**Educational Content on Confidence Intervals**Refer to the accompanying data display that results from a simple random sample of times (minutes) between eruptions of the Old Faithful geyser. A 95% confidence level was used. Complete parts (a) and (b) below.**TInterval Data:**- **Confidence Interval:** (85.74, 91.76)- **Sample Mean (x̄):** 88.75 minutes- **Sample Standard Deviation (Sx):** 8.897431411 minutes- **Sample Size (n):** 36(a) **The Point Estimate of μ is:** 88.75 minutes (Round to two decimal places as needed.)(b) **The Margin of Error (E) is:** ______ minutes (Round to two decimal places as needed.)*Instructional Note: The point estimate is the sample mean, which provides the best estimate of the population mean μ. The confidence interval gives a range around this estimate, reflecting the precision of the estimate. The margin of error, E, represents the maximum expected difference between the sample mean and the population mean.* **Diagram Explanation:**There is no diagram in this data display. The information shows numerical data related to a statistical analysis of eruption times. The confidence interval provides an estimate range, indicating where the true mean of the population is likely to fall with 95% confidence.

Given that Sample mean (x¯) = 88.75 Sample standard deviation (s) = 8.897431411 Sample size (n) =…

Answered: Refer to the accompanying data display…

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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The article “Americans are ‘Getting the Wrong Idea’on Alcohol and Health” (Associated Press, April 19, 2005)reported that observational studies in recent years thathave concluded that moderate drinking is associated with a reduction in the risk of heart disease may be mis-leading. The article refers to a study conducted by theCenters for Disease Control and Prevention that showedthat moderate drinkers, as a group, tended to be better educated, wealthier, and more active than nondrinkers. Explain why the existence of these potentially con-founding factors prevent drawing the conclusion that moderate drinking is the cause of reduced risk of heartdisease.
B. The test statistic, t, =____ C. The P-value is ____ D. State the conclusion for the test. Reject H0. or Fail to reject H0. E. Construct a confidence interval suitable for testing the claim that students taking nonproctored tests get a higher mean score than those taking proctored tests. ___<μ1−μ2<___ F. Does the confidence interval support the conclusion of the test? (Yes/ No) because the confidence interval contains (zero/ only positive value/ only negative values).
5.1 & 6.2 Question 13, 6.1.34 Part 1 of 2 Use the confidence interval to find the estimated margin of error. Then find the sample mean. HW Score: 44.29%, 15.5 of 35 O Points: 0 of 1 A store manager reports a confidence interval of (44.8,83.8) when estimating the mean price (in dollars) for the population of textbooks. The estimated margin of error is (Type an integer or a decimal.) Clear all his View an example Get more help - Q Search
Use the given data to find the 95% confidence interval estimate of the population mean ?μ. Assume that the population has a normal distribution. IQ scores of professional athletes: Sample size ?=10n=10Mean ?⎯⎯⎯=103x¯=103Standard deviation ?=12s=12 <?<<μ<
A statistics class is estimating the mean height of all female students at their college. They collect a random sample of 42 female students and measure their heights. The mean of the sample is 65.3 inches and the sample standard deviation is 5.4 inches. Find the 90% confidence interval for the mean height of all female students in their school? Assume that the distribution of individual female heights at this school is approximately normal. 1. The t value is 1.3025 2. The standard error has a value of -- 3. The margin of error has a value of 4. Find the 90% confidence interval for the mean height of all female students in their school
A researcher surveyed 12 men who lost their fathers earlier in their lives. His survey included the age of the subjects when their fathers died and their confidence that they would someday be happily married themselves (100 point scale – higher score = more confidence). The results are shown below. Mx=15 My =60 SSx=348 SSy=5198 SPxy=933 Age Confidence Rating 12 34 8 30 11 89 21 69 15 55 7 38 18 78 23 66 22 89 19 79 9 35 15 58 Can you estimate a confidence rating for a man who lost his father at 30 years old? Why or why not?
4
As a bonus assignment a former student checked if your professor gave a statisticallysignificant difference in grades between his male and female students. She based herstudy based on grades assigned in intermediate Econ courses (Econ 303, 305 and 317)and her sample included nm = 485 male students and nf = 264 female students. Theaverage grades received were xm = 84.6 and xf = 85.8 The population standad deviation were σ m = 12.0 and σ f = 11.4 8. From the same extra-credit study as in question 7 see above, this former student found that the proportion of female students in principle courses (Econ 203, 205) was ?̅? = 0.380, while the proportion of female students in intermediate courses (Econ 303, 305, 317) was ?̅? = 0.352. The principle courses sample size was np = 782, while the intermediate courses sample size was ni = 749. Test the hypothesis that female students are less in intermediate courses using a 90% confidence level and the p-value approach.
A researcher surveyed 12 men who lost their fathers earlier in their lives. His survey included the age of the subjects when their fathers died and their confidence that they would someday be happily married themselves (100 point scale – higher score = more confidence). The results are shown below. Mx=15 My =60 SSx=348 SSy=5198 SPxy=933 Age Confidence Rating 12 34 8 30 11 89 21 69 15 55 7 38 18 78 23 66 22 89 19 79 9 35 15 58 The effect size is equal to _____________ . This means it is a ______________ effect.
6.2.3 Find the critical value t, for the confidence levelc=0.99 and sample size n= 12. Click the icon to view the t-distribution table. (Round to the nearest thousandth as needed.)
6.2.3 Find the critical value t, for the confidence level c=0.80 and sample size n= 12. tents Click the icon to view the t-distribution table. Resources t.=(Round to the nearest thousandth as needed.)
Find the critical value to for the confidence level c = 0.99 and sample size n = 9. tc= Click the icon to view the t-distribution table. (Round to the nearest thousandth as needed.)

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