The one-sample t-statistic for a test of Ho: µ = 42 versus H1: µ < 42 based on n = 15 observations has the value 1 = -0.867, where Ho and Hjare the null and alternative hypotheses, respectively. The sample size is denoted by n. Use this t-table to determine which two P-values bracket the P-value of the test. Report your answers in decimal form as listed in the t-table. k P-value of the test <

The one-sample t-statistic for a test of Ho: µ = 42 versus H1: µ < 42 based on n = 15 observations has the value 1 = -0.867, where Ho and Hjare the null and alternative hypotheses, respectively. The sample size is denoted by n. Use this t-table to determine which two P-values bracket the P-value of the test. Report your answers in decimal form as listed in the t-table. k P-value of the test <

Name: The one-sample t-statistic for a test of Ho: µ = 42 versus H1: µ < 42
Uploaded: 2024-03-20T07:07:00.000Z
Channel: Bartleby
Description: The one-sample t-statistic for a test of Ho: µ = 42 versus H1: µ < 42 based on n = 15 observations has the valuet = -0.867, where Họ and Hjare the null and alternative hypotheses, respectively. The sample size is denoted by n.Use this t-table to det

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

With 0.05 significance level, H1: p > 0.5, and the value of test statistics is z = 1.25. Then the P-value is equal to 0.8944 -0.1056 0.1056 None
If the test statistic for a right sided test is z = 1.71, then the p-value isSelect one: A.0.9573 B.0.9564 C.0.0436 D.0.0427
The yield per tract in a certain region of farmland is known to have a populationstandard deviation of 13 bushels. Using the old seed variety, the population mean yield μwas 39 bushels. A farmer tests a new seed variety on a random sample of 36 tracts, andfinds a sample mean yield of 45.2 bushels. Is the population mean yield using the newseed variety different from 39? Use α= 0.10.a. State the null hypothesis . __________________State the alternative hypothesis.____________________b. Find the test statistic. _____ c. Find the P-value for the test. _____________
For each of the tests, what would be the conclusion if the test were carried out using the α = 0.05 level of significance? H0 : µ = 120, H1:µ ≠ 120, n = 36, S.D of population= 10.3, sample mean = 117.0
Got stuck, and wasn’t sure what to do.
I randomly sample 219 students and ask them to provide their USF GPA. I wish to determine if the average GPA of all USF students ex mu + 3.25 ceeds 3.25. Unfortunately, I generated the printout shown here: One-Sample T Test Null Hypothesis: H - 3.25 Alternative Hyp: u- 3.25 Variable Mean SE DF GPA 3.3121 0.0340 1.82 218 0.0694 Cases Included 219 What assumption is needed for this analysis to be valid? None. The Central Limit Theorem applies and no other assumptions are needed. The GPAS of all USF students need to be approximately normally distributed The mean GPA of all USF students need to be approximately normally distributed
In one study, researchers at McGill University recruited127 people with high cholesterol and split them into twogroups. One took a probiotic supplement twice a dayfor nine weeks, while the second group took a placebo.The probiotic group saw their total cholesterol drop by 9 percent and their LDL, or “bad cholesterol,” fall almost12 percent. In a different study, conducted in Britain,80 volunteers “were given probiotics for six weeks andthen switched, later on, to a placebo. That study found nodifference in cholesterol levels when the subjects took thesupplement versus the dummy pills.”Which of the following is true: a) The McGill study is a controlled randomized experi-ment, while the British study is an observational study. b) The McGill study is a completely randomized ex-periment, while the British study is a matched pairs experiment.c) The results from the McGill study are more validbecause it includes blinding while the British studydoes not.d) We have reason to believe that the…
Use the following R output to answer the question. Use the following R output to answer the question. >chisq.test(data) Pearsons Chi-square test data:data X-squared=4.7194, df=2, p-value=0.09445 If the significance level is 5%, what conclusion can you make? Group of answer choices A. fail to reject the alternative hypothesis B. cannot determine C. fail to reject the null hypothesis D. reject the null hypothesis and accept the alternative hypothesis
For each of the tests, what would be the conclusion if the test were carried out using the α = 0.05 level of significance? H0 : µ = 120, H1:µ ≠ 120, n = 36, S.D of population= 10.3, sample mean = 117.0 Should the null be rejected?
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…
In a 4-week study about the effectiveness of using magnetic insoles to treat plantar heel pain, 58 subjects wore magnetic insoles and 40 subjects wore nonmagnetic insoles. The results are shown at the right. At a= 0.06, can you support the claim that there is a difference in the proportion of subjects who feel better between the two groups? Assume the random samples are independent. Complete parts (a) through (e). Do you Feel Better? Magnetic Insoles I Yes 20 I No 38 Nonmagnetic Insoles Yes 17 No 23 (Use a comma to separate answers as needed. Type an integer or a decimal. Round to two decimal places as needed.) Identify the rejection region(s). Select the correct choice below and fill in the answer box(es) within your choice. (Round to two decimal places as needed.) O A. z O B. z> O D. z< C. (c) Find the standardized test statistic. (Round to two decimal places as needed.) (d) Decide whether to reject or fail to reject the null hypothesis Choose the correct answer below.
The recidivism rate for convicted sex offenders is 12%. A warden suspects that this percent is different if the sex offender is also a drug addict. Of the 322 convicted sex offenders who were also drug addicts, 48 of them became repeat offenders. What can be concluded at the a = 0.01 level of significance? a. For this study, we should use z-test for a population proportion V b. The null and alternative hypotheses would be: Ho: ? v Select an answer (please enter a decimal) H1: ? v Select an answer v (Please enter a decimal) c. The test statistic ? v = (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 ? V a f. Based on this, we should Select an answer v the null hypothesis. g. Thus, the final conclusion is that ... O The data suggest the population proportion is not significantly different from 12% at a = 0.01, so there is statistically insignificant evidence to conclude that the population proportion of…