Create and interpret thefollowing tests using thefollowing data: confidenceinterval, t-test, p-value and F-testY = weekly sales X =advertisement costModel: Y = 2.6911 -0.4795 XSE (0.1216) SE (0.1140)HO: b0 = 0 a = 0.05 tcrit =2.2622 RSS = 33.27H1: b0 + 0 n = 11 ESS= 8552.73

Given Information: Consider the given model: Y=2.6911-0.4795X Here, b0=2.6911b1=-0.4795…

Answered: Create and interpret the following…

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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Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week. A sample of 55 weekly reports showed a sample mean of 18.5 customer contacts per week. The sample standard deviation was 5.2. Provide 90% and 95% confidence intervals for the population mean number of weekly customer contacts for the sales personnel. 90% Confidence interval, to 2 decimals: 95% Confidence interval, to 2 decimals:
The M&Ms milk chocolate color distribution is the following: 25% blue, 35% orange, 23% red, and 17% brown. In a bag, I found 160 blue M&Ms, 224 orange ones, 169 red ones and 99 brown ones.I am wondering whether my bag of M&Ms is representative of the manufacturer's claim. Calculate the absolute value of the test statistic. number (rtol=0.01, atol=D0.0001) Calculate the absolute value of the test statistic that would correspond to a = 0.05. number (rtol=0.01, atol=D0.0001) Using a level of significance of a = 0.05, do the M&Ms in the bag match the required color distribution? State your decision. O (a) Reject HO O (b) Cannot Reject HO
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Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week. A sample of 85 weekly reports showed a sample mean of 18.5 customer contacts per week. The sample standard deviation was 5.4. Provide 90% and 95% confidence intervals for the population mean number of weekly customer contacts for the sales personnel. 90% confidence interval, to 2 decimals: 95% confidence interval, to 2 decimals:
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Salma claimed that the gas price in Blue Park neighborhood (mean of $3.65) is higher than that in Green Spring (mean of $3.59). She obtained the t-value of 2.223 after calculating data from 6 gas stations in Blue Park and 5 gas stations in Green Spring. Use the chart below and determine if Salma's claim is supported! Table 10-3. Critical values of "t" significance level p= 0.05 significance level p= 0.01 TP 1 12.706 63.657 2 4.303 9.925 3 3.182 5.841 2.776 4.604 5 2.571 4.032 6 2.447 3.707 7 2.365 3.499 2.306 3.355 9 2.262 3.250 10 2.228 3.169 11 2.201 3.106 Copy the following into your text box, fill in the blanks, and answer the questions A. The degree of freedom is: B. The t-calculated value is: and the t-critical value is : C. Comparison: t-calc t-crit (answer with ()) D. Is the null hypothesis accepted or rejected? Explain how you determined that! Also, explain what it means to reject or accept the hypothesis! E. Explain if Salma's claim is supported based on your t-value…

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Create and interpret the following tests using the following data: confidence interval, t-test, p-value and F- test Y = weekly sales X = advertisement cost Model: Y = 2.6911-0.4795 X SE (0.1216) SE (0.1140) HO: b0 = 0 a = 0.05 tcrit = 2.2622 RSS = 33.27 H1: b0 # 0 n = 11 ESS = 8552.73