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 youdecide 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 CompanyA, 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 permonth 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 claimat 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.

The objective of this question is to compute the value of the test statistic for the claim that…

Answered: Insurance Company A claims that its…

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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You hypothesize that people in stats classes are happier than people in all other classes. You compare happiness scores in three of your classes: Statistics, Developmental Psychology, and Social Psychology. In Statistics there are 15 students and the mean happiness rating is 23.1 with a standard deviation of 9.78. In Developmental Psychology there are 15 students and the mean happiness rating is 24.4 with a standard deviation of 2.92. In Social Psychology there are 15 students with a mean rating of 17.7 and standard deviation of 8.46. The sum of squares between groups is equal to 384. The sum of squares within groups is equal to 2461. Use the One-Way ANOVA calculation table to help calculate. What is the degrees of freedom within samples?
You are interested in seeing whether emotions impact decision making. You have three groups--a happy group, a sad group, and a neutral group. For the happy group there are 5 participants and the mean risky decision making score 5.0 with a standard deviation of 0.7. For the sad group there are 5 participants and the mean risky decision making score 5.4 with a standard deviation of 1.1. For the neutral group there are 5 participants and the mean risky decision making score 5.8 with a standard deviation of 2.8. The sum of squares between samples is equal to 1.6. The sum of squares within samples is equal to 38.0. What is the degrees of freedom within samples?
You are interested in seeing whether emotions impact decision making. You have three groups--a happy group, a sad group, and a neutral group. For the happy group there are 5 participants and the mean risky decision making score 5.0 with a standard deviation of 0.7. For the sad group there are 5 participants and the mean risky decision making score 5.4 with a standard deviation of 1.1. For the neutral group there are 5 participants and the mean risky decision making score 5.8 with a standard deviation of 2.8. The sum of squares between samples is equal to 1.6. The sum of squares within samples is equal to 38.0. What is the mean square between samples?
Suppose the graduate teaching assistant for Intro Stat at the University of Florida also collected data on whether the students sat in the front of the class, middle of the class or back of the lecture hall. How much of the variance in the percentage grade is explained by the study hours and the sitting position?
The psychology department is gradually changing its curriculum by increasing the number of online course offerings. To evaluate the effectiveness of this change, a random sample of n = 36 students who registered for Introductory Psychology is placed in the online version of the course. At the end of the semester, all students take the same final exam. The average score for the sample is M = 76. For the general population of students taking the traditional lecture class, the final exam scores form a normal distribution with a mean of μ = 71. If the final exam scores for the population have a standard deviation of σ = 18, does the sample provide enough evidence to conclude that the new online course is significantly different from the traditional class? Again, use a two-tailed test with α = .05. (Round z score to two decimal places.)
A phone company finds that the mean data usage for its residential customers is 10GBs, and the median for the same group is 4GBs. This indicates that the company's residential customer usage data has a right-skewed distribution True or False True False
Jason collects information about annual family income from each student at his college. He computes both the mean and median family income across all students and is surprised to find that the mean income is much higher than the median income. What is the most likely reason for the difference in the mean and the median? Select one: a. The sample size is too small. b. The median tends to produce lower estimates of central tendency. c. The median is not an appropriate estimate for annual family income's level of measurement. Incorrect d. The mean is more affected by a small number of extreme values.
You are interested in seeing whether emotions impact decision making. You have three groups--a happy group, a sad group, and a neutral group. For the happy group there are 5 participants and the mean risky decision making score 5.0 with a standard deviation of 0.7. For the sad group there are 5 participants and the mean risky decision making score 5.4 with a standard deviation of 1.1. For the neutral group there are 5 participants and the mean risky decision making score 5.8 with a standard deviation of 2.8. The sum of squares between samples is equal to 1.6. The sum of squares within samples is equal to 38.0. What is the mean square within samples?
A researcher thinks that people under forty have vocabularies different from those over sixty years of age. The researcher administers a vocabulary test to a group of 50 younger subjects and a group of 50 older subjects. Higher scores reflect better performance. The mean score for younger subjects was 9.0, and the standard deviation of younger subject's scores was 5.0. The mean score for older subjects was 20.0, and the standard deviation of older subject's scores was 8.0. Does this experiment provide evidence for the researcher's theory? As part of your answer: a. Please provide, in words, the null and alternative hypotheses (NO symbols). Note that you can choose whether to perform a one- or two-tailed test Answer: Null hypothesis: Alernative hypothesis: b. Provide a brief (one or two sentence(s) rationale for your decision to use a one or two tailed test Answer c. Calculate the Standard Error (SE) of the mean difference (demonstrate calculation steps to get an entire point) and…
One year Perry had the lowest ERA (earned-run average, mean number of runs yielded per nine innings pitched) of any male pitcher at his school, with an ERA of 2.57. Also, Rita had the lowest ERA of any female pitcher at the school with an ERA of 3.26. For the males, the mean ERA was 4.304 and the standard deviation was 0.894. For the females, the mean ERA was 4.158 and the standard deviation was 0.519. Find their respective z-scores. Which player had the better year relative to their peers, Perry or Rita? (Note: In general, the lower the ERA, the better the pitcher.) Perry had an ERA with a z-score of Rita had an ERA with a z-score of (Round to two decimal places as needed.) Textbook Get more help - $ 1 G Search or type URL % 31 6 & 7 * 8 + ( 9 0 Clear all + Check answer delete
Suppose you've just been hired as a statistics professor, but you've never taught undergrads before. While you're doing your best, you're not sure you're fully up to par with all the other excellent statistics professors out there. You decide to look at your 41 students' grades on their midterm exam. They averaged grades of 81 on this exam, with a standard deviation of 10. You reason that if they're higher than normal, you might be making things too easy for them. Likewise, if they're lower than normal, you might be making things too hard. Seeking a Goldilocks effect, you compare this with the national average of statistics midterm exam scores - this is 79. BEING SPECIFIC, what kind of statistical analysis should you perform, and WHY? Three parts of your answer will factor into the grade on this answer: the type of statistical test, one specific aspect of this test (hint- see the earlier questions), and your explanation as to why.
A researcher was interested in comparing the mean grade-point-average (GPA) of students at the University of Toronto Scarborough and Mississauga campuses. Independent random samples of 8 students from the Scarborough campus and 13 students from the Mississauga campus yielded the following GPAs. You can assume the data in each group follow a normal distribution. Do the data provide sufficient evidence to conclude that the mean GPA of students at the Scarborough campus different from the mean GPA of students at the Mississauga campus? A. What statistical test is most appropriate to answer this question? B. Identify the relevant assumptions and conditions. State how these have or have not been met. C. Write an appropriate null and alternative hypothesis for this question.

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