Use a 0.05 significance level to test the claim that there is a difference between the actual and reported heights, in inches, for 12-16 year old boys. The data is listed in the table below.Click the icon to view the data table of the reported heights.Let μ₁ denote the median of the first variable and μ₂ denote the median of the second variable. What are the null and alternative hypotheses?OA. Ho: H1 H₂H₁: H1 H₂O C. Ho: H₁ H2H₁:₁

The test is to check whether there is a difference between the actual and reported heights, in…

Answered: Let μ₁ denote the median of the first…

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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A random sample of 64 students was asked to respond on a scale from one (strongly disagree) to seven (strongly agree) to the proposition: "Cellphone helps raise our standard of living." The sample mean response was 4.60 and the sample standard deviation was 1.6. Find the minimum significance level at which we can show that the true mean is greater than 4. 1 =-T.INV(0.05,63) 2 =T.DIST(3,63) 3 =T.INV.2T(0.10,63) 4 =T.DIST.RT(2,63) 5 =T.DIST.RT(3,63)
The graph portrays the decision criterion for a one-mean z-test. The curve in the graph is the normal curve for the test statistic under the assumption that the null hypothesis is true. Use the graph to solve the problem. A graphical display of the decision criterion follows. Reject Ho 0.025 -3 -2 -1 -1.96 Do not reject Ho O-1.96 ≤ z ≤ 1.96 Oz≤-1.96 O z ≥ 0.025 Ozz-1.96 0 Determine the nonrejection region. 1 2 3 Z
U21T. Construct a 99% confidence interval estimate for the mean u using the given sample information. (Give your answers correct to two decimal plac n = 20, x = 16.8, and s = 2.1 |3D 15.6 X to 18 You may need to use the appropriate table in Appendix B to answer this question. Need Help? Read It Watch It Talk to a Tutor
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Choose the correct conclusion below A. Yes, market the wax as there is sufficient evidence to conclude the mean tine is less man 55 seconds. B. No do not market the wax as there is sufficient evidence to conclude the mean time is less than 55 seconds. C. Yes, market the wax as there is insufficient evidence to conclude the mean time is less than 56 seconds. D. No do not market the wax as there is insullicient evidence to conclude the mean time is less than 55 seconds
A psychiatrist is interested in finding a 98% confidence interval for the tics per hour exhibited by children with Tourette syndrome. The data below show the tics in an observed hour for 11 randomly selected children with Tourette syndrome. Round your decimal 3 places. 1,12,11,6,10,1,4,7,8,1,9 With 98% confidence the population mean number of tics per hour that children with Tourette syndrome exhibit is between ____ and ____ If many groups of 11 randomly selected children with Tourette syndrome are observed, then a different confidence interval would be produced from each group. About ____ percent of these confidence intervals will contain the true population mean number of tics per hour and about ____ percent will not contain the true population mean number of tics per hour.
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Use the given confidence interval to find the margin of error and the sample mean. (15.3,23.7) The sample mean is . (Type an integer or a decimal.) The margin of error is. (Type an integer or a decimal.)
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Q.19 Urgent question! I will thumbs up! In the final answer that you circle, please also include the brief approach / formula you used under "Answer". Thanks!
The state of CT claims that the average time on death row is 15 years. A random survey of 75 death row inmates revealed that the average length of time on death row is 16.6 years with a standard deviation of 5.8 years. Conduct a hypothesis to test the state of CT's claim. t test mean two tail test p value = ____________ What is the decision?
Access the hourly wage data on the below Excel Data File (Hourly Wage). An economist wants to test if the average hourly wage is less than $29. Assume that the population standard deviation is $5. Click here for the Excel Data File a. Select the null and the alternative hypotheses for the test. Hg: H = 29; HA: µ + 29 Ο Hg: μ 29 Hg: H 2 29; HA: H < 29 b-1. Find the value of the test statistic. (Negative value should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.) * Answer is complete but not entirely correct. Test statistic -2.40 8 b-2. Find the p-value. p-value < 0.01 O O p-value 2 0.10 O 0.05 s p-value < 0.10 O 0.025 s p-value < 0.05 O 0.01 s p-value < 0.025

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