Using the accompanying data of a TV viewing survey, conduct a hypothesis test to determine if married individualswatch less TV than single individuals. Assume that the population variances are unequal.Click the icon to view the TV viewing survey.Is there sufficient evidence at the 0.05 level of significance that the married individuals watch less TV than singleindividuals?***Determine the null hypothesis, Ho, and the alternative hypothesis, H₁. Let married people be population 1 and singles bepopulation 2.Ho:H₁:(Type integers or decimals. Do not round.)Compute the test statistic.(Round to two decimal places as needed.)Find the p-value for the test.(Round to three decimal places as needed.).State the conclusion.The p-value isthe chosen value of a, sothe null hypothesis. There isevidence to conclude that the mean number of hours spent watching TV per week is less for married people than for Survey dataMaritalStatusMarriedMarriedMarriedMarriedMarriedMarriedMarriedMarriedMarriedMarriedHrs of TV viewingper week81511189141782132MaritalStatusSingleSingleSingleSingleSingleSingleSingleSingleSingleSingleSingleSingleSingleSingleSingleHrs of TV viewingper week401615111311341956162191617

There is no evidence to conclude that the mean number of hours spent watching TV per week is less…

Answered: Using the accompanying data of a TV…

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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Suppose that you want to perform a hypothesis test for a population mean. Assume that the population standard deviation is unknown and that the sample size is relatively small. In each exercise, we have given the distribution shape of the variable under consideration. Decide whether you would use the t-test, the Wilcoxon signed-rank test, or neither. Explain your answers. Symmetric bimodal
Choose the correct answer.If the P-value of a test of significance is 0.999 then:A. the null hypothesis should be rejected.B. the null hypothesis should not be rejected.C. the alternative hypothesis provides a plausible explanation of the data.D. Both A and C are true.
Women athletes at UČLA have a long-term graduation rate of 96%. Over the past several years, a random sample of 40 women athletes at UCLA showed that 37 eventually graduated. Do the data indicate that the population proportion of women athletes who graduate from UCLA is now less than 96%? Use a 5% significance level. You must identify the following in your answer in order to receive credit. Upload your response 1. State Ho and H1 2. Find the test statistic z (Round to two decimal places) 3. Find the p-value (Round to four decimal places) 4. Indicate the calculator key you used. 5. Do you reject Ho or fail to reject Ho? 6. State your conclusion: Upload Choose a File « Previous Next
According to data released by the Organization for Economic Cooperation and Development (OECD) in 2018, the French spend more time eating their meals than anybody else. Perform a hypothesis test to determine if a difference exists between the average time an American spends eating lunch when compared to a person from France. The given data represents the time, in minutes, that random French and American diners spent at lunch. Assume that the population variances are equal. American 21 French 24 17 18 O A. tx=2.34 O B. t=1.41 O C. t = 1.63 O D. t = -2.20 X 17 20 20 28 25 16 18 29 17 20 16 If Population 1 is defined as French diners and Population 2 is defined as American diners, then what is the test statistic for this hypothesis test? Round to two decimal places as needed. Clear all Check answer
instructor thinks that the mean score is higher. She samples 10 statistics students and obtains the scores: Grades 63.9 68.4 | 69 | 96 64.3 | 63.9 65 69 62.7 68.4 Test grades are believed to be normally distributed. Use a significance level of 5%. A. State the alternative hypothesis: HA: Ou + 65 Ομ 65 B. State the mean of the sample: C. State the standard error of the sample means:
A Data analytics professor wants to know whether level of college education influences the grade earned by students in his course. Last semester, 400 students completed his class: 150 freshmen, 100 sophomores, 100 juniors, and 50 seniors. At end of the semester, 80 students earned an A, 148 earned a B, and 172 earned a C. The distribution of grades is shown in the following table. Do a chi-square test at alpha=0.05. What is the null hypothesis for this test? A. The letter grades of the students are significantly different. B. The letter grades of the students are dependent of the level of college education. C. The letter grades of the students are identical. D. The letter grades of the students are independent of the level of college education.
Which of the following represents the correct order of a statistical analysis? Formulate a hypothesis; Gather data; Determine rejection of the null hypothesis Formulate a hypothesis; Determine rejection of the null hypothesis; Gather data Determine rejection of the null hypothesis; Gather data; Formulate a hypothesis Gather data; Formulate a hypothesis; Determine rejection of the null hypothesis
Which one of the following tests is NOT used for hypothesis testing? a. Sign Test b. Regression c. T tests d. F Test
Periodically, customers of a financial services company are asked to evaluate the company's financial consultants and services. Higher ratings on the client satisfaction survey indicate better service, with 7 the maximum service rating. Independent samples of service ratings for two financial consultants are summarized here. Consultant A has 10 years of experience, whereas consultant B has 1 year of experience. Use a = 0.05 and test to see whether the consultant with more experience has the higher population mean service rating. Assume the population variances are unequal. Consultant A Consultant B n1 = 16 n2 10 X1 = 6,81 X2 = 6.25 %3D S1 = 0.68 S2 = 0.75 (a) State the null and alternative hypotheses. O Hoi H1 - H2 = 0 Hai H1- Hz>0 O Ho: H1 - 42 = 0 Hại H1 - H2# 0 O Ho: H1 - H2 > 0 %3D Ha: H1 - H2 = 0 O Ho: H1 - Hz # 0 Hai Hy - Hz = 0 Hoi H1 - H2 = Hai H1- H2 = 0 (b) Compute the value of the test statistic. (Round your answer to three decimal places.) 19.191 (c) What is the p-value?…
A statistician changes her level of significance from .001 to .05. What impact will this change have on her risk of making a Type I and Type II error? Is the change in the level of significance a good decision? Why or why not?

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