Consider a sample of 49 football games, where 27 of them were won by the home team. Use a 0.01 significance level to test the claim that the probability that the home team wins is greater than one-half.Identify the null and alternative hypotheses for this test. Choose the correct answer below.O A. Ho p=05H, p 0.5О В. Но р> 0. 5H, p=0.5O C. Ho p=0.5Hqip<0.5O D. Ho p=0.5H p>0.5Identify the test statistic for this hypothesis test.The test statistic for this hypothesis test is(Round to two decimal places as needed.)Identify the P-value for this hypothesis test.The P-value for this hypothesis test is(Round to three decimal places as needed.)Identify the conclusion for this hypothesis test.StatcrunchO Type here to search0 日問hpf4 IOIf8?f5f6f7f1of1f12insertprt scdeleteesc14411$&34backspacehome{EYpg uptabJKenterAcaps lockpause* shiftshift00%24米%23

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Answered: Consider a sample of 49 football games,…

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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Is Friday the 13th Unlucky? The table shows the number of hospital admissions for Friday the 6th and Friday the 13th. Assume that the two samples are randomly selected. At the 0.05 significance level, test the claim that there is no mean difference in the number of hospital admissions for Friday the 6th and Friday the 13th.(Be sure to subtract in the same direction). Friday 6th Friday 13th Difference 4 2 6 10 5 6 9 13 3 2 11 15 12 12 What are the correct hypotheses? (Select the correct symbols and values.)H0: Select an answer μ₁ μ(d) x̄₂ σ₁² x̄₁ s₁² p μ ? = < ≤ ≠ > ≥ Select an answer 0 μ₁ μ₂ x̄₁ p σ₁² s₁² x̄₂ μ H1: Select an answer x̄₂ μ₁ μ p μ₂ x̄₁ s₂² μ(d) σ₂² ? < ≥ ≠ = ≤ > Select an answer μ x̄₂ s₁² x̄₁ p 0 μ₂ σ₁² μ₁ Original Claim = Select an answer H₀ H₁ df = Based on the hypotheses, find the following: Test Statistic = (Round to three decimal places.)Critical value(s) = ±± (Round to three decimal places.)p-value = (Round…
Refer to the sample data for polygraph tests shown below. If one of the test subjects is randomly selected, what is the probability that the subject is not lying? Is the result close to the probability of 0.4720.472 for a negative test result? Did the Subject Actually Lie? No (Did Not Lie) Yes (Lied) Positive test results 1414 4343 Negative test results 3737 1414 The probability that a randomly selected polygraph test subject was not lying is _____.
A travel company is investigating whether the average cost of a hotel stay in a certain city has increased over the past year. The company recorded the cost of a one-night stay for a Friday night in January of the current year and in the previous year for 31 hotels selected at random. The difference in cost (current year minus previous year) was calculated for each hotel. Which of the following is the appropriate test for the company’s investigation? A one-sample zz-test for a population mean A A one-sample tt-test for a sample mean B A one-sample zz-test for a population proportion C A matched-pairs tt-test for a mean difference D A two-sample tt-test for a difference between means E
Explain the difference between statistical significance and practical significance. Choose the correct answer below. O A. Statistical significance means that the sample statistic is not likely to come from the population whose parameter is stated in the null hypothesis. Practical significance refers to whether the difference between the sample statistic and the parameter stated in the null hypothesis is large enough to be considered important in an application. O B. Statistical significance refers to how an unusual event is unlikely to actually appear in a real world application, such as every entry in a sample of size 50 having the same value. Practical significance refers to how an unusual event is likely to actually appear in a real world application, such as a rejection of a null hypothesis using data that looks feasible. O C. Statistical significance refers to the type of hypothesis test needed to analyze a population, with some tests being more important than Z tests. Practical…
Refer to the sample data for polygraph tests shown below. If one of the test subjects is randomly selected, what is the probability that the subject is not lying? Is the result close to the probability of 0.423 for a negative test result? Did the Subject Actually Lie? No (Did Not Lie) Yes (Lied) O Positive test results 15 41 Negative test results 30 11 The probability that a randomly selected polygraph test subject was not lying is (Type an integer or decimal rounded to three decimal places as needed.)
3
IDENTIFY IF IT IS| ONE-TAILED LEET TEST, ONE-TAILED RIGHT, NONE-TAILED, TWO-TAILED Given: n = 8 H.:µ = 30 x = 27 s = 3 HaiH 208 H = 208 3.
Ramos has a spinning wheel. The Spining wheel has 18 blue numbers 18 white numbers and 2 yellow numbers the table provided shows the frequency with which the ball landed on each color in 200 trials. color blue frequency 18 white frequency 18 yellow frequency 2 . At the 5% significance level do the data suggest that the wheel is out of balance?you will need to compute the expected counts to do your calculations remember that the expected counts can be computed with the formula E = np, where n is the number of trials and P is the probability of each color.
Consider a sample of 50 football games, where 28 of them were won by the home team. Use a 0.10 significance level to test the claim that the probability that the home team wins is greater than one-half. Identify the null and alternative hypotheses for this test. Choose the correct answer below. A. Upper H 0 : pequals 0.5 Upper H 1 : pless than 0.5 B. Upper H 0 : pgreater than 0.5 Upper H 1 : pequals 0.5 C. Upper H 0 : pequals 0.5 Upper H 1 : pnot equals 0.5 D. Upper H 0 : pequals 0.5 Upper H 1 : pgreater than 0.5 Your answer is correct. Identify the test statistic for this hypothesis test. The test statistic for this hypothesis test is 1.98 . (Round to two decimal places as needed.)
A researcher claims that 45% of U.S. adults think that the IRS is not aggressive enough in pursuing people who cheat on their taxes. You believe the true percentage is more than 45%. To test this, you take a simple random sample of 500 U.S. adults and find that 255 of them say the IRS is not aggressive enough. Test at .10 significance.Round to the fourth H0:Select an answer Select an answer HA:Select an answer Select an answer How many successes were there? Test Statistic: P-value: Did something significant happen? Select an answer Select the Decision Rule: There is is not enough evidence to conclude Select an answer
A union leader claims that absences on the different days of the week are equally likely. To test this claim, a company manager collects the data shown below. Use a 0.05 significance level to test the claim. Day Mon Tue Wed Thur Fri Absences 40 15 12 23 43 a. Before doing any calculations, do the data appear to support the union leader's claim? Why or why not? A. No, absences on The and Wed are much larger than on other days B. Yes, there are roughly the same number of absences on the different days of the week C. Yes, more absences on Mon and Fri are normal D. No, absences on Mon and Fri are much larger than on other days b. State Upper H 0 . A. At least one proportion is different than claimed B. p Subscript Mon Baseline equals 1 divided by 4 comma p Subscript Tue Baseline equals 1 divided by 4 comma p…
← In a random sample of males, it was found that 18 write with their left hands and 212 do not. In a random sample of females, it was found that 66 write with their left hands and 440 do not. Use a 0.05 significance level to test the claim that the rate of left-handedness among males is less than that among females. Complete parts (a) through (c) below. ✓ 1319 K H₁: P₁ OD. Ho: P₁ P₂ H₁: P₁ P₂ Identify the test statistic. Z= -2.06 (Round to two decimal places as needed.) Identify the P-value. P-value = 0.020 (Round to three decimal places as needed.) What is the conclusion based on the hypothesis test? w an example Get more help. P2 The P-value is less than the significance level of a = 0.05, so males is less than that among females. b. Test the claim by constructing an appropriate confidence interval. The 90% confidence interval is Clear all : ; I d F11 { + [ option = ? Check answer 11 I } * 1

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