The following table shows a sample dataset of observation values of an independent variable, x, and a dependent variable, y x 4 5 3 6 10 y 4 6 5 7 7 Compute the correlation coefficient between the two variables
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A: According to the question, the shoe print is x and the height is y.
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A: The coefficient of determination value is R-sq = 0.657.
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Q: The Minitab output shown below was obtained by using paired data consisting of weights (in Ib) of 31…
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Q: Police sometimes measure shoe prints at crime scenes so that they can learn something about…
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Q: Police sometimes measure shoe prints at crime scenes so that they can learn something about…
A: 1) The correlation value is obtained using EXCEL. The software procedure is given below: Enter the…
The following table shows a sample dataset of observation values of an independent variable, x, and a dependent variable, y
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- D Correlation coeff, r: 0.961961 Fifty-four wild bears were anesthetized, and then their weights and chest sizes were measured and listed in a data set. Results are shown in the accompanying Correlation Results display. Is there sufficient evidence to support the claim that there is a linear correlation between the weights of bears and their chest sizes? When measuring an anesthetized bear, is it easier to measure chest size than weight? If so, does it appear that a measured chest size can be used to predict the weight? Use a significance level of a=0.05. Critical r: +0.2680855 P-value (two tailed): 0.000 C -- there is not a linear correlation between the two. OB. Yes, it is easier to measure a chest size than a weight because measuring weight would require lifting the bear onto the scale. The chest size could not be used to predict weight because there is too much variance in the weight of the bears. OC. Yes, it is easier to measure a chest size than a weight because measuring weight…The Minitab output shown below was obtained by using paired data consisting of weights (in Ib) of 32 cars and their highway fuel consumption amounts (in mi/gal). Along with the paired sample data, Minitab was also given a car weight of 4500 lb to be used for predicting the highway fuel consumption amount. Use the information provided in the display to determine the value of the linear correlation coefficient. (Be careful to correctly identify the sign of the correlation coefficient.) Given that there are 32 pairs of data, is there sufficient evidence to support a claim of linear correlation between the weights of cars and their highway fuel consumption amounts? Click the icon to view the Minitab display. Minitab output The linear correlation coefficient is (Round to three decimal places as needed.) The regression equation is Highway = 50.9- 0.00539 Weight Predictor Сoef SE Coef P Constant 50.856 2.856 17.48 0.000 Weight - 0.0053916 0.0007634 - 7.59 0.000 S= 2.10918 R-Sq = 64.2% R-…Police sometimes measure shoe prints at crime scenes so that they can learn something about criminals. Listed below are shoe print lengths, foot lengths, and heights of males. Construct a scatterplot, find the value of the linear correlation coefficient r, and find the P-value of r. Determine whether there is sufficient evidence to support a claim of linear correlation between the two variables. Based on these results, does it appear that police can use a shoe print length to estimate the height of a male? Use a significance level of α=0.01. Shoe Print (cm) 29.0 29.0 31.2 31.7 27.1 Foot Length (cm) 26.3 24.9 28.0 26.0 25.7 Height (cm) 177.5 173.1 181.1 181.6 176.9
- 6. Police sometimes measure shoe prints at crime scenes so that they can learn something about criminals. Listed below are shoe print lengths, foot lengths, and heights of males. Construct a scatterplot, find the value of the linear correlation coefficient r, and find the P-value of r. Determine whether there is sufficient evidence to support a claim of linear correlation between the two variables. Based on these results, does it appear that police can use a shoe print length to estimate the height of a male? Use a significance level of = 0.05. 29.6 29.6 32.1 32.4 28.1 Shoe Print (cm) Foot Length (cm) Height (cm) 24.7 24.6 26.8 25.6 25.0 175.8 173.2 182.6 177.3 171.3 The test statistic is t = (Round to two decimal places as needed.) The P-value is (Round to three decimal places as needed.) Because the P-value of the linear correlation coefficient is (3)- the significance level, there (4) heights of males. sufficient evidence to support the claim that there is a linear correlation…Mail X Gene × b Succe x YouTube M Gmail V Books ✰ nightingale.instructure.com/courses/3367087/assignments/33979308?module_item_id=76315289 Kinnser Software Mail - Hayley Murra... The data shown below consists of the price (in dollars) of 7 events at a local venue and the number of people who attended. Determine if there is significant negative linear correlation between ticket price and number of attendees. Use a significance level of 0.01 and round all values to 4 decimal places. Week X ← → C Account Dashboard Courses 28 Groups looo HEI Calendar 凸 Inbox H History R Studio Logi X M Inbox x 2022-2 Home Announcements Syllabus Modules Grades People Office 365 Secure Exam Proctor (Proctorio) Library Make Tuition Payment Accredible Class Notebook Unicheck Type here to search Bookshelf Ambassa... Ticket Price 6 10 14 18 22 26 30 34 Ho: p = 0 Ha: p < 0 Find the Linear Correlation Coefficient Find the p-value p-value = Attendence 152 141 147 143 138 138 131 132 The p-value is O Greater than a…An instructor has given a short quiz consisting of two parts. For a randomly selected student, let X = the number of points earned on the first part and Y = the number of points earned on the second part. The accompanying table shows the number of students who obtained the indicated points for X (rows) and Y (column). The class is composed of 100 students. Y 10 15 6 2 10 15 20 10 10 1 15 14 1 Compute the correlation between the scores of students from the two parts of the quiz.
- find linear correlation coefficient, P value, and test statistic.A random sample of 7 countries gave the following figures for X = annual per capita income and Y = urbanization rate (percentage of people living in urban area): Country X (US dollars) Y (%) 1 3500 20 2 2700 24 3 2100 16 4 2400 19 5 3000 26 6 2000 19 7 1900 17 Calculate the correlation coefficient for the sample. Explain the results At the 5% significant level, test whether per capita income and urbanization rate are unrelated.For a data set of weights (pounds) and highway fuel consumption amounts (mpg) of twelve types of automobile, the linear correlation coefficient is found and the P-value is 0.022. Write a statement that interprets the P-value and includes a conclusion about linear correlation. The P-value indicates that the probability of a linear correlation coefficient that is at least as extreme is %, which is so there sufficient evidence to conclude that there is a linear correlation between weight and highway fuel consumption in automobiles. (Type an integer or a decimal. Do not round.)
- The Minitab output shown below was obtained by using paired data consisting of weights (in Ib) of 27 cars and their highway fuel consumption amounts (in mi/gal). Along with the paired sample data, Minitab was also given a car weight of 4500 lb to be used for predicting the highway fuel consumption amount. Use the information provided in the display to determine the value of the linear correlation coefficient. (Be careful to correctly identify the sign of the correlation coefficient.) Given that there are 27 pairs of data, is there sufficient evidence to support a claim of linear correlation between the weights of cars and their highway fuel consumption amounts? Click the icon to view the Minitab display. Minitab output The linear correlation coefficient is. (Round to three decimal places as needed.) The regression equation is Highway = 50.4- 0.00535 Weight %3D Predictor Сoef SE Coef T Constant 50.383 2.732 17.75 0.000 Weight - 0.0053502 0.0007856 -7.55 0.000 S= 2.17876 R-Sq = 65.1% R-…The Minitab output shown below was obtained by using paired data consisting of weights (in lb) of 31 cars and their highway fuel consumption amounts (in mi/gal). Along with the paired sample data, Minitab was also given a car weight of 4000 lb to be used for predicting the highway fuel consumption amount. Use the information provided in the display to determine the value of the linear correlation coefficient. (Be careful to correctly identify the sign of the correlation coefficient.) Given that there are 31 pairs of data, is there sufficient evidence to support a claim of linear correlation between the weights of cars and their highway fuel consumption amounts? Click the icon to view the Minitab display. The linear correlation coefficient is (Round to three decimal places as needed.) Is there sufficient evidence to support a claim of linear correlation? Yes O No Minitab output The regression equation is Highway = 50.8 -0.00508 Weight Predictor Coef SE Coef T P Constant 50.772 2.793…The Minitab output shown below was obtained by using paired data consisting of weights (in lb) of 26 cars and their highway fuel consumption amounts (in mi/gal). Along with the paired sample data, Minitab was also given a car weight of 3000 lb to be used for predicting the highway fuel consumption amount. Use the information provided in the display to determine the value of the linear correlation coefficient. (Be careful to correctly identify the sign of the correlation coefficient.) Given that there are 26 pairs of data, is there sufficient evidence to support a claim of linear correlation between the weights of cars and their highway fuel consumption amounts? Click the icon to view the Minitab display. The linear correlation coefficient is (Round to three decimal places as needed.) Minitab output The regression equation is Highway = 50.3 -0.00539 Weight Predictor Coef SE Coef Constant 50.288 2.998 Weight -0.0053868 0.0007773 |S=2.11773 R-Sq=64.0% R-Sq(adj) = 60.9% Predicted Values…
