis tool to analyze the last 4 years of quarterly data (i.e., n = 16) with the following results: Regression StatisticsMultiple R 0.802R Square 0.643Adjusted R Square 0.618Standard Error SYX 0.9224Observations 16 ANOVA df SS MS F Sig.FRegression 1 21.497 21.497 25.27 0.000Error 14 11.912 0.851Total 15 33.409 Predictor Coef StdError tStat P-valueIntercept 3.962 1.440 2.75 0.016Industry 0.040451 0.008048 5.03 0.000 Durbin-Watson Statistic 1.59 Referring to Table 13-5, the prediction for a quarter in which X = 80 is Y -hat = ________.Recent research published by Frumin and colleagues (2011) in the journal Science addresses whether females' tears have an effect on males. Imagine that exposure to tears lowered self-rated sexual arousal by 1.27 points, with a margin of error of ±0.32 points. The point estimate is: Question 7 options: [0.95, 1.59]. 1.27. 1.27 +/– 0.32. 0.32.
- A. There is no statistical difference in recycling behavior between people who have pessimistic vs. optimistic attitudes toward eco-friendly behaviors. B. People who have pessimistic attitudes toward eco-friendly behaviors recycled significantly more often than those who have optimistic attitudes. C. People who have optimistic attitudes toward eco-friendly behaviors recycled significantly more often than those who have pessimistic attitudes. Conclusion = Effect Size. Calculate the value of the effect size using Cohen's d (NOT eta-squared). d = (use 2 dec places) Is the effect size small, medium, or large? =Based on the SPSS output given below is the Intention to Purchase Branded Shoes, BrandAwareness, Brand Association, Brand Perception and Brand Loyalty on customers in Kangar,Perlis for the year 2018. Model Summaryb Model R R Square Adjusted RSquare Std. Error of theEstimate Durbin-Watson 1 .218a .047 .036 1.08730 1.158 ANOVAa Model Sum ofSquares df Mean Square F Sig.1 Regression 18.972 1 4.743 4.012 .003bResidual 380.672 328 1.182Total 399.645 329Coefficientsa Model Unstandardized Coefficients StandardizedCoefficients B Std. Error Beta t Sig.1 (Constant) 2.7 .412 7.050 .000Brand Awareness -.107 .060 -.101 -1.774 .077Brand Association -.073 .054 -.077 -1.351 .178Brand Perception -11.228 5.552 -7.996 -2.022 .044Brand Loyalty 11.415 5.565 8.110 2.051 .041i. Identify dependent and independent variablesii. Develop a regression model to predict the Intention to Purchase Branded Shoes.iii. Interpret the value of 2.902.iv. Explains the value of R and Adjusted R Squared.v. Determine which…A city's transportation committee has conducted research on traffic and car accidents for downtown streets. There is a 9% chance of being involved in a car accident when 15 or more cars are driving on Johnson Street and a 7.5% chance when 20 or more cars are driving on Dublin Street. The two streets intersect to form a risk area with a radius of 0.1 mi. Approximate the density of cars per square mile in the risk area when 17 cars are on Johnson Street and 21 cars are on Dublin Street. Use 3.14 for pi. There are approximately 1,210 cars per square mile in the risk area. There are approximately 2,674 cars per square mile in the risk area. There are approximately 2,165 cars per square mile in the risk area. There are approximately 1,082 cars per square mile in the risk area.