P-value=(Round to three decimal places as needed.)What is the conclusion based on the hypothesis test?The P-value isis higher than the cure rate for those given a placebo.▼the significance level of α = 0.05, sob. Test the claim by constructing an appropriate confidence interval.The 90% confidence interval is < (P₁-P₂) <-(Round to three decimal places as needed.)What is the conclusion based on the confidence interval?Because the confidence interval limitsappears that the cure rate isthe null hypothesis. There0, it appears that the two cure rates arefor the oxygen treatment than for the placebo.✔ evidence to support the claim that the cure rate with oxygen treatment▼Because the confidence interval limits include▼values, it In a study of treatments for very painful "cluster" headaches, 153 patients were treated with oxygen and 157 other patients were given a placebo consisting of ordinary air. Among the 153 patients inthe oxygen treatment group, 120 were free from headaches 15 minutes after treatment. Among the 157 patients given the placebo, 20 were free from headaches 15 minutes after treatment. Use a0.05 significance level to test the claim that the oxygen treatment is effective. Complete parts (a) through (c) below.a. Test the claim using a hypothesis test.Consider the first sample to be the sample of patients treated with oxygen and the second sample to be the sample of patients given a placebo. What are the null and alternative hypotheses forthe hypothesis test?O A. Ho: P1 P2H₁: P₁ = P2O D. Ho: P1 P2H₁: P₁ P2Identify the test statistic.z=0(Round to two decimal places as needed.)OB. Ho: P1 P2H₁: P₁P2O E. Ho: P1H₁: P₁P2P2OC. Ho: P1 P2H₁: P₁P₂OF. Ho: P1H₁: P₁P2P2

There are two independent samples which are oxygen and placebo. We have to test whether the oxygen…

Answered: In a study of treatments for very…

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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Polyunsaturated fatty acids in the diet favorably affect several risk factors for cardiovascular disease. The principal dietary polyunsaturated fat is linoleic acid. Suppose we randomly selected twelve adults for our study. We took their blood pressure measurements (systolic blood pressure) at the start and recorded them for each adult. Then we placed required these adults to consume 23g/day of safflower oil which is high in linoleic acid for four weeks. At the end of the four weeks, we took another systolic blood pressure(SBP) measurement. The paired data is provided below: Subject 1 2 3 4 5 6 7 8 9 10 11 12 Baseline SBP 119.67 100.00 123.56 109.89 96.22 133.33 115.78 126.39 122.78 117.44 111.33 92.39 1-month SBP 117.33 98.78 123.83 107.67 95.67 128.89 113.22 121.56 126.33 110.39 107.00 93.06 Use the appropriate nonparametric test at alpha level of 0.05, to test whether the diet heavy in linoleic acid increases systolic blood pressure (SBP).
Two professors at a local college developed a new teaching curriculum designed to increase students' grades in math classes. In a typical developmental math course, 53% of the students complete the course with a letter grade of A, B, or C. In the experimental course, of the 15 students enrolled, 10 completed the course with a letter grade of A, B, or C. Is the experimental course effective at the a = 0.05 level of significance? Complete parts (a) through (g). (b) Verify that the normal model may not be used to estimate the P-value. npo (1 - Po) : (Round to one decimal place as needed.) Весause = 3.7 < 10, the normal model may not be used to approximate the P-value. (c) Explain why this is a binomial experiment. There is a fixed number of trials with two mutually exclusive outcomes. The trials are independent and the probability of success is fixed at 0.53 for each trial. (Type an integer or a decimal. Do not round.) (d) Determine the P-value using the binomial probability distribution.…
A small pilot study is conducted to investigate the effect of a nutritional supplement on total body weight. Six participants agree to take the nutritional supplement. To assess its effect on body weight, weights are measured before starting the supplementation and then after 6 weeks. The data are shown below. Is there a significant increase in body weight following supplementation? Run the test at a 5% level of significance. Demonstrate the five steps for hypothesis testing and explain your results. Subject Initial Weight Weight after 6 Weeks 1 155 157 2 142 145 3 176 180 4 180 175 5 210 209 6 125 126
According to a certain government agency for a large country, the proportion of fatal traffic accidents in the country in which the driver had a positive blood alcohol concentration (BAC) is 0.38. Suppose a random sample of 105 traffic fatalities in a certain region results in 51 that involved a positive BAC. Does the sample evidence suggest that the region has a higher proportion of traffic fatalities involving a positive BAC than the country at the a = 0.01 level of significance? Because npo (1 - Po 10, the sample size is 5% of the population size, and the sample the requirements for testing the hypothesis satisfied. (Round to one decimal place as needed.) can be reasonably assumed to be random, What are the null and alternative hypotheses? is given to be random, cannot be reasonably assumed to be random, Ho: versus H,: (Type integers or decimals. Do not round.) is given to not be random, Find the test statistic, zo. Z0 = (Round to two decimal places as needed.) Find the P-value.…
Two professors at a local college developed a new teaching curriculum designed to increase students' grades in math classes. In a typical developmental math course, 54% of the students complete the course with a letter grade of A, B, or C. In the experimental course, of the 15 students enrolled, 12 completed the course with a letter grade of A, B, or C. Is the experimental course effective at the a = 0.01 level of significance? Complete parts (a) through (g). (a) State the appropriate null and alternative hypotheses. Ho: (Type integers or decimals. Do not round.) versus H,:
Polyunsaturated fatty acids in the diet favorably affect several risk factors for cardiovascular disease. The principal dietary polyunsaturated fat is linoleic acid. Suppose we randomly selected twelve adults for our study. We took their blood pressure measurements (systolic blood pressure) at the start and recorded them for each adult. Then we placed required these adults to consume 23g/day of safflower oil which is high in linoleic acid for four weeks. At the end of the four weeks, we took another systolic blood pressure(SBP) measurement. The paired data is provided below: Subject 1 2 3 4 5 6 7 8 9 10 11 12 Baseline SBP 119.67 100.00 123.56 109.89 96.22 133.33 115.78 126.39 122.78 117.44 111.33 92.39 1-month SBP 117.33 98.78 123.83 107.67 95.67 128.89 113.22 121.56 126.33 110.39 107.00 93.06 Use the appropriate nonparametric test at alpha level of 0.05, to test whether the diet heavy in linoleic acid increases systolic blood pressure (SBP). Use Steps 0-5
A researcher wanted to examine whether type of dog food (freeze dried meat vs generic brand kibble) would affect dogs’ attention span. The researcher hypothesized that dogs who ate freeze dried meat would have longer attention spans than those who ate generic brand kibble. The researcher used a single sample of 13 subjects, that took part in both treatment conditions. The researcher set the alpha level at .01. The mean of the difference scores (MD)was 7.0 and the SSD for the difference in attention spans was 144. What is the calculated test statistic to the nearest thousandth? Be sure to put the positive or negative sign in front of it. Note: SHOW ALL WORK!!! Based on your work above, do you reject or retain the null? reject retain Based on your conclusion regarding the null, was there a treatment effect? Yes No
Suppose IQ scores were obtained for 20 randomly selected sets of siblings. The 20 pairs of measurements yield x=99.75, y=101, r=0.921, P-value=0.000, and y=2.6+0.99x, where x represents the IQ score of the older child. Find the best predicted value of y given that the older child has an IQ of 102? Use a significance level of 0.05.
Select all that apply: We should reject H0. We should not reject H0. At the 1% significance level, the data do not provide sufficient evidence to conclude that the amount of time American teenagers watch television per week is different from 14.1 hours. At the 1% significance level, the data provide sufficient evidence to conclude that the amount of time American teenagers watch television per week is different from 14.1 hours.
The data in the attached image represents blood pressure based on a new trial medication used in an experiment involving a total of 50 male and female subjects. There were two levels of treatment: 0 = placebo and 1 = treatment group. Family History Blood Pressure = Fam-Hist-BP; from the paternal side = PS; from the maternal side = MS; from both sides = BS; none on both sides or not known = N.; Blood pressure before the experiment = BP-Before-Exp; Blood pressure after the experiment = BP-After-Exp. Task: Construct a histogram to represent the family BP history, the subjects' BP before treatment and the subjects' BP after treatment.
Find the value of the test statistic and find the 5% critical value
Clothes dried outside smell better than those dried in a dryer because of a process called photolysis. Sunlight breaks down compounds that cause odor. A trained smeller is asked to smell a random sample of 20 towels. The smeller does not know it, but all of the 20 towels have been dried in the sun. The smeller is blindfolded and is not allowed to touch the towels but leans forward to smell the towel then rates the towel on a smell-scale of 1 to 10, where 1 is the an extremely unpleasant smell and 10 is a very good smell. A smell rating of 8 is typically attributed to a towel that is dried in the dryer. The mean smell rating for these 20 towels is approximately normal with a mean of 8.45 and the standard deviation is 0.887. A statistician performs a test of Ho μ = 8 versus H₁ μ> 8 where is the true mean smell rating for all towels dried in the sun. This test yields a P-value of 0.0176. What conclusion would you make at the significance level a = 0.05? Because the P-value of 0.0176 < α =…

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