Week 7 Test
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Feb 20, 2024
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Week 7 Test - Results X Attempt 1 of 2 Written Oct 16, 2023 10:37 PM - Oct 17, 2023 12:40 AM Attempt Score 18 /20-90% Overall Grade (Highest Attempt) 18 /20 -90 % Question 1 1/ 1 point The least squares regression line for a data set isy"= -4.6+1.56x and the standard deviation of the residuals is .52 Does a case with the values x = -1.12, y = -8 qualify as an outlier? ) Cannot be determined with the given information O NG v Yes W Hide question 1 feedback Plug in -1.12 for x. y=-46+ 1.56(-1.12) y=-6.3472 Residual is y-given - y-predicted. -8 - (-6.3472)
-8 + 6.3472 = -1.6528 -> this is the residual value. To see if it is an outlier take -2 and multiply it by .52 -2*.52 =-1.04 -1.6528 is less than -1.04, Yes, it is an outlier because if it outside of the -2 to 2 range. Question 2 1/ 1 point A vacation resort rents SCUBA equipment to certified divers. The resort charges an up-front fee of $25 and another fee of $12.50 an hour. How much is it if you rent the SCUBA equipment for 45 minutes? O %23 - $375 () $35 v() $34.38 ¥ Hide question 2 feedback The equation is Dollars = $12.5(hour) + $25 45 minutes = .75 of an hour Dollars = $12.5(.75) + $25 Question 3 1/ 1 point The following data represent the weight of a child riding a bike and the rolling distance achieved after going down a hill without pedaling.
Rollin Weight ns (Ibs) Distance (m.) 59 26 83 43 97 49 56 20 103 65 87 44 88 48 91 42 52 39 63 33 71 39 100 49 89 55 103 53 99 42 74 33 75 30 89 30 102 40 103 33 99 33 102 35 86 37 85 37 Using the regression line for this problem, the approximate rolling distance for a child on a bike that weighs 99 Ibs. is: ) 43.982 587213 v 443761
() 45.6723 w Hide question 3 feedback Copy and paste the data into Excel. Then use the Data Analysis Toolpak and run a Regression. The y-variable is the distance and the x-variable is the weight. How far the bike will travel will depend on the weight of the child. You want to predict the distance of the bike. Once you get the Regression output, look under the Coefficients to find the values to use for the regression equation. y = 10.3364819 + 0.343834842 (x) Plug 99 in for x and solve. y = 10.3364819 + 0.343834842 (99) y=44.37613122 Question 4 1/ 1 point The following data represent the weight of a child riding a bike and the rolling distance achieved after going down a hill without pedaling. Rolli Weight c.> ns Distance (Ibs.) (m.) 59 26 83 43 97 49 56 20 103 65 87 44 88 48
921 42 52 39 63 33 71 39 100 49 89 55 103 53 99 42 74 33 75 30 89 30 102 40 103 33 99 33 102 35 86 37 85 37 Find the 99% prediction interval for rolling distance when a child riding the bike weighs 99 1bs. (round to 4 decimal places) <y< Answer for blank # 1: 19.3556 (50 %) Answer for blank # 2;: 69.3966 (50 %) w Hide question 4 feedback Copy and paste the data into Excel. Then use the Data Analysis Toolpak and run a Regression. The y-variable is the distance and the x-variable is the weight. How far the bike will travel will depend on the weight of the child. You want to predict
the distance of the bike. Once you get the Regression output, look under the Coefficients to find the values to use for the regression equation. y = 10.3364819 + 0.343834842 (x) Plug 99 in for x and solve. y = 10.3364819 + 0.343834842 (99) y =44.37613122, this is our y-hat value. This is the equation to use for the prediction interval 1 (33() — 21_3)2 yEt*x (SE) 14+ — + J (SE) n (n—1)SDx? T-Critical Value =T.INV.2T(.01, 22) = 2.818756061 The SE we get from the Regression output and you can use Excel to find the Average and SD of the Weight variable. LL =44.37613122- 2.818756061%8.573835284" 1 (99 — 85.6667)2 1+— + : 24 (24 —1) % 16.00724 UL =44.37613122+ 2.818756061*8.573835284" 1 (99 — 85.6667)? 1+ — + . 24 (24 — 1) % 16.00724
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Related Questions
Part A of question one asks for the variance (sigma squared), not the equation of regression line. Is the variance 2.79 as I listed on my answer sheet?
Maybe my image didn't go through, but I already had the answer to subparts B and C to the question, which you list as #2 and #3.
But I need help with parts D, E, and F.
I have the following answers for D.
d. Y hat (fitted value = 426.46), with residual of -1.62.
e. r squared value = 1.00
g. 99% confidence interval = (-11.63, -1.05)
Thanks
thanks
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Which of the following are correct statements for linear regression?
1. If the scatter plot of the residual and the predicted response of a linear regression model shows a relationship between the two, then the model is good.
2. Linear regression is not sensitive to outliers.
3. LASSO could be used for variable selection.
4. If the Pearson's correlation coefficient between feature X1 and response Y is small, then there is weak evidence of correlation between X1 and Y.
5. If one uses a large penalty in ridge regression, then the bias in the model is high.
6. If the training data size is increased, then the bian and variance of the model increase.
7. One can use a maximum likelihood estimator in the case of linear regression under the assumption that the errors are independent and identically distributed.
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a. Draw/sketch a SCATTER PLOT with regression line and regression equation.
b. State the null and alternative hypotheses.
Ho (Null Hypothesis):
Ha ( Alternative Hypothesis)
c. Level of Significance
d. Determine the degrees of freedom (df) and t critical value.
e. Correlation coefficient (Pearson r).
f. t statistic
g. Compute the slope of regression line.
h. Compute the mean of x and y
i. Compute the intercept of regression line.
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Question 23 options:
Conduct a simple linear regression between the outcome and the treatment group assignment.
Conduct an ANOVA followed by post-hoc Bonferroni adjustment.
Conduct a Kruskal-Wallis test
Calculate within-group Spearman correlations.
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Which of the following is not a plot of residuals typically used in multiple regression analysis?
Select one:
None of these
Residuals versus correlation coefficients
Residuals versus X1
Residuals versus time
Residuals versus X2.
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