Calculate the correlation coefficient from the following results : N=10,sigmax= 350,Sigma Y= 310 Sigma (X – 35) ^2 = 162, Sigma (Y – 31) ^ 2 = 222, Sigma(X – 35) (Y– 31) = 92, Also find the regression line of Y on X
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- Respiratory Rate Researchers have found that the 95 th percentile the value at which 95% of the data are at or below for respiratory rates in breath per minute during the first 3 years of infancy are given by y=101.82411-0.0125995x+0.00013401x2 for awake infants and y=101.72858-0.0139928x+0.00017646x2 for sleeping infants, where x is the age in months. Source: Pediatrics. a. What is the domain for each function? b. For each respiratory rate, is the rate decreasing or increasing over the first 3 years of life? Hint: Is the graph of the quadratic in the exponent opening upward or downward? Where is the vertex? c. Verify your answer to part b using a graphing calculator. d. For a 1- year-old infant in the 95 th percentile, how much higher is the walking respiratory rate then the sleeping respiratory rate? e. f.The following table provides values of the function f(x,y). However, because of potential; errors in measurement, the functional values may be slightly inaccurately. Using the statistical package included with a graphical calculator or spreadsheet and critical thinking skills, find the function f(x,y)=a+bx+cy that best estimate the table where a, b and c are integers. Hint: Do a linear regression on each column with the value of y fixed and then use these four regression equations to determine the coefficient c. x y 0 1 2 3 0 4.02 7.04 9.98 13.00 1 6.01 9.06 11.98 14.96 2 7.99 10.95 14.02 17.09 3 9.99 13.01 16.01 19.02Olympic 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?
- The slope of a regression line (y on x) is -0.24, we know that SDy=3.2 and SDx=4.0, then the correlation coefficient between the two variables is:A. 0.3B. -0.3C. 0.8D. -0.810) A regression was run to determine if there is a relationship between hours of TV watched per day (x) and number of situps a person can do (y).The results of the regression were:y=ax+b a=-0.767 b=31.009 r2=0.609961 r=-0.781 Use this to predict the number of situps a person who watches 7.5 hours of TV can do (to one decimal place)A regression was run to determine if there is a relationship between hours of TV watched per day (x) and number of situps a person can do (y).The results of the regression were:y=ax+b a=-1.38 b=39.555 r2=0.693889 r=-0.833 Assume the correlation is significant, and use this to predict the number of situps a person who watches 7 hours of TV can do (to one decimal place)
- A regression was fun to determine if there is a relationship between hours of TV watched per day (x) and number of situps a person can do (y). The results of the regression were: y-ax+b a-0.753 b-31.112 -0.439569 -0.663 Assume the correlation is significant, and use this to predict the number of situps a person who watches 12 hours of TV can do (to one decimal place) 1 0 Submit Question Details 5 JEDc) Show that the coefficient of determination, R², can also be obtained as the squared correlation between actual Y values and the Y values estimated from the regression model where Y is the dependent variable. Note that the coefficient of correlation between Y and X is Eyixi r = And also that ỹ = ŷ (18.75)mean of x = 2.882, standard deviation = 1.634 Mean of y = 2.588, standard deviation = 0.246 The correlation coefficient is 0.159 1. for the data above, Find the slope coefficient for the regression line (assume that GPA (y) is a function of HOURS (x)).
- Calculate the correlation coefficient from the following results N-10, ΣX= 350 Σ Υ- 310 Σ (X-35= 162, Σ (Y- 31)" = 222, E (X- 35) (Y- 31) = 92. %3D %3D Also find the regression line of Y on XConsider the model Ci= B0+B1 Yi+ ui. Suppose you run this regression using OLS and get the following results: b0=-3.13437; SE(b0)=0.959254; b1=1.46693; SE(b1)=0.0697828; R-squared=0.130357; and SER=8.769363. Note that b0 and b1 the OLS estimate of b0 and b1, respectively. The total number of observations is 2950. The number of degrees of freedom for this regression is A. 2950 OB. 2948 OC. 2952 OD. 2A particular article presented data on y = tar content (grains/100 ft³) of a gas stream as a function of x₁ = rotor speed (rev/min) and x₂ = gas inlet temperature (°F). The following regression model using X₁, X2, X3 = ×₂² and ×4 = X₁X₂ was suggested. (mean y value) = 86.5 – 0.121x₁ +5.07x2 - 0.0706x3 + 0.001x4 (a) According to this model, what is the mean y value (in grains/100 ft³) if x₁ = 3,400 and x₂ = 55. grains/100 ft³ (b) For this particular model, does it make sense to interpret the value of ₂ as the average change in tar content associated with a 1-degree increase in gas inlet temperature when rotor speed is held constant? Explain. Yes, since there are no other terms involving X2. O Yes, since there are other terms involving X₂. ● No, since there are other terms involving X2. O No, since there are no other terms involving X2.
