In this regression, which predictor(s) is/are significant? Coefficients: (Intercept) time pub Estimate Std. Error 43082.4 982.9 121.8 t value Pr(>|t|) 3099.5 13.900 9.26e-09 452.1 2.174 0.0504 149.7 0.814 0.4317
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- The discharge of suspended solids from a phosphate mine is normally distributed, with a mean daily discharge of 27 milligrams per liter (mg/l) and a standard deviation of 14 mg/l. On what proportion of days will the daily discharge exceed 60 mg/l? (Round your answer to four decimal places.) X USE SALTfitting a straight line to a set of data yields the following prediction line. Predict the mean value of Y for x=4Find the indicated IQ score. The graph to the right depicts IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15. 3/2 Click to view page 1 of the table, Click to view page 2 of the table. pr 0.55 =/2 pr /2 or The indicated IQ score, x, is (Round to one decimal place as needed.) /2 /2 or 2 or on 2 on ? Enter your answer in the answer box.
- Using your favorite statistics software package, you generate a scatter plot that displays a linear form. You find a regression equation and the standard deviation for both variables. The standard deviation for x is 1.12, and the standard deviation for y is 4.12. The regression equation is reported as y=-4.7 +1.41x What fraction of the variation in y can be explained by the variation in the values of x? (Enter your answer as a decimal between 0 and 1.)Using your favorite statistics software package, you generate a scatter plot which displays a linear form. You find a regression equation and the standard deviation for both variables. The standard deviation for x is 2.03, and the standard deviation for y is 5.2. The regression equation is reported as - 5.5 + 1.84x What fraction of the variation in y can be explained by the variation in the values of x? (Enter your answer as a decimal between 0 and 1.)Which of the following applies the concept of “layering” to increase the sensitivity of a research analysis? Group of answer choices a. Chi-square Contingency Tables b. Binary Logistic Regression c. Regression d. Answers 1 and 2 only e. Answers 1, 2 and 3
- IQ tests supposedly measure our intelligence. They typically are ‘normed’ so that they have a mean of 100 and a standard deviation of 15. (Higher scores show higher intelligence.) Use the graph and use the z-score and percents from columns b and c in Table A, when relevant. If you scored 132, what is your z-score? (Interpret.) Use table A X, z(a), (b), and (c) X z (a) (b) (c) and use the graph! Write down or save your response to use in Part 2 of this quiz. Group of answer choices -2.53 z. I'm at the 98.34 percentile rank and 1.66% of people scored smarter than me. 2.13 z. I'm at the 98.34 percentile rank and 1.66% of people scored smarter than me. -2.13 z. I'm at the 1.66 percentile rank and 98.34% of people scored smarter than me. 2.53 z. I'm at the 0.57 percentile rank and 99.43% of people scored smarter than me. 2.13 z. I'm at the 1.66 percentile rank and 98.34% of people scored smarter than me. 2.53 z. I'm at the 99.43 percentile rank…Given the below output, interpret R-squared. Note: the response variable is Charges Coefficients: (Intercept) age bmi regionnorthwest -979.86 regionsoutheast 64.83 regionsouthwest -1562.19 Estimate Std. Error t value Pr(>|t|) -5510.16 1814.88 -3.036 0.00244 ** 22.30 10.893 < 2e-16 *** 53.62 5.994 2.64e-09 *** 893.36 -1.097 0.27291 897.84 0.072 0.94245 896.58 -1.742 0.08167 . 242.97 321.37 signif. codes: 0 ***' 0.001 **' 0.01 ** 0.05 .' 0.1'1 Residual standard error: 11380 on 1332 degrees of freedom Multiple R-squared: 0.1203, F-statistic: 36.44 on 5 and 1332 DF, p-value: < 2.2e-16 Adjusted R-squared: 0.117 O a) 11.7% of the variation in Charges can be explained by the regression model b) 12.03% of the variation in Charges can be explained by the regression model c) 12.03% of the variation in Charges can be explained by the variation of the regression model d) 11.7% of the variation in Charges can be explained by the variation of the regression modelThe Z-score tells you the number of standard deviations away from the mean (and in what direction) a data value is. The formula: Z = *- can be used to find the Z-score for a single member of the population. Notice X - μ tells you the signed distance the data value (X) is from the mean. When you divide that by the size of each chunk (the standard deviation) you are measuring the distance in units of the size of the standard deviations (how many standard deviations fit in the distance between the data value and the mean) - which gives you the Z-score. The formula X = μ+Z.o can be used to find the value in the population when given the Z-score (signed number of standard deviations). Notice that Z. tells you how far from the mean the data value is (since Z is the number of standard deviations and is the size of each standard deviation the product tells the total distance). So adding the distance from the mean to the mean gives the location along the number line for the data value. A…
