Conscientiousness is a tendency to show self-discipline, act dutifully, andaim for achievement. The trait shows a preference for planned rather thanspontaneous behavior. A random sample of 650 students is asked to fill outthe Hogan Personality Inventory (HPI) to measure their level ofconscientiousness. The 350 graduate students had a mean score of 153with a standard deviation of 21. The 300 undergraduate students scoredan average of 148 with a standard deviation of 16. We are interested indetermining whether graduate students score higher, on average, on theHPI than undergraduate students. What are the appropriate null andalternative hypotheses?$4RO Ho: HG- HUG # 0 and Ha: HG- HUG = 0O Ho: HG- HUG = 0 and Ha: HG- HUGOO Ho: HG HUG = 0 and H₂: HG- HUG #0O Ho: HG- HUG = 0 and H₂: HG- HUG > Otv%Q Search or enter address5eTA6MacBook ProYA&7U*8↑-(9O)0

Given that Sample sizes n1=350 , n2=300 Sample means X̄1=153 , X̄2=148 s1…

Answered: Conscientiousness is a tendency to show…

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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Insurance Company A claims that its customers pay less for car insurance, on average, than customers of its competitor, Company B. You wonder if this is true, so you decide to compare the average monthly costs of similar insurance policies from the two companies. For a random sample of 13 people who buy insurance from Company A, the mean cost is $150 per month with a standard deviation of $19. For 9 randomly selected customers of Company B, you find that they pay a mean of $157 per month with a standard deviation of $16. Assume that both populations are approximately normal and that the population variances are equal to test Company A's claim at the 0.05 level of significance. Let customers of Company A be Population 1 and let customers of Company B be Population 2. Step 2 of 3: Compute the value of the test statistic. Round your answer to three decimal places.
Previously, an organization reported that teenagers spend 8.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Twenty-three randomly chosen teenagers were asked how many hours per week they spend on the phone, with the following results (in hours): 8.3, 8.5, 10.5, 2.9, 6.7, 7.7, 13.8, 7.8, 8.7, 11.8, 12, 6.1, 5.4, 5.8, 8.5, 7.9, 10.1,11.1, 11.6, 12.9, 3.6, 6.4, 11.2 Perform a hypothesis test using a 3% level of significance. Step 1: State the null and alternative hypotheses. Ho: ??v H: ? v ? v (So we will be performing a Select an answer v test.) Step 2: Assuming the null hypothesis is true, determine the features of the distribution of point estimates using the Central Limit Theorem. v with By the Central Limit Theorem, we know that the point estimates are Select an answer and distribution standard deviation distribution mean Step 3: Find the p-value of the point estimate. P? v? v = P(| ? ♥| ? v
According to the College Board, scores on the math section of the SAT Reasoning college entrance test for the class of 2010 had a mean of 516 and a standard deviation of 116. Assume that they are roughly normal.One of the quartiles of the scores from the math section of the SAT Reasoning test is 438. The other quartile is _______.
Insurance Company A claims that its customers pay less for car insurance, on average, than customers of its competitor, Company B. You wonder if this is true, so you decide to compare the average monthly costs of similar insurance policies from the two companies. For a random sample of 7 people who buy insurance from Company A, the mean cost is $150 per month with a standard deviation of $16. For 12 randomly selected customers of Company B, you find that they pay a mean of $160 per month with a standard deviation of $14. Assume that both populations are approximately normal and that the population variances are equal to test Company A's claim at the 0.10 level of significance. Let customers of Company A be Population 1 and let customers of Company B be Population 2. Step 2 of 3: Compute the value of the test statistic. Round your answer to three decimal places.
Insurance Company A claims that its customers pay less for car insurance, on average, than customers of its competitor, Company B. You wonder if this is true, so you decide to compare the average monthly costs of similar insurance policies from the two companies. For a random sample of 13 people who buy insurance from Company A, the mean cost is $150 per month with a standard deviation of $19. For 9 randomly selected customers of Company B, you find that they pay a mean of $157 per month with a standard deviation of $16. Assume that both populations are approximately normal and that the population variances are equal to test Company A's claim at the 0.05 level of significance. Let customers of Company A be Population 1 and let customers of Company B be Population 2. Step 1 of 3: State the null and alternative hypotheses for the test. Fill in the blank below. Ho: M₁ = μ₂ Ha:M₁ •H₂
Insurance Company A claims that its customers pay less for car insurance, on average, than customers of its competitor, Company B. You wonder if this is true, so you decide to compare the average monthly costs of similar insurance policies from the two companies. For a random sample of 15 people who buy insurance from Company A, the mean cost is $154 per month with a standard deviation of $13. For 11 randomly selected customers of Company B, you find that they pay a mean of $159 per month with a standard deviation of $16. Assume that both populations are approximately normal and that the population variances are equal to test Company A’s claim at the 0.02 level of significance. Let customers of Company A be Population 1 and let customers of Company B be Population 2. Step 2 of 3 : Compute the value of the test statistic. Round your answer to three decimal places.
Insurance Company A claims that its customers pay less for car insurance, on average, than customers of its competitor, Company B. You wonder if this is true, so you decide to compare the average monthly costs of similar insurance policies from the two companies. For a random sample of 12people who buy insurance from Company A, the mean cost is $153 per month with a standard deviation of $16. For 15 randomly selected customers of Company B, you find that they pay a mean of $160 per month with a standard deviation of $10. Assume that both populations are approximately normal and that the population variances are equal to test Company A’s claim at the 0.10 level of significance. Let customers of Company A be Population 1 and let customers of Company B be Population 2. Step 1 of 3: State the null and alternative hypotheses for the test. Fill in the blank below. H0: μ1=μ2 Ha: μ1_____μ2 Step 2 of 3: Compute the value of the test statistic. Round your answer to three decimal places Step 3 of…
Previously, an organization reported that teenagers spend 12.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is lower. Twenty-three randomly chosen teenagers were asked how many hours per week they spend on the phone, with the following results (in hours): 10.4, 10.4, 17.3, 11.2, 10, 15.2, 9.5, 5.8, 12.5, 11.8, 7.3, 10.6, 11.5, 16.2, 16.5, 14.3, 12.3, 9.9, 13.7, 8.6, 11.3, 13, 16.5 Perform a hypothesis test using a 10% level of significance. Step 1: State the null and alternative hypotheses. Ho: pv 12.5 12.5 (So we will be performing a left-tailed Vv test.) Step 2: Assuming the null hypothesis is true, determine the features of the distribution of point estimates using the Central Limit Theorem. By the Central Limit Theorem, we know that the point estimates are t-distributed with distribution mean and distribution standard deviation Step 3: Find the p-value of the point estimate. = P(tv sv p-value = Step 4: Make a Conclusion About the nulI…
The Survey of Study Habits and Attitudes (SSHA) is a psychological test that measures students' study habits and attitudes toward school. The survey yields several scores, one of which measures student attitudes toward studying. The mean student attitude score for college students is about 30, and standard deviation is about 12. A researcher in the Philippines is concerned about the declining performance of college graduates on professional licensure and board exams. She suspects that poor attitudes of students are partly responsible for the decline and that the mean for college seniors who plan to take professional licensure or board exams is less than 30. She gives the SSHA to an SRS of 144 college seniors in the Philippines who plan to take professional licensure or board exams. Suppose we know that the student attitude scores in the population of such students are Normally distributed with standard deviation σ = 12. (a) We seek evidence against the claim that μ = 30. What is the…
Insurance Company A claims that its customers pay less for car insurance, on average, than customers of its competitor, Company B. You wonder if this is true, so you decide to compare the average monthly costs of similar insurance policies from the two companies. For a random sample of 1313 people who buy insurance from Company A, the mean cost is $151$⁢151 per month with a standard deviation of $16$⁢16. For 99 randomly selected customers of Company B, you find that they pay a mean of $158$⁢158 per month with a standard deviation of $19$⁢19. Assume that both populations are approximately normal and that the population variances are equal to test Company A’s claim at the 0.050.05 level of significance. Let customers of Company A be Population 1 and let customers of Company B be Population 2. Step 2 of 3: Compute the value of the test statistic. Round your answer to three decimal places.
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In the US, 45.9% of all people have type O blood, 39.7% have type A blood, 11.2% have type B blood and 3.2% have type AB blood. A researcher wants to see if the distribution of blood type is different for millionaires. The table below shows the results of a random sample of 569 millionaires. What can be concluded at the significant level of a = 0.01. The test-statistic for this data = The p-value for this sample =

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