2 Five samples of cloth are cut in half and one half of each sample is treated with a3.strengthening compound. Both the treated and untreated pieces are pulled by atesting machine till they break and the weights in kg to break them are givenbelow in Table 1. From previous experiments, it is known that the variance ofeach population of possible weights is 0.5 kg. Explain your test adopted todetermine whether or not the use of the strengthening compound has had anyeffect on the breaking strength at the 10 per cent significance level.Table 1. Breaking strength in kg weight for treated and untreated clothesBreaking strength in kg-weightSample clothes12345Treated pieces20.521.319.818.420.6Untreated pieces20.120.819.718.520.084

Given the data as Sample clothes Treated pieces Untreated pieces 1 20.5 20.1 2 21.3 20.8 3…

Answered: 2 Five samples of cloth are cut in half…

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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Q4
Two groups of mice were injected with a measured amount of tumor pulp. The first group of 27 mice was given a high dosage of chemotherapy while the second group of 30 mice was given a low dosage of chemotherapy. After forty days, the first group had an average tumor size of 0.51cc with a variance of 0.10; the second group had an average tumor size of 0.64cc with a variance of 0.045. Were the tumor sizes of group 1 significantly smaller than those of group 2? What is the observed test statistic?
A random sample of 20 newly-born baby boys showed an average weight of 7.40 pounds while a sample of 25 newly-born baby girls showed a mean weight of 6.50 pounds. If the variance of all newly-born babies are 1.25 pounds, can we say that newly-born baby boys are heavier than newly-born baby girls using the 0.05 level?
Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 7.6 parts/million (ppm). A researcher believes that the current ozone level is not at a normal level. The mean of 1313 samples is 7.2 ppm with a variance of 0.49 Assume the population is normally distributed. A level of significance of 0.02 will be used. Make the decision to reject or fail to reject the null hypothesis.
We are running a hypothesis test to see if blood pressure of African zebras differs significantly from Arabian horses. We assume blood pressure in both populations is normally distributed and both populations have the same population variance in blood pressure which is unknown to us, so we use the sample variances to estimate the standard error for our calculations. We have a sample of size 9 for the African zebras and a sample of size 16 for the Arabian horses. How many degrees of freedom does our test statistic have?
The data from an independent-measures research study produce a sample mean difference of 4 points, and each sample has a variance of 10. If there are n = 20 scores in each sample, then what is the estimated standard error of the difference between means?
Professor Hayashi has implemented a new book for his Fall 2013 statistics course. He wants to see if the new book has an effect on the exam average for Exam 3. He knows that the Exam 3 mean for all students who have taken his statistic course over the past 10 years is 77.2. He collects some data from a sample of 9 students and finds that the Exam 3 mean for his Fall 2013 statistics students is 81.7 with a variance of 32.49.Measure effect size using percentage of variance
The variability in the amounts of impurities present in a batch of chemical used for a particular process depends on the length of time the process is in operation. A manufacturer using two production lines, 1 and 2, has made a slight adjustment to line 2, hoping to reduce the variability as well as the average amount of impurities in the chemical. Samples of 25 and 25 measurements from the two batches yield 1.04, respectively 0.51 variances. Do the data present sufficient evidence to indicate that the process variability is less for line 2? Use 5% significance level. Conduct a hypothesis test of the airline executive’s belief. [α = 0.05]
A nutrition store in the mall is selling "memory booster" which is a concoction of herbs and minerals that is intended to improve memory performance. To test the effectiveness of the herbal mix, a researcher obtains a sample of n=16 participants and has each person take the suggested dosage each day for 4 weeks. At the end of the four week period, each individual takes a standard memory test. The scores from the participants produced a M=26 with a sample variance of s2=64. In the general population, the standardized test is known to have a μμ=20. Do the sample data support the conclusion that memory booster has a significant effect? Use a two-tailed test with αα=.05. Compute Cohen's d to measure the size of the treatment effect.
A certain producer of chocolate has some problems with the variability of five suppliers of cocoa. Since the cocoa quality is critical to the final chocolate product, the company has decided to perform a DOE to compare these suppliers. According to experiences with customers, the capacity of detection of any variation is 3.5 units. Estimate the sample size needed to get at least a power of 85%, given that the historical variance of the process is 2.5 units.
5. We don't want predictor variables that are too highly correlated with each other or they are redundant (say ≥±.70) or multicollinear (2±.90), which means the variables are measuring pretty much the same thing so why have them all in the equation. What was the finding for redundancy between SPI Scientist and SPI Practitioner? a. No issue with redundancy since the shared variance between the two variables was only 4% (r2 = .04). b. There was an issue with redundancy between the two variables shared 62% of the variance in the relationship.
2. The average zone of inhibition (in mm) for mouthwash L as tested by the medical technology students has been known to be 9mm. A random sample of 10 mouthwash L was tested and the test yielded an average zone of inhibition of 7.5mm with a variance of 25 mm. Is there enough reason to believe that the anti-bacterial property of the mouthwash has decreased? Test the hypothesis that the average zone of inhibition of the mouthwash is no less than 9mm using 0.05 level of significance. A. State the hypotheses. B. Determine the test statistic to use. C. Determine the level of significance, critical value, and the decision rule. D. Compute the value of the test statistic. E. Make a decision. F. Draw a conclusion.

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