A credit score is used by credit agencies (such as mortgage companies and banks) to assess the creditworthiness of individuals. Values range from 300 to 850, with a credit score over 700 considered tobe a quality credit risk. According to a survey, the mean credit score is 709.3. A credit analyst wondered whether high-income individuals (incomes in excess of $100,000 per year) had higher creditscores. He obtained a random sample of 31 high-income individuals and found the sample mean credit score to be 722.3 with a standard deviation of 81.1. Conduct the appropriate test to determineif high-income individuals have higher credit scores at the a= 0.05 level of significance.State the null and alternative hypotheses.Hoμ= 709.3H₁ μ709.3(Type integers or decimals. Do not round)Identify the t-statisticto = (Round to two decimal places as needed.)

Population mean, Sample size, Sample mean, Sample standard deviation, s = 81.1

Answered: A credit score is used by credit…

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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Socially conscious investors screen out stocks of alcohol and tobacco makers, firms with poor environmental records, and companies with poor labor practices. Some examples of "good," socially conscious companies are Johnson and Johnson, Dell Computers, Bank of America, and Home Depot. The question is, are such stocks overpriced? One measure of value is the P/E, or price-to-earnings ratio. High P/E ratios may indicate a stock is overpriced. For the S&P Stock Index of all major stocks, the mean P/E ratio is ? = 19.4. A random sample of 36 "socially conscious" stocks gave a P/E ratio sample mean of x = 17.7, with sample standard deviation s = 5.2. Does this indicate that the mean P/E ratio of all socially conscious stocks is different (either way) from the mean P/E ratio of the S&P Stock Index? Use ? = 0.05. (a) What is the level of significance?State the null and alternate hypotheses. H0: ? = 19.4; H1: ? < 19.4H0: ? ≠ 19.4; H1: ? = 19.4 H0: ? = 19.4; H1: ? ≠ 19.4H0: ?…
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