Match the calculated correlations to the corresponding scatterplot. (1) (2) (3) (4) DI
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need figure 1, 2, 3, 4
![Match the calculated correlations to the corresponding scatterplot.
(1)
(2)
(3)
(4)
Plot (1)
V [ Choose ]
r = - 0.70
r = 0.06
Plot (2)
r = 0.92
r = 0.45](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3223603e-5fb3-4909-8357-9342b29f73c2%2Fb083f992-a503-49c7-b9c8-639abdd85564%2Fijxmzz_processed.jpeg&w=3840&q=75)
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- For the following exercises, use Table 4 which shows the percent of unemployed persons 25 years or older who are college graduates in a particular city, by year. Based on the set of data given in Table 5, calculate the regression line using a calculator or other technology tool, and determine the correlation coefficient. Round to three decimal places of accuracyFor the following exercises, consider the data in Table 5, which shows the percent of unemployed in a city ofpeople25 years or older who are college graduates is given below, by year. 41. Based on the set of data given in Table 7, calculatethe regression line using a calculator or othertechnology tool, and determine the correlationcoefficient to three decimal places.Olympic 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?
- b.Find the linear correlation coefficient, r, then determine whether there is sufficient evidence to support the claim of a linear correlation between the two variables. c. find the critical valuesWhen the correlation coefficient is 0 A. There is a moderately weak relationship between variables B. The coefficient of determination is 1 C. The slope is 0A researcher found a linear correlation between course grades and the average number of hours spent on a mobile phone each day. The line of best fit has equation y ^ = 3.7 − 0.786 x with correlation coefficient r = − 0.84. a. Does this prove that spending too much time on a phone causes students to get lower grades? b. What does the regression line predict your final grade in a course will be if you spend an average of 150 minutes per day on your phone?
- A negative correlation means that decreases in the X variable tend to be accompanied by decreases in the Y variable. T or FThe value of correlation coefficient between variables X and Y lies between:A. 1 and 2B. 0 and 1C. -1 and +1D. -1 and 0An r value of –0.9 indicates a strong negative correlation
- How is the magnitude of the standard error of estimate related to the value of the correlation? The larger the correlation (closer to ... ) a)r b)0.00 c)infinity sign d)1.00 or -1.00 the a) larger b)smaller is the standard error of estimatewhich implies a stronger linear relationship a correlation of +0.4 or a correlation of -0.6?Q1 - Returns on stocks X and Y are listed below: Period 1 2 3 4 5 6 7Stock X 6% 5% -2% 10% 3% 8% -4%Stock Y 11% 7% 10% -2% 3% 5% -1% What is the correlation of returns on the two stocks?Please round your answer to the nearest hundredth.Note that the correct answer will be evaluated based on the full-precision result you would obtain using Excel.
