The standard treatment for a disease works in 0.65 of all patients.A new treatment ispatients (of similar general health) isproposed. Is it better? An initial clinical trial of n = 100conducted: 73 people are cured. Assume the selection of patients is random(a) Perform a significance test to test whether the treatment is better at a 10% significance level.lp Z-=l.v5.= 73 = .731 oowe are go corfident that the populaticr.9rofortienlbe be(b) Find the power of the test under the alternative ofHa 0.810 ,812-neu PP.(I-PtoYl

Answered: The standard treatment for a disease…

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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Women are recommended to consume 1770 calories per day. You suspect that the average calorie intake is different for women at your college. The data for the 11 women who participated in the study is shown below: 1916, 1992, 1708, 1764, 1685, 1667, 1905, 2078, 1922, 1816, 1786 Assuming that the distribution is normal, what can be concluded at the αα = 0.05 level of significance? The null and alternative hypotheses would be: H0:H0: ? p μ Select an answer < ≠ = > Correct H1:H1: ? p μ Select an answer ≠ < = > Correct The test statistic ? t z Correct = (please show your answer to 3 decimal places.) The p-value = (Please show your answer to 4 decimal places.)
Q4
15% of all Americans suffer from sleep apnea. A researcher suspects that a lower percentage of those who live in the inner city have sleep apnea.Of the 352 people from the inner city surveyed, 39 of them suffered from sleep apnea.Test the calim at the level of significance of αα = 0.01? What type of test should be run? t-test of a mean z-test of a proportion The alternative hypothesis indicates a two-tailed test one-tailed test
Answer it correctly please. Explain . I will rate accordingly with multiple votes.
Researchers wanted to know if therapy could help smokers quit smoking. Of the 71 smokers they tested, 39 were smoking one year after treatment. Use a 0.05 significance level to test the claim that the majority are smoking one year after treatment. What is/are the critical value(s)? ттT Arial 3 (12pt) v TE -E · -
Use formulas to solve this and also with explaination
Is the proportion of wildfires caused by humans in the south higher than the proportion of wildfires caused by humans in the west? 367 of the 526 randomly selected wildfires looked at in the south were caused by humans while 390 of the 594 randomly selected wildfires looked at the west were caused by humans. What can be concluded at the a = 0.01 level of significance? a. For this study, we should use Select an answer b. The null and alternative hypotheses would be: Ho: Select an answer > Select an answer > Select an answer O (please enter a decimal) H j: Select an answer Select an answer > Select an answer O (Please enter a decimal) c. The test statistic (please show = your answer to 3 decimal places.) d. The p-value (Please show your answer to %3D 4 decimal places.)
Mark performed a two-sample z-test for proportions to test the hypothesis that there was no difference in the proportion who support increasing student fees between male and female students at a particular university. Mark obtained a z-statistic of 0. Based on this information, which of the following is always true. The sample proportions for male and females will be the same as the hypothesized proportions for male and female students at this university. The sample proportions for both male and female students at this university will be 0. All male and female students in the sample will feel the same way about supporting the increase in student fees. The sample proportions will be the same for both male and female students at this university. Test whether p1>p2 The sample data are x1=116,n1=258,x2=135,n2=311. Conduct a test at the α=0.10 level of significance. Determine the correct null and alternative hypothesis below. Determine the test statistic. Find the P-value: What is…
A new medication is being developed to treat severe nausea and vomiting among chemotherapy patients. The information below was generated from a small sample size (n = 11) and shows the number of times a chemotherapy patient vomited both before and after treatment due to the side effects of chemotherapy. Set up and interpret the results of a sign test at the alpha equals .05 level to determine if there is a difference in symptoms before and after treatment. Chemotherapy Patient Before Treatment After Treatment Difference Sign 1 9 3 2 12 5 3 14 7 4 9 4 5 8 2 6 11 3 7 15 7 8 12 8 9 7 8 10 9 5 11 8 7 A) We fail to reject H0, which states the median difference in the number of vomiting episodes in a day before and after treatment are equal because 1 is less or equal to critical value 1 for a sign test…
Suppose IQ scores were obtained for 20 randomly selected sets of couples. The 20 pairs of measurements yield x = 101.02, y = 100.75, r = 0.852, P-value = 0.000, and y = 5.93 + 0.94x, where x represents the IQ score of the husband. Find the best predicted value of y given that the husband has an lQ of 100? Use a significance level of 0.05. %3D Click the icon to view the critical values of the Pearson correlation coefficient r. The best predicted value of y is Critical values of the pearson correlation coefficient r (Round to two decimal places as needed.) Critical Values of the Pearson Correlation Coefficient r X = 0.05 a = 0.01 INOTE: To test Ho: p=0 n Jagainst H,: p+0, reject Ho if the absolute value of r is greater than the critical value in the table. 4 0.950 0.990 0.878 0.959 0.811 0.917 7 0.754 0.875 0.707 0.834 0.666 0.798 10 0.632 0.765 11 0.602 0.735 12 0.576 0.708 13 0.553 0.684 14 0.532 0.661 15 0.514 0.641 16 0.497 0.623 17 0.482 0.606 18 0.468 0.590 19 0.456 0.575 20 0.444…
Refer to the below table. Using an alpha = 0.05, test the claim that IQ scores are the same for children in three different blood lead level groups: low lead level, medium lead level, and high lead level). One-Way Analysis of Variance Summary Table for IQ Measurements for Children among Three Blood Lead Level Groups: Low Lead Level, Medium Lead Level, and High Lead Level. Source df SS MS F p Between-group (treatment) 2 469.1827 2677.864 2.30 0.104 Within-group (error) 118 203.6918 11745.05 Total 120
A clinical study was conducted on a medicine and the data obtained as follows: calculate Sensitivity Diseased •True Positive: 1000 •True negative: 50 Non diseased •True positive: 100 •True negative: 800

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