The survey of 523 students showed that 141 first year engineering students werewilling to enroll ECON810 in Summer term. However, the survey of 477 studentsshowed that only 81 interested in enrolling ECON810 in Summer term.Using a =.01, test for a significant difference between the enrollment and state yourconclusion.(You NEED to use BOTH the Critical-value approach and P-value approach toget a full mark).Do NOT use EXCELO a) Do not reject Ho.Ob None of the answers are correct.O c) Reject Ho.-

Given n1=523, n2=477, X1=141, X2=81 , level of significance ɑ=0.01

Answered: The survey of 523 students showed that…

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 website Data USA stated that in 2016, Texas had the lowest number of adults with a serious mental illness with an average of M=3.26% of the population affected. The state of Texas wants to test if they have a significantly decreased number of mental illness patients compared with the states with the (n=8) highest level of mental illness found. In 2016 the following states had the highest level of adult patients with mental illness: Arkansas= 5.45%, Montana=5.34%, Vermont=5.32%, Oregon=5.19%, Utah=5.2%, Ohio=5.13, Kentucky=5.25% West Virginia=5.18%. The alpha level for the study was set to α=.05 for the criteria. a. State the hypotheses of the study. b. Find the critical t value for this study c. Compute the one-sample t test. (Again calculate sample SD or s - and get SEM before calculating t.) d. Compute the Cohen's d for the data (even if not significant) e. State the results - is the sample significantly higher? Do you reject the null hypothesis? Report the t test results in APA…
The scores of a test for an engineering class of 30 students are shown here. (Due to the nature of this problem, do not use rounded intermediate values in your calculations-including answers submitted in WebAssign.) Scores: 81, 54, 55, 56, 96, 86, 98, 61, 90, 52, 92, 76, 66, 84, 79, 70, 71, 88, 62, 74, 53, 65, 55, 78, 79, 58, 60, 72, 51,80 (a) Determine the frequency (f), midpoint (x), and xf for each range and fill in the table below. (First, organize the scores into the following ranges: 50-59, 60-69, 70-79, 80-89, and 90-99.) (b) Using the following equation, determine the mean value (x) and fill in the table below. Σ(XA) n X = (c) Calculate (x − x) for each range and fill in the table below. (d) Calculate (x - x)²f for each range and fill in the table below. Range Frequency f Midpoint x 50-59 60-69 70-79 80-89 90-99 S = Enter a number. S = (e) Using the following equation, compute the standard deviation of the class scores. Σ(x-x) ²f n - 1 Σ(x-x) ²4 n - 1 = xf 29 X X - X (x - x)²f
A past survey claimed that at least 15% of sports activities per day. In a sample of 160 students, 25 said they spend more than 2 hours on sports activities per day. Test the claim at a = 0.05. students spend more than 2 hours on
In 1995, it was determined that 78% of US citizens supported a ban on household aerosols. Thinking that this proportion may have changed over the past few decades, a sample will be taken to test if a change has occurred. In a SRS of 400 citizens taken this year, 332 supported the ban. Determine the value of the appropriate test statistic to perform this test. Round to 3 decimal places.
In a survey of 1150 adult males, 805 said they use the Internet. In a survey of 1050females, 746 said they use the Internet. Atα= 0.05, can you reject the claim that the proportion of Internet users are the same for both groups?
Historically, the proportion of students entering a university who finished in 4 years or less was 63%. To test whether this proportion has decreased, 120 students were examined and 52% had finished in 4 years or less. To determine whether the proportion of students who finish in 4 year or less has statistically significantly decreased (at the 5% level of signficance), what is the p-value? (please round your answer to 4 decimal places)
A student conducted a test at 0.05, and she found the 95% CI to be [4.79,7.25]. The mean for the sample is 6.12. Is the conclusion for the test that there is a significant difference or that there is not a significant difference? You should fully explain how you can use this information to draw a conclusion.
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A time study analyst observed a repetitive work cycle performed by a worker. During the observation, the analyst made a record of the work elements. She also made a note of the worker's performance and rated it as 95%. Back in the office, the analyst listed the basic motion elements for the task and the sum of the normal times for the basic motion elements was 1.38 minutes. The company uses a PFD allowance factor of 15%. What is the standard time for the task? Tstd 1.225 minutes/unit Tstd 1.587 minutes/unit Tstd 0.625 minutes/unit Tstd 1.959 minutes/unit =

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