People age, the number of hours per week spent on exercise tends to decrease. Here is a random sample comparing are and hours per week of exercise.
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People age, the number of hours per week spent on exercise tends to decrease. Here is a random sample comparing are and hours per week of exercise. Assume both variables follow a
- Test the claim of
correlation at the α = 0.05 level of significance. Find the P-value - Will the null hypothesis be rejected?
- Round the regression equation to one decimal place and use it to
Age | 21 | 36 | 41 | 45 | 31 | 25 | 50 | 61 |
---|---|---|---|---|---|---|---|---|
Hours | 15 | 14 | 6 | 12 | 10 | 20 | 5 | 4 |
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- Find the equation of the regression line for the following data set. x 1 2 3 y 0 3 4Olympic Pole Vault The graph in Figure 7 indicates that in recent years the winning Olympic men’s pole vault height has fallen below the value predicted by the regression line in Example 2. This might have occurred because when the pole vault was a new event there was much room for improvement in vaulters’ performances, whereas now even the best training can produce only incremental advances. Let’s see whether concentrating on more recent results gives a better predictor of future records. (a) Use the data in Table 2 (page 176) to complete the table of winning pole vault heights shown in the margin. (Note that we are using x=0 to correspond to the year 1972, where this restricted data set begins.) (b) Find the regression line for the data in part ‚(a). (c) Plot the data and the regression line on the same axes. Does the regression line seem to provide a good model for the data? (d) What does the regression line predict as the winning pole vault height for the 2012 Olympics? Compare this predicted value to the actual 2012 winning height of 5.97 m, as described on page 177. Has this new regression line provided a better prediction than the line in Example 2?Use the shoe print lengths and heights shown below to find the regression equation, letting shoe print lengths be the predictor (x) variable. Then find the best predicted height of a male who has a shoe print length of 28.5 cm. Would the result be helpful to police crime scene investigators in trying to describe the male? Use a significance level of α=0.05. Shoe Print (cm) 29.1 29.1 31.8 31.9 27.5 Foot Length (cm) 25.7 25.4 27.9 26.7 25.1 Height (cm) 175.4 177.8 185.2 175.4 173.2 The best predicted height is enter your response here cm. (Round to two decimal places as needed.) Would the result be helpful? A. No, because the description would be the same regardless of shoe print length. B. Yes, because the description would be based on an actual shoe print length. C. Yes, because the correlation is strong, so the predicted…
- The coefficient of correlation in a simple regression analysis is = -0.6. The coefficient of determination for this regression would be 0.36 - 0.36 0.6 0.13 O 0.6 or + 0.6Listed below are systolic blood pressure measurements (in mm Hg) obtained from the same woman. Find the regression equation, letting the right arm blood pressure be the predictor (x) variable. Find the best predicted systolic blood pressure in the left arm given that the systolic blood pressure in the right arm is 85 mm Hg. Use a significance level of 0.05. Right Arm 100 99 91 76 76 5 Left Arm 175 170 146 147 146 Click the icon to view the critical values of the Pearson correlation coefficient r The regression equation is y = +x (Round to one decimal place as needed.) Given that the systolic blood pressure in the right arm is 85 mm Hg, the best predicted systolic blood pressure in the left arm is mm Hg. (Round to one decimal place as needed.)Listed below are systolic blood pressure measurements (in mm Hg) obtained from the same woman. Find the regression equation, letting the right arm blood pressure be the predictor (x) variable. Find the best predicted systolic blood pressure in the left arm given that the systolic blood pressure in the right arm is 85 mm, Hg. Use a significance level of 0.05. Right Arm 100 99 91 79 80 Left Arm 177 172 143 146 147 1. The regression equation is y= __ + __ x 2. Given that the systolic blood pressure on the right arm is 85 mm Hg, the best predicted systolic blood pressure in the left arm __ mm Hg.
- a ift = The table below gives the age and bone density for five randomly selected women. Using this data, consider the equation of the regression line, bo + b₁x, for predicting a woman's bone density based on her age. Keep in mind, the correlation coefficient may or may not be statistically significant for the data given. Remember, In practice, it would not be appropriate to use the regression line to make a prediction if the correlation coefficient is not statistically significant. tab ** Answer How to enter your answer (opens in new window) Step 2 of 6: Find the estimated y-Intercept. Round your answer to three decimal places. esc 1945 ! 1 q alt a Z ebook x360 ng from any angle. @ 2 W S →>> X # 3 e d C $ 4 C r f Age Bone Density 359 % 5 V t g 36 51 63 66 70 357 328 314 310 Oll A 6 b y hp h & 7 O n u j * 8 N O i m ( 9 k O 100- 4 Tables ctrl { [ Keypad Keyboard Shortcuts Previous step answers Submit Answer D + = 11 : 90 ; ? Copy Data O SEAR Dec 2 Table ] USE YOUR SMARTPHONE FOR…Write down the estimated regression equation. What is the value of R square in this regression model? Compared with the R-square in Question 1, what is the additional contribution of gender to the percentage of variance explained? Interpret the meaning of the regression coefficient of Year in College (how does year in college affect salary?). Does this variable have a significant impact on income level at α= .05? How does one more year in college affect salary if the gender is the same? How do you get this conclusion?Listed below are systolic blood pressure measurements (in mm Hg) obtained from the same woman. Find the regression equation, letting the right arm blood pressure be the predictor (x) variable. Find the best predicted systolic blood pressure in the left arm given that the systolic blood pressure in the right arm is 90 mm Hg. Use a significance level of 0.05 Right arm - 102; 101; 94; 79; 80 Left arm - 177; 172; 143; 144; 143 The regression equation is y(carety)= ___+___x. Given that the systolic blood pressure in the right arm is 90mm Hg, the best predicted systolic blood pressure in the left arm is _____ mm Hg.
- Listed below are systolic blood pressure measurements (in mm Hg) obtained from the same woman. Find the regression equation, letting the right arm blood pressure be the predictor (x) variable. Find the best predicted systolic blood pressure in the left arm given that the systolic blood pressure in the right arm is 85 mm Hg. Use a significance level of 0.05. Right Arm 101 100 92 79 79 Left Arm 175 168 181 142 144 LOADING... Click the icon to view the critical values of the Pearson correlation coefficient r The regression equation is y=enter your response here+enter your response herex. (Round to one decimal place as needed.) Given that the systolic blood pressure in the right arm is 85 mm Hg, the best predicted systolic blood pressure in the left arm is enter your response here mm Hg. (Round to one decimal place as needed.)The data show the chest size and weight of several bears. Find the regression equation, letting chest size be the independent (x) variable. Then find the best predicted weight of a bear with a chest size of 51 inches. Is the result close to the actual weight of 595 pounds? Use a significance level of 0.05. Chest size (inches) Weight (pounds) Click the icon to view the critical values of the Pearson correlation coefficient r. 2 What is the regression equation? y=+x (Round to one decimal place as needed.) What is the best predicted weight of a bear with a chest size of 51 inches? The best predicted weight for a bear with a chest size of 51 inches is pounds. (Round to one decimal place as needed.) Is the result close to the actual weight of 595 pounds? OA. This result is close to the actual weight of the bear. OB. This result is exactly the same as the actual weight of the bear. OC. This result is very close to the actual weight of the bear. OD. This result is not very close to the actual…A football coach is looking for a way to identify players that are "under weight". The coach decides to get data for each player's height (x, in inches) and weight (y, in pounds), then does a linear regression. The results are: 58+3.7x, r = 0.86 and the standard error is Se = 10.4. Since there is a strong linear correlation the coach, who also majored in Statistics, decides to identify all "outliers" in the data. Obviously, any player whose weight is above the regression line is not "under weight". So the only outliers the coach is interested in are those that are below the regression line. What is the lowest weight possible for a 75 inch player to not be considered "under weight"? Do not round. Submit Question ctor GSearch or type URL & % 24 6 7