t.test(improved,local, var.equal=TRUE, paired=TRUE)Paired t-testdata:improved and localt = 2.3294, df = 24, p-valuealternative hypothesis: true difference in means is not equal to 095 percent confidence interval:= 0.0285843.55653 720.66977sample estimates:mean of the differences382.1131t.test (diff,mu=0, alternative ="greater")One Sample t-testdata:difft = 2.3294, df = 24, p-valuealternative hypothesis: true mean is greater than 095 percent confidence interval:= 0.01429101.4644Infsample estimates:mean of x382.1131Select the correct output, and using the selected output:(1)Write the null and alternative hypotheses for the situation under question.(ii)ference in this case. Assume that all those conditions are satisfied and proceed further.Write all the conditions/assumptions necessary for drawing statistical in- The improved crop varieties are evaluated at farmers fields before their release for alarge scale production. The claim made by a chickpea breeder is that, on average,her improved variety produces higher yield of chickpea compared to the local varietygrown by farmers in her region.To examine her claim, the improved variety of chickpea was evaluated at 25 randomlyselected farmer-fields (farms) in that region. Each field was divided into two parts,one part of the field was randomly assigned to the improved variety to be grown undera crop management practice recommended for this variety, and the other part to thelocal variety with an ongoing practice followed for it. You may note there were 25different fields and the two varieties were grown within each of them.The chickpea yields of the two varieties were measured, in kg per hectare, from each ofthe fields. The following R- code outputs contain the information that you may use toanswer the questions. The data vectors, "improved" and "local" , hold the observedyields of the improved variety and local variety, respectively, and are of same length25. The "diff =improved - local". The row i of these vectors are the values for thefield i, i = 1,..., 25.t.test (improved,local, var.equal=TRUE, paired=FALSE )Two Sample t-testdata:improved and localt = 2.6664, df48, p-value = 0.01042alternative hypothesis: true difference in means is not equal to 095 percent confidence interval:93.97748 670.24882sample estimates:mean of x mean of y2496.0332113.919t.test (improved,local, var.equal=TRUE, paired=FALSE, alternative =c("greater"))Two Sample t-testimproved and local48, р-valuedata:t = 2.6664, df == 0.005209alternative hypothesis: true difference in means is greater than 095 percent confidence interval:141.7572Infsample estimates:mean of x mean of y2496.0332113.919

From the given information, The claim made by a chickpea breeder is that, on average her improved…

Answered: t.test (improved,local, var.equal=TRUE,…

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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& Exit Submit my work mode : This shows what is correct or incorrect for the work you have completed so far. It does not indicate completion. Return to question Of 45 bank customers depositing a check, 16 received some cash back. (a) Construct a 90 percent confidence interval for the proportion of all depositors who ask for cash back. (Round your answers to 4 decimal places.) The 90% confidence interval is from to (b) May normality of p be assumed? Yes O No Next > 10 of 10 < Prev
Q16 Use t distribution table and normal standard distribution table
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In a test of hypothesis the p-value was 0.09 . If the level of significance is 5 % . What is the conclusion ? Not enough information Reject the alternative hypothesis Fail to reject the null hypothesis Reject the null hypothesis
best point estimate for 0.678>p>0.788 with a 95% confidence interval
A repeated measures study is done to measure the change in IQ test scores taken on Monday versus those taken on Friday. There were n=9 participants in the study. The mean difference was MD =-5 points and the standard error for the mean difference was SMD=0.51. Construct a 95% confidence interval to estimate the size of the population mean difference.
erproduction of uric acid in the body can be an Indication of cell breakdown. This may be an advance indication of illness such as go adult male patlent has taken eleven blood tests for uric acid. The mean concentration was x = 5.30 mg/dl. The distribution of uric rmal, with o = 1.95 mg/dI. a) Find a 95% confidence interval for the population mean concentration of uric acid in this patient's blood. What is the margin of ei lower limit upper limit margin of error (b) What conditions are necessary for your calculations? (Select all that apply.) normal distribution of uric acid uniform distribution of uric acid o is known On is large Oo is unknown (c) Interpret your results in the context of this problem. O The probability that this interval contains the true average uric acid level for this patient is 0.95. O We are 5% confident that the true uric acid level for this patient falls within this interval. O We are 95% confident that the true uric acid level for this patient falls within…
One Sample T-test and Confidence Interval The average monthly cell phone bill was reported last year to be $53.88 by the U.S. Wireless Industry. Random sampling of a large cell phone company found the current monthly cell phone charges (in dollars) in the table on the left. At the 0.05 level of significance, can it be concluded that the average cell phone bill has increased? Also find an 85% confidence interval estimate of the current mean monthly cell phone bill. Cell Phone Bills 55.83 60.47 58.60 49.88 52.45 51.29 32.98 49.20 70.42 50.02 75.04 41.88 35.00 61.58 58.99 61.25 62.00 80.21 73.60 71.52 42.80 38.65 49.54 61.88 46.30 74.10 48.63 63.45 44.43 36.77 71.96 77.79 42.13 63.24 67.56 74.44 31.77 76.91 73.04 79.92 67.10 78.13 37.95 45.60 68.88 54.47 43.47 54.57 76.12 80.45 38.08 57.21 54.28 a. State the random variable and the parameter in words. b. State the null and…
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attached a picture! need help with the 99% confidence interval!

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t.test (improved,local, var.equal=TRUE, paired=TRUE) Paired t-test data: improved and local t = 2.3294, df = 24, p-value = 0.02858 %3D alternative hypothesis: true difference in means is not equal to 95 percent confidence interval: