Find the regression equation, letting overhead width be the predictor (x) variable. Find the best predicted weight of a seal if the overhead width measured from a photograph is 1.7 cm. Can the prediction be correct? What is wrong with predicting the weight in this case? Use a significance level of 0.05. Overhead Width (cm) 8.1 Weight (kg) 8.3 218 9.5 264 8.8 207 9.4 8.9 O 240 179 255 9 Click the icon to view the critical values of the Pearson correlation coefficient r. The regression equation is y =D•D. (Round to one decimal place as needed.) The best predicted weight for an overhead width of 1.7 cm is kg (Round to one decimal place as needed.) Can the prediction I correct? What is wrong with predicting the weight in this case? O A. The prediction cannot be correct because a negative weight does not make sense. The width in this case is beyond the scope of the available sample data. O B. The prediction cannot be correct because a negative weight does not make sense and because there is not sufficient evidence of a linear correlation. OC. The prediction cannot be correct because there is not sufficient evidence of a linear correlation. The width in this case is beyond the scope of the available sample data. OD. The prediction can be correct. There is nothing wrong with predicting the weight in this case.

(2 Find the regression equation, letting the diameter be the predictor (x) variable. Find the best predicted circumference of a marble with a diameter of 1.2 cm. How does the result compare to the actual circumference of 3.8 cm? Use a significance level of 0.05. Baseball Basketball Golf Soccer Tennis Ping-Pong Volleyball Diameter 7.5 24.2 4.4 21.7 6.9 4.0 21.4 Circumference 23.6 76.0 13.8 68.2 21.7 12.6 67.2 Click the icon to view the critical values of the Pearson correlation coefficient r. The regression equation is y=+x. (Round to five decimal places as needed.) The best predicted circumference for a diameter of 1.2 cm is cm. (Round to one decimal place as needed.) How does the result compare to the actual circumference of 3.8 cm? O A. Since 1.2 cm is beyond the scope of the sample diameters, the predicted value yields a very different circumference. O B. Even though 1.2 cm is within the scope of the sample diameters, the predicted value yields a very different circumference. O C.…

Researchers studying anthropometry collected body girth measurements and skeletal diameter measurements, as well as age, weight, height and gender, for 507 physically active individuals. Use the table to answer the questions. Min 147.2 Q1 164.3 Median 171.3 Mean 172.5 SD 8.7 Q3 178.1 Max 198.1 a) What is the point estimate for the average height of active individuals? ______________ cmb) What is the median height of these active individuals? ______________ cmc) What is the point estimate for standard deviation of heights of active individuals? ____________ cmd) The IQR extends from ______________ cm to ______________ cm, giving an IQR of ______________ cm.e) Is a person who is 1m 90cm (190cm) tall considered unusually tall? And is a person who is 1m 50cm (150cm) considered unusually short? __No, they are not unusual since these heights are within two standard deviations of the mean __No people can come in all sizes, we cannot define usual…

Use the of to calculate the 80% of a who studied The and test for 6 in Phelp 2. Use the margin of error to calculate the 80% prediction interval of a student's score who just 2 hours. 7th grade math class. Test Score y Hours Studied r 65 5. 71 9. 78 7. 82 8. 86 6. 94 1. Regression Equation: ŷ = 5.543x + 37.762 The standard error of estimate Se = 1.13 %3D355 (7) pue g = (x) The estimated score will be between A. 48.848 and 51.796 with a margin of error of 2.838 B. 46.010 and 51.686 with a margin of error of 2.838 C. 40.357 and 46.253 with a margin of error of 2.948 D. 45.900 and 48.848 with a margin of error of 2.768 E. 45.900 and 51.796 with a margin of error of 2.948

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Based on the photo below and a study of 20 men, please solve and calculate the pvalue of the slope.

Data was collected on the average number of yards of a golfer’s length of drive  and game score. Excel returned the following analysis.   Regression Statistics Multiple R 0.856 R Square 0.733 Adjusted R Square 0.643 Standard Error 1.657 Observations 5       Coefficients Std Error Intercept 93.399 7.918 X Variable 1 -0.089 0.031   How many data pairs formed the sample State the value of the coefficient of determination  State the value of the correlation coefficient  Write the regression equation in the form  What percent of variation in score is NOT explained by the golfer's drive length in the regression equation

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Data were gathered and displayed on the scatter plot. 15 y 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Which of the following is the best equation to model the data? ŷ = 11.5(0.81)* Oŷ = 8.1(0.115)* Oŷ= -0.115x+8.1 Oŷ=-0.81x+11.5