5. 24 obese female adults participated in the trial to examine the efficacy of isometric and isotonicexercise methods: Twelve are chosen at random and given one exercise method; the other twelveare given the other exercise approach. The decreases in abdomen measurements are measuredafter eight weeks and are reported in millimetres. The results are as follows:IsotonicIsometric4.63.22.11.5MeanStandard Deviationi. State any necessary assumption about the underlying distribution of the data (assumethat both variabilities of the methods are equal).ii. Does the evidence support the idea that the isotonic approach is more successful? Testwith a = 0.05.Solution

Let X1 be the random variable that the efficacy of the isotonic method. X2 is the random variable…

Answered: i. State any necessary assumption about…

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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Violation of which assumption below for the two-factor ANVOA is not a cause for concern with large sample sizes? a. The populations from which the samples are selected must have equal variances. b. The populations from which the samples are selected must be normal. c. A violation of any assumption below would be a concern, even with large sample sizes. d. The observations within each sample must be independent.
According to a recent study in New York, 15% of shoppers are unfairly being charged sales tax in non-taxable items. Mr. Hypothesis feels this proportion is too low. He tests his belief by conducting a survey of 200 random shoppers. Which of the following would be the alternate hypothesis? a. The percent of shoppers unfairly being charged sales tax on non-taxable items < .15 b. The percent of shoppers unfairly being charged sales tax on non-taxable items > .15 c. The percent of shoppers unfairly being charged sales tax on non-taxable items = .15
A marriage counselor has traditionally seen that the proportion p of all married couples for whom her communication program can prevent divorce is 80%. After making some recent changes, the marriage counselor now claims that her program can prevent divorce in more than 80% of married couples. In a random sample of 215 married couples who completed her program, 180 of them stayed together. Based on this sample, is there enough evidence to support the marriage counselor’s claim at the 0.05 level of significance? Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places. A. State the null hypothesis Hoand the alternative hypothesis H1. Ho: H1: B. Find the value of the test statistic. (Round to three or more decimal places.) C. Find the critical value. (Round to three or more decimal places.) D. Is there enough evidence to support the marriage counselor's claim that the proportion of married couples for whom her program…
In a previous poll, 39% of adults with children under the age of 18 reported that their family ate dinner together seven nights a week. Suppose that, in a more recent poll, 422 of 1145 adults with children under the age of 18 reported that their family ate dinner together seven nights a week. Is there sufficient evidence that the proportion of families with children under the age of 18 who eat dinner together seven nights a week has decreased? Use the x = 0.01 significance level. Because npo (1-Po) = 272.4 > 10 and the sample size is the hypothesis are satisfied. (Round to one decimal place as needed.) What are the null and alternative hypotheses? Ho: P = versus H₁: ▼ V (Type integers or decimals. Do not round.) Find the test statistic, Zo- C less than 5% of the population size, and the sample is given to be random, the requirements for testing Zo = (Round to two decimal places as needed.) Find the P-value. P-value = (Round to three decimal places as needed.) Is there sufficient…
Do the t-test and test the null hypothesis that the two means are equal at the 95% confidence interval?
The Substance Abuse and Mental Health Services Administration (SAMHSA, 2017) estimates that 10.9% of the population of the United States age 18–24 had an episode of depression in the previous 12 months. Suppose a researcher takes a random sample of 225 United States adults aged 18–24. Determine the value of the sample proportion, p, such that 5% of samples of 225 United States adults age 18–24 are greater than p. You may find software or a z-table useful. Give your answer precise to three decimal places. p = II
How would i find: the 90% confidence interval is enter your response here <(p1−p2)< enter your response here.
(2) Choose the one alternative that best completes the statement or answers the question. Assume two independent random samples are available which provide sample proportions. For the first sample assume n₁ = 100 and x₁= 39. For the second sample, assume n₂= 100 and x₂= 49. Test the null hypothesis that the population proportions are equal versus the alternative hypothesis that the proportions are not equal at the 90% confidence level. Frame the test statistic by subtracting the proportion for population 1 from that for population 2. Pick an appropriate z value, p-value and conclusion. Round your answer to the nearest thousandth. Note that this is a two- tailed test. z-value = -1.425 p-value= 0.077 statistically significant z-value = 1.425 p-value= 0.077 statistically significant z-value = -1.425 p-value= 0.077 not statistically significant z-value = 1.425 p-value= 0.154 statistically not significant
Using ANOVA, a null hypothesis could look like this: Group of answer choices H0: µ1 =µ2 =µ3 H0: µ1 > µ2 > µ3 H0: µ1 = µ2 > µ3 H0: µ1 ≠ µ2 ≠ µ3

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i. State any necessary assumption about the underlying distribution of the data (assume that both variabilities of the methods are equal). ii. Does the evidence support the idea that the isotonic approach is more successful? Test with = 0.05