### Analysis of Advertisement Position and Time Effects**(c) Significance Test of Advertisement Position**To test the significance of advertisement position effects with a significance level of α = 0.05:- **F statistic:** 149.14- **p-value:** Less than 0.001- **Decision:** Reject the null hypothesis- **Conclusion:** The position of the ads is important.**(d) Pairwise Comparisons Using Tukey's Method for Time of Day**Make pairwise comparisons of the morning, afternoon, and evening times using Tukey simultaneous 95% confidence intervals. Negative amounts should be indicated by a minus sign. Round your answers to two decimal places.- **Tukey q:** 3.53- **Mean Square Error (MSE):** 8.917Comparisons:- \( \mu_M - \mu_A = \) [Blank]- \( \mu_M - \mu_E = \) [Blank]- \( \mu_A - \mu_E = \) [Blank]**(e) Pairwise Comparisons Using Tukey's Method for Advertisement Positions**Make pairwise comparisons of the four ad positions using Tukey simultaneous 95% confidence intervals. Negative amounts should be indicated by a minus sign. Round your answers to two decimal places.Comparisons:- \( \mu_1 - \mu_2 = \) [Blank]- \( \mu_1 - \mu_3 = \) [Blank]- \( \mu_1 - \mu_4 = \) [Blank]- \( \mu_2 - \mu_3 = \) [Blank]- \( \mu_2 - \mu_4 = \) [Blank]- \( \mu_3 - \mu_4 = \) [Blank]**(f) Maximizing Consumer Response**Identify which time of day and advertisement position maximizes consumer response. Compute a 95% (individual) confidence interval for the mean number of calls placed for this optimal time and ad position. Round your answers to two decimal places.- **Confidence Interval:** - Lower Bound: [Blank] - Upper Bound: [Blank] **ANOVA Analysis of Telemarketing Response Rates Based on Time of Day and Advertisement Position**A telemarketing firm conducted a study to understand how two factors affect responses to television advertisements. The first factor is the time of day the ad is aired, while the second is the ad's position within the hour. Data is collected based on the number of calls to an 800 number after a sample broadcast.### Telemarketing Data Table#### Time of Day vs. Position of Advertisement- **10:00 (Morning)** - On the Hour: 42, 37, 41 - On the Half-Hour: 36, 38, 38 - Early in Program: 62, 64, 64 - Late in Program: 51, 48, 47- **4:00 (Afternoon)** - On the Hour: 60, 58, 60 - On the Half-Hour: 60, 55, 56 - Early in Program: 85, 80, 81 - Late in Program: 60, 67, 66- **9:00 (Evening)** - On the Hour: 102, 99, 97 - On the Half-Hour: 96, 91, 92 - Early in Program: 127, 120, 126 - Late in Program: 105, 101, 107### ANOVA: Two-Factor with Replication#### Summary- **Count**: 3 for each time block- **Sum and Average Responses**: - **Morning**: Sum = 575, Average = 47.92 - **Afternoon**: Sum = 799, Average = 66.58 - **Evening**: Sum = 1279, Average = 106.58 - **Total Average**: 73.72- **Variance**: - **Morning**: 16.58 - **Afternoon**: 132.45 - **Evening**: 106.58 - **Total Variance**: 625.03### ANOVA Output- **Source of Variation** - Sample: SS = 21506.89, df = 2, MS = 10780

a) Test of significance of time of day effects with α = .05. F-statistic = 1209.02 p-value =…

Answered: (c) Test the significance of position…

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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We consider the differences between the reading and writing scores of a random sample of 200 students who took the High School and Beyond Survey. The mean and standard deviation of the differences are Xread = -0.531 and 8.858 points. write (a) Calculate a 95% confidence interval for the average difference between the reading and writing scores of all students. (Round your answers to two decimal places.) (b) Interpret this interval in context. (Round your answers to two decimal places.) We are 95% confident that on the reading test students score, on average, points lower to You may need to use the appropriate table in the Appendix of Tables to answer this question. points higher than they do on the writing test. (c) Does the confidence interval provide convincing evidence that there is a real difference in the average scores? Explain. Since 0 is included in the interval, the confidence interval does not provide convincing evidence that there is a real difference in the average scores.
Suppose you sample six people on their number of visits to Chick-fil-A in a month given by variable Y. Y = 3,6,7,10,15,19 (A) What is the mean of Y? Standard Deviation of Y?(B) What is the Standard Error of Y?(C) Construct 90% and 95% Confidence Intervals for Y.
The data below were collected in a clinical trial to compare a new drug to a placebo for its effectiveness in lowering total serum cholesterol. Generate a 95% confidence interval for the difference in mean total cholesterol levels between treatments. Generate a 95% confidence interval for the proportion of all patients with total cholesterol < 200. New Drug Placebo Total Sample (n=75) (n=75) (n=150) Mean (SD) Total Serum Cholesterol 185.0 (24.5) 204.3 (21.8) 194.7 (23.2) % Patients with Total Cholesterol < 200 78.00% 65.00% 71.50% a. 95% CI ( , )
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 random sample of high school seniors were asked whether they were applying for college. The resulting confidence interval for the proportion of students applying for college is (0.65,0.69). What is the margin of error?
Researchers are studying whether the proportion of homeschooled children in district 1 (considered 1) is less than the proportion of homeschooled children in neighboring district 2 (considered as 2). To test the theory info was gathered: district 1 had 600 students of which 90 were homeschooled. District 2 had 550 students of which 99 were homeschooled. What is the 95% confidence interval for the difference proportion between district 1 and 2? A. (-0.086, 0.026) B. (0.143, 0.217) C. (0.113, 0.187)
Calcium is essential to tree growth. In 1990, the concentration of calcium in precipitation in a certain area was 0.11milligrams per liter (mg/L).A random sample of 10 precipitation dates in 2018 results in the following data table. Complete parts (a) through (c) below. (B) With 98% confidence, the mean concentration of calcium in precipitation in this area in 2018 is between
2/8
Refer to the accompanying stayed outside and construct a 90% confidence interval as many of the mean pulse rate of adult females then do the same for adult males compare the results.
A random sample of 75 customers was polled. .626 and the 95% confidence interval
True or False A 95% confidence level in a mean number of minutes that it takes for an amublance to arrive after a 911 call is (4.2, 7.4). Then there is 95% chance that a randomly selected emergency will have an ambulance response time between 4.2 and 7.4 minutes.

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