1. Fred and Martina, senior agents at an airline security checkpoint, cary outadvanced screening procedures for hundreds of randomly selected passengers perday. For a randomsample of 30 passengers recently processed by Fred, the meanprocessing time was 124.5 seconds, with a standard deviation of 20.4 seconds. For arandom sample of 36 passengers recently processed by Martina, the corespodingmean and standard deviation were 133.0 seconds and 38.7 seconds, respectively.Using the 0.05 level of significance, can we conclude that the population meanprocessing times for Fred and Martina could be the same? Use Unequal-VariancesT-Test.

In order to determine whether the population mean processing times for individual F and individual M…

Answered: 1. Fred and Martina, senior agents at…

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 National Coalition on Healthcare suggests that the mean annual premium that a health insurer charges an employer for a health plan covering a family of four averaged $13,000 in 2009. A sample of 30 families of four yields a mean annual premium paid by their employer to be $13,500 with a sample standard deviation of $300. We are interested in whether the mean annual premium that a health insurer charges an employer for a health plan covering a family of four is different from $13,000 using a significance level of 0.10. Select one: a. tdata = 9.129, do not reject Ho. b. tdata = 9.129, reject Ho. c. tdata = -9.129, do not reject Ho. O d. tdata = 4.156, reject Ho.
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. You calculated the f-ratio above. The corresponding p-value for the f-ratio you calculated is p = 0.78. Is this finding significant?
12. (Two Samples) According to a survey, the mean price of gasoline in the United States was $1.20 per gallon in 1995 and $1.10 per gallon in 1994. (U.S. News & World Report, June 12, 1995). Suppose these means were based on random samples of 100 gasoline stations for 1995 and 120 gasoline stations for 1994. Also, assume that the sample standard deviations were $0.11 for 1995 and $0.10 for 1994. a. Construct a 98% confidence interval for the difference between the mean gasoline prices for 1995 & 1994. b. Test at the 1% significance level if the mean gasoline price was higher in 1995 than in 1994.
1) According to Consumer Expenditures, the average consumer spent $1729 on apparel and services in 2017. In 2018, a random sample of 36 consumers had a mean of $1667.11 and a standard deviation of $351.69. Can it be said that at the 5% level of significance that the mean level of consumption differed in 2018 with respect to 2017?
A successful basketball player has a height of 6 feet 11 inches, or 211 cm. Based on statistics from a data set, his height converts to the z score of 5.17. How many standard deviations is his height above the mean?
Lucille, Rocio, and Cristina are all taking College Mathematics but from different professors. They all recently took a statistics exam and the exam scores were normally distributed in each of their classes. • Lucille's class had an average grade of 78 and the standard deviation was 7. Lucille earned 83.95 on the exam. • Rocio's class had an average grade of 82 and the standard deviation was 3. Rocio earned 86.5 on the exam. • Cristina's class had an average grade of 76 and the standard deviation was 4. Cristina earmed 82.8 on the exam. Since some professors give harder exams than others, these students decided to use statistics to figure out who did the best on their exam. Use the Cumulative Z-Score Table to answer the following questions. Write your answers as a percent with two decimal places. Hint • Lucille's score places her above % of her class. • Rocio's score places her above % of her class. • Cristina's score places her above % of her class.
new drug is being proposed for the treatment of migraine headaches. The FDA will not approve the new drug if it does not show a statistically significant improvement as compared to a placebo. In an experiment 40 people who suffer from migraine headaches receive the new drug and 18 of them report an improvement in their migraine pain (group 2). Another 40 people were place in the placebo group (group 1). Of these 13 reported an improvement in their migraine pain. Help the FDA, by forming a conclusion based on the following supporting statistical evidence. Sample Data Sample X N Sample Proportion Placebo (group 1) 13 40 0.325 New Drug (group 2) 18 40 0.450 Difference = p1 - p2 95% Confidence Interval for Difference p1-p2: (-0.337, 0.087) P-value = 0.247 The estimated difference between the placebo group and the new drug treatment group is -0.125. A. True B. False
One year Hank 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 3.09. Also, Karen had the lowest ERA of any female pitcher at the school with an ERA of 2.98. For the males, the mean ERA was 4.635 and the standard deviation was 0.576. For the females, the mean ERA was 4.469 and the standard deviation was 0.944. Find their respective z-scores. Which player had the better year relative to their peers, Hank or Karen? (Note: In general, the lower the ERA, the better the pitcher.) Hank had an ERA with a z-score of Karen had an ERA with a z-score of (Round to two decimal places as needed.) Which player had a better year in comparison with their peers?
A recent study found that 61 children who watched a commercial for potato chips featuring a celebrity endorser ate a mean of 39 grams of potato chips as compared to a mean of 28 grams for 51 children who watched a commercial for an alternative food snack. Suppose that the sample standard deviation for the children who watched the celebrity-endorsed commercial was 21.2 grams and the sample standard deviation for the children who watched the alternative food snack commercial was 12.7 grams. What is the test statistic? tSTAT= __________ (Round to two decimal places as needed.)
Records from the Washington Department of Licensing show that 47 former SFCC business students have incorporated their own businesses in the last five years. Eight of these students are selected at random and contacted. Each is asked how many people are employed by their corporation. The responses are: 7, 12, 13, 17, 10, 5, 10, and 6 employees. What is your best estimate of the standard deviation of the number of employees for all 47 corporations?
A professor believes that, for the introductory art history classes at his university, the mean test score of students in the evening classes is lower than the mean test score of students in the morning classes. He collects data from a random sample of 250 students in evening classes and finds that they have a mean test score of 85.6. He knows the population standard deviation for the evening classes to be 4.6 points. A random sample of 150 students from morning classes results in a mean test score of 86.7. He knows the population standard deviation for the morning classes to be 8.3 points. Test his claim with a 99 % level of confidence. Let students in the evening classes be Population 1 and let students in the morning classes be Population 2. Step 3 of 3: Draw a conclusion and interpret the decision. Answer 围 Tables E Keypad Keyboard Shortcuts We fail to reject the null hypothesis and conclude that there is sufficient evidence at a 0.01 level of significance to support the…
A professor believes that, for the introductory art history classes at his university, the mean test score of students in the evening classes is lower than the mean test score of students in the morning classes. He collects data from a random sample of 250 students in evening classes and finds that they have a mean test score of 86.6. He knows the population standard deviation for the evening classes to be 7.2 points. A random sample of 200 students from morning classes results in a mean test score of 87.8. He knows the population standard deviation for the morning classes to be 4.7 points. Test his claim with a 90% level of confidence. Let students in the evening classes be Population 1 and let students in the morning classes be Population 2. Step 2 of 3 : Compute the value of the test statistic. Round your answer to two decimal places.

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