4) The results of a random sample of children with pain from musculoskeletal injuries treatedwith medicine A, B or C are shown in the table.Medicine AMedicine BMedicine CSignificant568260ImprovementSlight441840Improvement(a) At a = 0.01, is there enough evidence to conclude that the treatment and result areindependent?(b) Find the p-value for the test in (a) using Chi-square table.(c) From the sample data above, it seems that there is no difference in significant improvementon helping children with pain from musculoskeletal injuries using Medicine A or Medicine C.Conduct an appropriate test on it using 5% level of significance.

Given that Two varaibles: Improvement and treatment We have 2×3 contingency table The data is…

Answered: 4) The results of a random sample of…

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 study was conducted to investigate drinking problems among college students. In 1983, a group of students were asked whether they had ever driven an automobile while drinking. In 1987, after the legal drinking age was raised, a different group of college students was asked the same question. The results are as shown in the table below. Test to compare the proportions of students with a drinking problem at 0.05 level of significance, 1987 versus 1983; and provide the chi-squared statistic value below (in 2 decimal places). Year Drove while drinking Yes No 1983 1,258 1,355 1987 994 1,643
You wish to determine the average VO2max (maximum rate of oxygen consumption, often associated with athletic performance) of a sample of 100 Olympic athletes, and to use a quantitative metric to assess your confidence in the average (mean) value. You calculate a mean of 71 ml/kg/min and a variance of 26.06 from your sample. Question: What are the standard error of the mean (SEM) and the approximate 95% CI of the mean?
I want to know the answer and detail step by step
Á study was conducted that examined the relationship between alcohol consumption and risk of heart attack. In all, 417 men, age 40 or older, were tracked over ten years and the number of that experienced a heart attack was recorded. The participants were grouped by those who totally abstained from alcohol and those that drank moderately. For the results given in the table below, use the chi-squared test to find the p value for NHST at the .05 level. (In SPSS use the Yates continuity correction value.) Heart attack Alcohol Crosstabulation Count Alcohol Abstainer Moderate Total Heart attack No 197 192 389 Yes 9. 19 28 Total 206 211 417 Op= .090 Op= 121 !! Op=.067 Op = .048
mean of the process is the estimate developed in part (a). Twenty-five samples of 100 items each were inspected when a process was considered to be operating satisfactorily. In the 25 samples, a total of 135 items were found to be defective. What is an estimate of the proportion defective when the process is in control? b. a. What is the standard error of the proportion if samples of size 100 will be used for statistical process control? Compute the upper and lower control limits for the control chart. C.
You are part of a trivia team and have tracked your team's performance since you started playing, so you know that your scores are normally distributed with u = 88 and a = 15 Recently, a new person joined the team, and you think the scores have changed Sample data is: n = 28, sample mean = 72 Assuming a two tailed test, What is the test statistic (Z)?
The manager of a political party wants to find out the association between the campaign spending (Large, medium and small) and the vote obtained (Majority, Average and minority). Use Chi-squared hypothesis test for association to test that there is an association between the variable in campaign spending and the variable in vote obtained with five percent level of significance. The manager uses MS excel to calculate the Chi-squared stat value and the result is 8.874. Critical value: At 1% = 13.277 :At 5% = 9.488 :At 10% = 7.779 1. State the decision rule 2. Conclusion and decision 3. State the alternative hypothesis and the null hypothesis
An experiment was carried out to compare the sound distortion produced by 4 different types of coating on magnetic tape. The ANOVA table for this data is given below, where X indicates a missing value. Source Degrees of Freedom Sum of Squares Mean Squares IF Treatment X 14.49 X Error X X X Total 49 165.69 Compute the F statistic. Give your answer to 2 decimal places.
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Results of a research claim to predict wind speed at a place with a variance of 4 km/hr or less. A random sample of 24 measurements provides a sample variance of s2 = 4.9. Assuming that the population distribution of wind speed is approximately normal, check the validity of the claim at 5% level of significance using chi-square test.
help with 3
Much is still to be learned about the relationship between sound frequency and loudness. One way to study this relationship is to have listeners perform loudness judgments for tones of different frequencies. For each listener, the output of these judgments is a number, measured in sones, that gives the loudness of the tone relative to the loudness of a reference tone. You have in front of you data from an experimental study in which 54 listeners were asked to perform such loudness judgments for tones of various intensities and frequencies. Of these 54 listeners, 18 had normal hearing, 18 used hearing aids, and 18 had some hearing loss at low frequencies (but didn't use hearing aids). Here is a summary of the data obtained for a 50 dB SPL, 500-Hz tone (the data are the sone values corresponding to this tone). Sample size Sample Sample variance Groups mean Normal hearing 18 2.52 0.3 Hearing aid 18 2.23 0.3 Some hearing loss 18 1.97 0.2 Send data to calculator Send data to Excel

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