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 to be a quality credit risk. According to a survey, the mean credit score is 706.3. A credit analyst wondered whether high-income individuals (incomes in excess of $100,000 per year) had higher credit scores. He obtained a random sample of 45 high-income individuals and found the sample mean credit score to be 725.6 with a standard deviation of 81.6. Conduct the appropriate test to determine if high-income individuals have higher credit scores at the α = 0.05 level of significance.State the null and alternative hypotheses.H₀: μ [≤] [706.3]H₁: μ [>] [706.3](Type integers or decimals. Do not round.)

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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The preeminent basketball philosopher Rasheed Wallace popularized the phrase, "Ball don't Lie." This phrase means that if a foul was called incorrectly, the free-throw shooter is more likely to miss their shots. (We can think of this as a sort of Basketball Karma.) Due to the recent "Last Two Minute Report," we now have some data to test this claim. Of the 177 free throws awarded incorrectly, the shooter made 78.5%. Use this information to test the claim, with a .05 level of significance, that the percentage of made free throws after an incorrect foul call is less than 80%, which is the FT Percentage for an average NBA player. Round your answers to three decimal places or to the nearest tenth of a percent. (e.g. .021 or 2.1%) For credit, the hypothesis test must contain: 1) Claim in terms of the appropriate parameter (e.g. LaTeX: \mu=\:4.5\:miles) 2) Null and alternative hypothesis 3) Sample evidence used 4) Test stat (z or t) , p-value and direction of test. (right, left or…
Sleep apnea is a condition in which the sufferers stop breathing momentarily while they are asleep. This condition results in lack of sleep and extreme fatigue during waking hours. A current estimate is that 9.79.7 million out of the 312.7312.7 million Americans suffer from sleep apnea, or approximately 3.1%3.1%. A safety commission is concerned about the percentage of commercial truck drivers who suffer from sleep apnea. They do not have any reason to believe that it would be higher or lower than the population’s percentage. To test the claim that the percentage of commercial truck drivers who suffer from sleep apnea is not 3.1%3.1%, a simple random sample of 406406 commercial truck drivers is examined by a medical expert, who concludes that 55 suffer from sleep apnea. Does this evidence support the claim that the percentage of commercial truck drivers who suffer from sleep apnea is not 3.1%3.1%? Use a 0.050.05 level of significance. Step 2 of 3 : Compute the value of the…
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In the United States, standardized testing is used to compare students' knowledge retention across the country. States will often use the results of standardized testing to allocate funding to districts within the state. One state in particular wanted to determine whether there was a difference in scores between schools on the eastern side of the state and schools on the western side of the state. A random sample of 30 schools on the eastern side and 35 schools on the western side were chosen. The mean score for the eastern schools was x1 = 109 with a standard deviation of 5.3. The mean score for the western schools was x2 = 104 with a standard deviation of 4.4. We would like to estimate the difference in the mean standardized scores from the eastern schools and western schools and determine the margin of error of this estimate with 98% confidence. (a) Determine the point estimate for the difference in mean standardized scores for the eastern and western sides of the state, ?1 −…
The U.S. Census Bureau reported that in 2015,the proportion of adult Americans age 25 and olderwho have a bachelor’s degree or higher is 0.325(“Educational Attainment in the United States:2015,” census.gov, retrieved January 22, 2017).Consider the population of all adult Americans age25 and over in 2015 and define 户to the proportionof people in a random sample from this populationwho have a bachelor’s degree or higher.a. Would 卢based on a random sample of only 10people from this population have approximatelya normal distribution? Explain why or why not.b. What 征e the mean value and standard deviation of户based on a random sample of size 400?c. Suppose that the sample size is n = 200 ratherthan n = 400. Does the change in sample sizeaffect the mean and standard deviation of thesampling distribution of 卢？ If so, what are thenew values for the mean and standard deviation?If not, explain why not.
An article in the Washington Post on March 16, 1993 stated that nearly 45 percent of all Americans have brown eyes. A random
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A newspaper infographic titled "Social Media Jeopardizing Your Job?" summarized data from a survey of 1,850 recruiters and human resource professionals. The infographic indicated that 54% of the people surveyed had reconsidered a job candidate based on his or her social media profile. Assume that the sample is representative of the population of recruiters and human resource professionals in the United States. USE SALT (a) Use the given information to estimate the proportion of recruiters and human resource professionals who have reconsidered a job candidate based on his or her social media profile using a 95% confidence interval. (Round your answers to three decimal places.) Give an interpretation of the interval in context. O There is a 95% chance that the true proportion of recruiters and human resource professionals who have reconsidered a job candidate based on his or her social media profile falls directly in the middle of this interval. O There is a 95% chance that the true…
Based on the Nielsen ratings, the local CBS affiliate claims its 11 p.m. newscast reaches 52% of the viewing audience in the area. In a survey of 100 viewers, 47% indicated that they watch the late evening news on this local CBS station. What is the test statistic?
he Computer Anxiety Rating Scale (CARS) measures an individual’s level of computer anxiety, on a scale from 20 (no anxiety) to 100 (highest level of anxiety). Researchers at Miami University administered CARS to 172 business students. One of the objectives of the study was to determine whether there are differences in the amount of computer anxiety experienced by students with different majors. They found the following: Major n Mean Marketing 19 44.37 Management 11 43.18 Other 14 42.21 Finance 45 41.80 Accountancy 36 37.56 MIS 47 32.21 (A) Complte the following ANOVA table: Source of Variation SS df MS F-ratio F-crit Between groups 3172 5 634.4 4.9567 2.2685 Within groups 21246 166 127.988 Total 24418 171 (A) At the 0.05 level of significance, is there evidence of a difference in the mean computer anxiety…
The mean birthweight of an infant is thought to be associated with the smoking status of the mother during the first trimester of the pregnancy. Consider the following data from mothers divided into four categories according to smoking habits, and the corresponding birthweights of their children. Group 1: Mother is a non smoker (NON): 7.5, 6.2, 6.9, 7.4, 9.2, 8.3, 7.6 Group 2: Mother is an ex-smoker (EX): 5.8, 7.3, 8.2, 7.1, 7.8 Group 3: Mother is a current smoker, smokes less than 1 pack per day (CUR < 1) 5.9, 6.2, 5.8, 4.7, 8.3, 7.2, 6.2 Group 4: Mother is a current smoker, smokes more than or equal to 1 pack per day (Cur ≥ 1) 6.2, 6.8, 5.7, 4.9, 6.2, 7.1, 5.8, 5.4 Using Stata test whether or not the mean birthweights are equal across all four groups at the alpha = 0.05 level. What is the corresponding p-value? O 0.0284 0.0137 O 0.05 O 0.737
A report stated that based on observing a large number of vehicle occupants, the estimated percentage of drivers and passengers in a certain country who wear seatbelts is 85.7%. The report also gave information on seat belt use by sex and by whether the vehicle is traveling in an urban or rural area. The information in the following table is consistent with summary values given in the report. Seat belt No Seat belt Urban Areas Male Female 851 119 908 62 Rural Areas Male Female 750 220 817 153 Assume that these data are representative of drivers and passengers in the country. Consider the following conclusion. Females are more likely to wear seat belts than males in both urban and rural areas. The difference in the percentage of females and the percentage of males who wear seat belts is greater for rural areas than for urban areas. Provide a justification for this conclusion. Use the information in the table to calculate relevant probabilities. The differences in the proportions for…

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