In analyzing a factorial experiment, what is this type of plot used for? mean of yield 90 80 70 09 T 50 ntz A1 a. detecting interactions b. comparing means c. comparing variances d. checking normality A2 A3 pesticide A4 variety B3 B2 B1
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- Using a sample of 46 college students, we want to determine if there is a significant correlation between weight (in lbs.) and weekly exercise (in minutes). The results of a correlation and regression analysis are indicated in the Excel output below. The mean weight (the independent variable) was 166.80 lbs., and the mean weekly exercise time (the dependent variable) was 158.83 minutes. SUMMARY OUTPUT Regression Statistics Multiple R 0.027082077 R Square 0.000733439 Adjusted R Square -0.021977165 Standard Error 79.41761298 Observations 46 ANOVA df SS MS F Significance F Regression 1 203.6896 203.6896 0.032295 0.858207 Residual 44 277514.9 6307.157 Total 45 277718.6 Coefficients Standard Error t Stat P-value Lower 95%…Two companies produce widgets of varying quality. An important measure of quality is the number of days the widgets can go before they need to be replaced. Do the following data provide evidence that the quality of widgets from the two companies differ? Round the p-value to three decimal places. Do not assume equal variances. Widget Life in Days from Company 1 13.0 13.2 43.2 40.7 55.0 57.3 43.8 49.9 31.7 Widget Life in Days from Company 2 57.1 57.7 59.1 58.7 51.6 57.9o pis pe Suppose a doctor measures the height, x, and head circumference, y. of 11 children and obtains the data below. The correlation coefficient is 0.899 and the least squares regression line is y = 0.185x+ 12.276. Complete parts (a) and (b) below. Height, x Head Circumference, y 17.4 17.1 17.2 16.9 17.4 17.1 17.1 17.3 17.3 17.3 17.4 27.75 25.75 26.75 25.75 28 26.5 25.75 26.75 27 27.25 27.25 (a) Compute the coefficient of determination, R?. R2 = % (Round to one decimal place as needed) (b) Interpret the coefficient of determination k Approximately % of the variation in (Round to one decimal place as needed.) is explained by the least-squares regression model. Enter your answer in each of the answer boxes. Save for Later DUE RTTT:DY PIVT Score: o:5SYO Tor Tv attempts 3:35 PA- a 2/22/202 Type here to search insert prt sc 4+ backsp %23 6. 8.
- Suppose a doctor measures the height, x, and head circumference, y, of 11 children and obtains the data below. The correlation coefficient is 0.866 and the least squares regression line is y = 0.143x + 13.465. Complete parts (a) and (b) below. Height, x Head Circumference, y 17.3 17.0 17.2 17.1 27.25 25 26.5 25 27.75 26.5 26 26.75 26.75 27 27 17.4 17.3 17.1 17.4 17.4 17.4 17.3 (a) Compute the coefficient of determination, R?. R2 = % (Round to one decimal place as needed.) (b) Interpret the coefficient of determination. Approximately % of the variation in is explained by the least-squares regression model. (Round to one decimal place as needed.)Students in an experimental psychology class did research on depression as a sign of stress. A test was administered to a sample of 30 students. The scores are given. Complete parts (a) through (c) below. 44 28 45 28 57 31 A. 15 55 56 (a) Find the mean and median of the data. The mean is 34.767. (Round to three decimal places as needed.) 1 22345678 2 223456788 3|1234578 4 1368 5 167 The median is 32. (Type an integer or a decimal.) (b) Draw a stem-and-leaf plot for the data using one row per stem. Which stem-and-leaf plot below shows the data? 42 32 48 B. 41 23 27 1 68 2 1368 3|1234588 4 223456678 522345777 14 32 52 O C. 47 32 14 1123466788 213567 3 26 4 23458 5 223457788 16 23 23 36 13 38 O D. 57 58 16 1344566 2 333788 3 122268 4 124578 5 256778Exercise 6. Regression Fallacy. Historically, scores on the two midterms had a correlation of 0.48. Suppose that Jeri scored 2.1 standard deviations below the mean on the first midterm. (a) How many standard deviations [above or below?] the mean would you predict for her second midterm?
- You run a PPM correlation on sprint speed and vertical jump height in NFL running backs and find an r value of 0.89. Your sample has a mean sprint speed of 11 m/s ± 0.3 m/s and a mean jump height of 0.49 m ± 0.3 m. What is the effect size estimate? 2.96 0.89 -35.033 35.033 0.267We give the total variation, the unexplained variation (SSE), and the least squares point estimate b1 . Total variation = 13.459; SSE = 2.806; b1 = 2.6652 Click here for the Excel Data File Using the information given, find the explained variation, the simple coefficient of determination (r2), and the simple correlation coefficient (r). Interpret r2. (Round your answers to 3 decimal places. Round your percent to 1 decimal place.) Explained variation r2 r % of the variation in demand can be explained by variation in price differential.Justin wants to know whether a commonly prescribed drug does improve the attention span of students with attention deficit disorder (ADD). He knows that the mean attention span for students with ADD who are not taking the drug is 2.3 minutes long. His sample of 12 students taking the drug yielded a mean of 4.6 minutes. Justin can find no information regarding σx , so he calculated s2x =1.96. a. Identify the independent and dependent variables. b. In a sentence, state the null hypothesis being tested. c. Using symbols, state the null and alternative hypotheses. d. Determine the critical region using a one-tailed test with alpha = .05. e .Conduct the hypothesis test (Do the math and compare the t-critical and t-obtained values). f. State your conclusions in terms of H0 (Should you reject the H0 or fail to reject/accept the H0). g. Based on your analysis, is there a relationship between the drug and attention span?
- Data was collected about the number of text messages a sample of college students received in the last day. In this particular sample, most students received few text messages, but a small number of students received a very large number of text messages. Two measures of center were computed for this sample: 15.6 and 20.2. One of these measures is the mean and one is the median. Which measure must be the median? 1 The median is 20.2 because the distribution of number of text messages would be skewed to the left. 2 The median is 20.2 because the distribution of number of text messages would be skewed to the right. 3 The median is 15.6 because the distribution of number of text messages would be skewed to the left. 4 The median is 15.6 because the distribution of number of text messages would be skewed to the right.Two companies produce widgets of varying quality. An important measure of quality is the number of days the widgets can go before they need to be replaced. Do the following data provide evidence that the quality of widgets from the two companies differ? Round the p-value to three decimal places. Do not assume equal variances. Widget Life in Days from Company 1 Widget Life in Days from Company 2 12.0 52.2 14.0 45.8 48.4 44.6 47.0 57.5 50.4 44.2 57.4 50.8 30.8 43.0 30.0The correlation coefficient between variables X and Y is 0.62. What does this signify? Question 8Answer a. When Y increases 1, X increases by 0.62 b. When X increases by 1, Y increases by 0.62 c. The variance shared by X and Y is less 20% d. The variance shared by X and Y is 62%