A study was carried out to relate sales revenue y (in thousands of dollars) to advertising expenditure x (also in thousands of dollars) for fast-food outlets during a 3 month period. A sample of 15 outlets yieldedthe accompanying summary quantities.== 1438.30Σ - - 14.30 ΣΥ - 1438.30 Σχ2 = 13.95XΣγ2 = 140,355 Σχ= 1388.20-2401.87 -2 - 561.45Σε - γ2 = 24Σε(a) What proportion of observed variation in sales revenue can be attributed to the linear relationship between revenue and advertising expenditure? (Give the answer to three decimal places.)0.766(b) Calculate s₂ and sы. (Give the answers to four decimal places.)SeSb= 6.5718= 11.6661(c) Obtain a 90% confidence interval for b, the average change in revenue associated with a $1000 (that is, 1 unit) increase in advertising expenditure. (Give the answers to three decimal places.)(32.977×74.296× )

(a) The proportion of observed variation in sales revenue attributed to the linear relationship…

Answered: A study was carried out to relate sales…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Cardiologists use the short-range scaling exponent α1, which measures the randomness of heart rate patterns, as a tool to assess risk of heart attack. The article “Applying Fractal Analysis to Short Sets of Heart Rate Variability Data” compared values of α1 computed from long series of measurements (approximately 40,000 heartbeats) with those estimated from the first 300 beats to determine how well the long-term measurement (y) could be predicted the short-term one (x). Following are the data (obtained by digitizing a graph). Short Long 0.54 0.55 1.02 0.79 1.4 0.81 0.88 0.9 1.68 1.05 1.16 1.05 0.82 1.05 0.93 1.07 1.26 1.1 1.18 1.19 0.81 1.19 0.81 1.2 1.28 1.23 1.18 1.23 0.71 1.24 Note: This problem has a reduced data set for ease of performing the calculations required. This differs from the data set given for this problem in the text. Find a 95% confidence interval for the mean long-term measurement for those with short-term measurements…
Cardiologists use the short-range scaling exponent α1, which measures the randomness of heart rate patterns, as a tool to assess risk of heart attack. The article “Applying Fractal Analysis to Short Sets of Heart Rate Variability Data” compared values of α1 computed from long series of measurements (approximately 40,000 heartbeats) with those estimated from the first 300 beats to determine how well the long-term measurement (y) could be predicted the short-term one (x). Following are the data (obtained by digitizing a graph). Short Long 0.54 0.55 1.02 0.79 1.4 0.81 0.88 0.9 1.68 1.05 1.16 1.05 0.82 1.05 0.93 1.07 1.26 1.1 1.18 1.19 0.81 1.19 0.81 1.2 1.28 1.23 1.18 1.23 0.71 1.24 Note: This problem has a reduced data set for ease of performing the calculations required. This differs from the data set given for this problem in the text. A. Compute the least-squares line for predicting the long-term measurement from the short-term measurement.…
A sample of ?=14n=14 observations is drawn from a normal population with ?=1030μ=1030 and ?=200σ=200. Find each of the following: A. ?(?¯>1110)P(X¯>1110) Probability = B. ?(?¯<944)P(X¯<944) Probability = C. ?(?¯>997)P(X¯>997) Probability =
To compare the dry braking distances from 30 to 0 miles per hour for two makes of automobiles, a safety engineer conducts braking tests for 35 models of Make A and 35 models of Make B. The mean braking distance for Make A is 43 feet. Assume the population standard deviation is 4.6 feet. The mean braking distance for Make B is 46 feet. Assume the population standard deviation is 4.5 feet. At α=0.10, can the engineer support the claim that the mean braking distances are different for the two makes of automobiles? Assume the samples are random and independent, and the populations are normally distributed. The critical value(s) is/are Find the standardized test statistic z for μ1−μ2.
Calculate the Chi-square following instructions either in Excel or Google sheets
Every year, the students at a school are given a musical aptitude test that rates them from 0 (no musical aptitude) to 5 (high musical aptitude). This year's results were: Aptitude Score Frequency 0 1 2 3 4 233 1 32 5 2 The mean () aptitude score: to 1 decimal place.) The median aptitude score: (Please show your answer The mode aptitude score: (Please separate your answers by ',' in bimodal situation. Enter DNE if there is no mode.)

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