Is a weight loss program based on exercise less effective thana program based on diet? The 54 overweight people put on astrict one year exercise program lost an average of 27 poundswith a standard deviation of 5 pounds. The 47 overweightpeople put on a strict one year diet lost an average of 29pounds with a standard deviation of 6 pounds. What can beconcluded at the a = 0.10 level of significance?a. For this study, we should useSelect an answerb. The null and alternative hypotheses would be:Ho: Select an answerSelect an answerSelect an answer (please enter a decimal)H₁: Select an answerSelect an answerSelect an answer (Please enter a decimal)c. The test statistic ? ◊=your answer to 3 decimal places.)d. The p-value =4 decimal places.)(please show(Please show your answer to

Given that Sample sizes n1=54 , n2=47 Sample means X̄1=27 , X̄2=29 s1 =5…

Answered: a program based on diet? The 54…

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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Use the following scenario to answer the next 6 questions. An engineer has designed a valve that will regulate water pressure on an automobile engine. The valve was tested on 30 engines and the mean pressure was 4.8 psi with a standard deviation of 1 psi. If the valve was designed to produce an overall mean pressure of 5.0 psi, is there sufficient evidence at the 10% level of significance that the valve performs below specifications? In this scenario, the decision to use a two-tailed versus a one-tailed test affected our ability to reject the null (and thus our conclusion). Group of answer choices True False
A clinical researcher performs a clinical trial on 12 patients to determine whether a drug treatment has an effect on serum glucose. The sample mean glucose of the patients before and after the treatment are summarized in the following table. The sample standard deviation of the differences was 6. Sample mean glucose (mg/dL) Before treatment 114 After treatment 108 What type of hypothesis test should be performed?Select: Left-tailed z-test Unpaired t-test Two-tailed z-test Paired t-test What is the test statistic? What is the number of degrees of freedom? Does sufficient evidence exist to support the claim that the drug treatment has an effect on serum glucose at the α=0.05 significance level? Select: _____Yes OR _____No
According to a large oil company, the average cost per barrel of oil last year was $47. You believe that due to the current oil crisis in Russia and Iran that the cost of a barrel of oil has decreased. You collect a simple random sample of 60 barrels of oil and find an average price of $42 and a standard deviation of $12. Use this information to answer the following questions and conduct the hypothesis test with a level of significance of a = .01. You may use Statcrunch or a calculator. Write all non-calculation answers in complete sentences. a. Which test would you use and why? b. Write the null and alternative hypothesis in complete sentences. (Make sure you mark which is the null and which is the alternative.) c. Write the null and alternative hypothesis as symbols. d. State the requirements that have been met to run the test. e. What is the test statistic and the p-value? (Make sure you use the right notation.) f. Should you reject or fail to reject the null hypothesis? g. State…
Suppose the national average dollar amount for an automobile insurance claim is $670.543. You work for an agency in Michigan and you are interested in whether or not the state average is different from the national average. The hypotheses for this scenario are as follows: Null Hypothesis: μ = 670.543, Alternative Hypothesis: μ ≠ 670.543. A random sample of 27 claims shows an average amount of $668.645 with a standard deviation of $68.6867. What is the test statistic and p-value for this test? Question 8 options: 1) Test Statistic: -0.144, P-Value: 0.44345 2) Test Statistic: -0.144, P-Value: 0.5566 3) Test Statistic: -0.144, P-Value: 1.5566 4) Test Statistic: 0.144, P-Value: 0.8869 5) Test Statistic: -0.144, P-Value: 0.8869
The average salary for American college graduates is $44,300. You suspect that the average is different for graduates from your college. The 61 randomly selected graduates from your college had an average salary of $44,099 and a standard deviation of $4,460. What can be concluded at the a = 0.10 level of significance? a. For this study, we should useSelect an answer b. The null and alternative hypotheses would be: Họ: vSelect an answer H: ? vSelect an answer c. The test statistic (please show your answer to 3 decimal places.) d. The p-value = (Please show your answer to 4 decimal places.) e. The p-value is va the null hypothesis. f. Based on this, we should Select an answer g. Thus, the final conclusion is that ... O The data suggest that the population mean is not significantly different from 44,300 at a = 0.10, so there is statistically insignificant evidence to conclude that the population mean salary for graduates from your college is different from 44,300. The data suggest that…
1 Jump to level 1 A pipes manufacturer makes pipes with a length that is supposed to be 11 inches. A quality control technician sampled 32 pipes and found that the sample mean length was 11.09 inches and the sample standard deviation was 0.2 inches. The technician claims that the mean pipe length is not 11 inches. What type of hypothesis test should be performed? Select What is the test statistic? Ex: 0.123 What is the number of degrees of freedom? Ex: 20 Does sufficient evidence exist at the a= 0.05 significance level to support the technician's claim? Select Check HIN 60°F
We want to conduct a hypothesis test of the claim that for middle-aged adults the population's mean of their cholesterol levels is more than 199.3 mgdL. We choose a random sample of such levels. The sample has a mean of 197.3 mgdL and a standard deviation of 19.5 mgdL. For each of the following sampling scenarios, choose an appropriate test statistic for our hypothesis test on the population mean. Then calculate that statistic. Round your answers to two decimal places. (a) The sample has size 105, and it is from a non-normally distributed population with a known standard deviation of 19.2. z = t = It is unclear which test statistic to use. (b) The sample has size 12, and it is from a normally distributed population with an unknown standard deviation. z = t = It is unclear which test statistic to use.
Which is cheaper: eating out or dining in? The mean cost of a steak, salad, and rice bought at the grocery store is $13.04. A sample of 30 neighborhood restaurants showed a mean price of $12.15 and a standard deviation of $2 for a comparable restaurant meal. 1.1) Develop appropriate hypotheses for a test to determine whether the sample data support the conclusion that the mean cost of a restaurant meal is less than $13.04. 1.2) At alpha = 0.05, what is your conclusion? Show a detailed solution to support your conclusion. (Find p-values at: https://www.socscistatistics.com/pvalues/tdistribution.aspx)
Americans are about 1 inch taller but nearly 25 pounds heavier than in 1960. The average BMI increased from approximately 25 to 28 in 2002. Boston is considered one of americas healthiest cities. Is the weight gain since 1960 similar in boston? Sample of n=25 and mean increase of 17 with standard deviation of 8.6. Is boston statistically significantly different in terms of weight gain since 1960? Run the appropriate test at 5%level of significance. A.state the null and the research hypotheses, b. Idebtify critical values of z 9r t used to define the decision rule, c. Calculate the test statistic, d. Report the p-value, e. Report your findings
Given two dependent random samples with the following results: Population 1 35 34 45 30 49 25 34 Population 2 44 24 40 42 38 31 43 Use this data to find the 90% confidence interval for the true difference between the population means. Assume that both populations are normally distributed. Step 2 of 4 : Calculate the sample standard deviation of the paired differences. Round your answer to six decimal places. please highlight the answer
Two treatments for a new infectious disease have been developed to reduce hospitalizations. Each treatment was given to different groups of patients. Treatment 1 was given to 65 people. Subjects of treatment 1 had a mean hospital stay of 5.6 days with a standard deviation of 2.6 days. Treatment 2 was given to 54 people. Subjects of treatment 2 had a mean hospital stay of 8.4 days with a standard deviation of 3.0 days. Use the method of hypothesis testing to test the claim that the mean time in the hospital for treatment 1 is shorter than the mean time in the hospital for treatment 2 assuming a 0.05 significance level. (a) List the null and alternative hypotheses for this test. (b) Compute the value of the test statistic (c) Determine the rejection region.(d) Determine whether or not we can reject the null hypothesis.
A sample of 42 observations is selected from one population with a population standard deviation of 4.5. The sample mean is 100.5. A sample of 56 observations is selected from a second population with a population standard deviation of 3.8. The sample mean is 98.5. Conduct the following test of hypothesis using the 0.10 significance level. H₂ H₁ H₂ H₁ H₁ H₂ a. Is this a one-tailed or a two-tailed test? O One-tailed test O Two-tailed test b. State the decision rule. (Negative values should be indicated by a minus sign. Round your answers to 2 decimal places.) The decision rule is to reject H0 if z is the interval (

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