It is recommended that adults get 8 hours of sleep each night. A researcher hypothesized college students got less than the recommended number of hours of sleepeach night, on average. The researcher randomly sampled 50 college students and calculated a sample mean of 7.9 hours per night. The researcher performed ahypothesis test. What is the null hypothesis?.....Choose the correct answer below.O A. Ho: Hx <7.9 hours per nightВ. Но: Их= 7.9 hours per nightO C. Ho: Hx < 8 hours per nightO D. Ho: Hx = 8 hours per night

We want to find the null hypothesis

Answered: It is recommended that adults get 8…

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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A grassroots group opposed to a proposed increase in the gas tax claimed that the increase would hurt working-class people the most, since they commute the farthest to work. Suppose that the group randomly surveyed 24 individuals and asked them their daily one-way commuting mileage. The results are below. Using a 5% significance level, test the hypothesis that the three mean commuting mileages are the same. (Let 1 = Working-class, 2 = Professional (middle incomes), and 3 = Professional (wealthy).) Part (a) Part (b) Part (c) Part (d) Part (e) Part (f) Working-class 18.1 27.5 49.7 10.1 65.8 47.2 20.1 51.8 Professional (middle incomes) 17.0 17.5 22.7 7.6 10.2 2.9 7.3 14.5 What is the test statistic? (Round your answer to two decimal places.) Part (g) What is the p-value? (Round your answer to four decimal places.) Professional (wealthy) 9.1 7.2 4.9 12.8 11.6 29.0 16.3 9.5
A marketing agency wanted to determine if a new commercial changed people's opinion about a company. They recruited n=30 participants to watch the commercial. First, everyone was givien a survey to determine their attitudes toward the company. Then, everyone watched the new commercial. Finally, they were given the same survey again and asked about their attitudes toward the comapny after seeing the commercial. The agency wanted to see if attitudes were changed by the commercial. What statistical test should they use?
One study claimed that 88 % of college students identify themselves as procrastinators. A professor believes that the claim regarding college students is too high. The professor conducts a simple random sample of 272 college students and finds that 231 of them identify themselves as procrastinators. Does this evidence support the professor's claim that fewer than 88 % of college students are procrastinators? Use a 0.02 level of significance. Step 2 of 3: Compute the value of the test statistic. Round your answer to two decimal places.
Past studies have indicated that the percentage of smokers was estimated to be about 30%. Given the new smoking cessation programs that have been implemented, you now believe that the percentage of smokers has reduced. You randomly surveyed 2414 people and found that 680 smoke. Use a 0.05 significance level to test the claim that the percentage of smokers has reduced. a) Identify the null and alternative hypotheses? Hp: ? v H:? v b) What type of hypothesis test should you conduct (left-, right-, or two-tailed)? O left-tailed O right-tailed O two-tailed c) Identify the appropriate significance level. Check Answer
A hypothesis regarding the weight of newborn infants at a community hospital is that the mean is 6.6 pounds. A sample of seven infants is randomly selected and their weights at birth are recorded as 9.0, 7.3, 6.0, 8.8, 6.8, 8.4, and 6.6 pounds. When evaluated using the 5% level of significance, what is the decision and interpretation of the null hypothesis? Show all steps.
The Bureau of Labor Statistics reports that the official unemployment rate for Black people was 10.4% and 4.7% for White people in February 2015. Select all correct answers for this question. O The samples of white and black people are independent. The explanatory variable is the unemployment rate. The response variable is the unemployment rate. The response variable is race.
A sociologist claims that the students’ college GPA’s in unrelated to their families’ incomes. For a random sample of 20 students the coefficient of correlation was computed as 0.40. Using the 0.01 significance level, do the formal hypothesis test.Remember to show your steps (including all relevant information) show the steps on how to solve it
The US Department of Energy reported that 48% of homes were heated by natural gas. A random sample of 331 homes in Oregon found that 132 were heated by natural gas. Test the claim that proportion of homes in Oregon that were heated by natural gas is different than what was reported. Use a 1% significance level. Give answer to at least 4 decimal places. What are the correct hypotheses? (Select the correct symbols and use decimal values not percentages.) H0: H1: Based on the hypotheses, compute the following:Test Statistic = p-value = Based on the above we choose to The correct summary would be: that the proportion of homes in Oregon that were heated by natural gas is different than what the DOE reported value of 48%.
According to the most recent General Social Survey, 70% of adults watch TV nightly. A researcher believes that the proportion of college students who watch TV nightly is less than this. To test this claim the research samples 400 college students and finds that 276 watch TV nightly. At the 0.05 level of significance does this data provide evidence to suggest that the proportion of college students who watch TV nightly is less than 70%? Select the correct hypothesis test for this problem:
Z?
The coach of a very popular men’s basketball team claims that the average distance the fans travel to the campus to watch a game is 35 miles. The team members feel otherwise. A sample of 16 fans who travel to games was randomly selected and yielded a mean of M= 36 miles and s= 5 miles. Test the coach’s claim at the 5% (.05) level of significance. one-tailed or two-tailed test: State the hypotheses: df= tα or t value for the critical region = sM = t (test statistic)= Decision:

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