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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Please provide a clear solution and kindly solve it using excel. Screenshot your solution using excel. Thank you
please answer all parts
Answer (1.32, 10.68), (1, 11)
How do I construct a confidence interval of the requested level with the given data below: x=28, n=64, confidence level 93%
In a study of the use of hypnosis to relieve pain, sensory ratings were measured for 8 subjects and the mean sensory ratings was found to be 2.8 and standard deviation to be 0.3. The sensory ratings are normally distributed. The 80% confidence interval for the mean sensory ratings of the population is (9.35,11.65) b. (3.58, 4.64) (2.65, 2.95) d. (3.09, 3.91) (6.27, 8.73) a. C. e.
A repeated measures study is done to measure the change in IQ test scores taken on Monday versus those taken on Friday. There were n=9 participants in the study. The mean difference was MD =-5 points and the standard error for the mean difference was SMD=0.51. Construct a 95% confidence interval to estimate the size of the population mean difference.
Decide whether it describes a confidence interval for a proportion or a confidence interval for a mean. (b) Apply the correct formula to calculate the confidence interval in the format of an estimate + margin of error. During the 2016 season, Mark Williams played in 18 golf matches and used his driver club 126 times, hitting his drives an average of 278 yards per drive, with a standard deviation of 15 yards per drive. Calculate a 95% confidence interval for the distance he was able to hit his driver club in that season.
A poll of 1070 U.S. adults found that 563 U.S. adults, or 52.617% regularly used Facebook as a news source. Find the margin of error and confidence interval for the percentage of U.S. adults who regularly use Facebook as a news source, at the 90% level of confidence. Express all answers as percentages rounded to 2 decimal places. Do not write the percent sign. Margin of Error (as a percentage): Confidence Interval: Margin of Error (as a percentage): Confidence Interval: Now find the margin of error and confidence interval for the percentage of U.S. adults who regularly use Facebook as a news source, at the 95% level of confidence. Margin of Error (as a percentage): Confidence Interval: % to O increases O stays roughly the same O decreases % to Finally, find the margin of error and confidence interval for the percentage of U.S. adults who regularly use Facebook as a news source, at the 99% level of confidence. and the precision of our estimate O decreases O increases O stays roughly the…
One Sample T-test and Confidence Interval The average monthly cell phone bill was reported last year to be $53.88 by the U.S. Wireless Industry. Random sampling of a large cell phone company found the current monthly cell phone charges (in dollars) in the table on the left. At the 0.05 level of significance, can it be concluded that the average cell phone bill has increased? Also find an 85% confidence interval estimate of the current mean monthly cell phone bill. Cell Phone Bills 55.83 60.47 58.60 49.88 52.45 51.29 32.98 49.20 70.42 50.02 75.04 41.88 35.00 61.58 58.99 61.25 62.00 80.21 73.60 71.52 42.80 38.65 49.54 61.88 46.30 74.10 48.63 63.45 44.43 36.77 71.96 77.79 42.13 63.24 67.56 74.44 31.77 76.91 73.04 79.92 67.10 78.13 37.95 45.60 68.88 54.47 43.47 54.57 76.12 80.45 38.08 57.21 54.28 a. State the random variable and the parameter in words. b. State the null and…
The mean score on the math portion of the ACT for the year 2000 is μ = 20.7. A sample of 20 students in a local school district had a mean of 22.95. Using the sample, a 95% confidence interval estimate for the mean is (20.76, 25.14). State the point estimate for the mean ACT score for students. Based on this sample information , do you believe that the local students have a mean different from the general population? Why? 22.95 is in the center of the interval. The means are definitely not different. 22.95 is just outside the 95% confidence interval. Can't say the means are really different. 22.95 is not equal to 20.7. The means are definitely different. 20.7 is just outside the 95% confidence interval. Can't say the means are really different.
A company that has a 10% market share (they sell their product to 10% of all potential buyers) launches a marketing campaign. At the end of the campaign period, the company conducts a survey in order to assess whether its market share has increased. The result of the survey issued to 500 customers was 12.5%.h0= p=0.10, h1= p>0.10test statistic= 1.86P value= 0.0314 a. State your conclusion using ΅ = 0.05.b. What would be the potential Type I and Type II errors in this problem? Explain these possible errors in context.

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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