A study was conducted to determine how people get jobs. The table lists data from 400 randomly selected subjects. Construct a Pareto chart that corresponds to the given data. If someone would like to get a job, whatseems to be the most effective approach?Choose the correct Pareto chart.A.300-250-200-150-100-50-0-Ν Η Ε ΜB.B. Help-wanted ads (H)C. Executive search firms (E)D. Mass mailing (M)300-250-200-150-100-50-0-HEN MIf someone would like to get a job, what seems to be the most effective approach?O A. Networking (N)300-250-200-150-100-50-0-Η Ε ΝMO D.300-250-200-150-100-50-0-NHE MJob SourcesHelp-wanted ads (H)Executive search firms (E)Networking (N)Mass mailing (M)Frequeno604526827

Given data job sources frequency Help wanted ads(H)…

Answered: A study was conducted to determine how…

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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A research center claims that 30% of adults in a certain country would travel into space on a commercial flight if they could afford it. In a random sample of 800 adults in that country, 34% say that they would travel into space on a commercial flight if they could afford it. At α=0.05, is there enough evidence to reject the research center's claim? Complete parts (a) through (d) below. (a) Identify the claim and state H0 and Ha. Let p be the population proportion of successes, where success is an adult in the country who would travel into space on a commercial flight if they could afford it. State H0 and Ha. Select the correct choice below and fill in the answer boxes to complete your choice. b) Use technology to find the P-value. Identify the standardized test statistic. Identify the P-value. (c) Decide whether to reject or fail to reject the null hypothesis and (d) interpret the decision in the context of the original claim.
A sunscreen company is attempting to improve upon their formula so that it lasts in water longer. They have 4 lead scientists who each came up with a different formulas. In order to see if there is a difference in the time the sunscreen lasts the CEO collects a random sample of each of the four sunscreens the data is shown below. Test the claim that at least one sunscreen has a different lifespan in water at a 0.05 level of significance. Sunscreen A Sunscreen B Sunscreen C Sunscreen D 81 75 52 79 77 34 40 87 64 39 32 72 47 43 61 58 57 47 48 46 80 43 38 63 The hypotheses for this ANOVA test would be: H0:μA=μB=μC=μDH0:μA=μB=μC=μD HA:HA: At least one mean is different. (claim) α=0.05α=0.05 Complete the ANOVA table below: (round answers to 3 decimal places) SS df MS F p-value Between Within The decision of the test is to: reject H0H0 do not reject H0H0 The final conclusion is: There is not enough evidence to reject the claim that…
A soft-drink bottling company fills and ships soda in plastic bottles with a target volume of 354 . The filling machinery does not deliver a perfectly consistent volume of liquid to each bottle, and the three quartiles for the fill volume are O1=353 , Q2=354 , and Q3=356 (A) Find the IQR (B) Find the outlier limits. Round your answer to one decimal place. Lower limit=? Upper limit=? (C) A fill volume of 348 mL is considered low. Would a fill volume of 348 mL be considered an outlier? Explain. A. No, since 348 falls between the lower and upper limits. B. Yes, since 348 is below the lower outlier limit. C. Yes, since 348 is above the upper outlier limit.
At any age, about 20% of American adults participate in physical conditioning activities at least twice a week. However, these activities change as people get older, and occasionally participants cease to be older as they age. In a local survey of n = 100 adults over 40 years of age, a total of 15 people indicated that they participated in these activities at least twice a week. Does this data indicate that the percentage of participation for adults over 40 years of age is considerably less than the 20% figure? Find the p-value and use it to draw the appropriate conclusions.
A student decides to spin a dime and determine the proportion of times it lands on heads. The student spins the dime 25 times and records that it lands on heads 17 times. Let p = the true proportion of times the dime would land on heads when spun. Under the assumption that the true proportion is 0.5, 100 simulated proportions for samples of size 25 is shown in the dotplot. Using the dotplot, is there evidence that the proportion of times a spun dime lands on heads is greater than 0.5? A) Yes, a proportion of 0.68 proves that the true proportion of heads is greater than 0.5. B) Yes, a proportion of 0.68 only occurred once out of 100 simulated proportions; therefore, there is sufficient evidence that the true proportion of heads is greater than 0.5. C) No, a proportion of 0.68 is only 0.18 more than 0.5; therefore, there is insufficient evidence that the true proportion of heads is greater than 0.5. D) No, a proportion of 0.68 or more occurred 7 times out of 100 simulated…
A recent poll found that 664 out of 1026 randomly selected people in a particular country felt that colleges and universities with big sports programs placed too much emphasis on athletics over academics. Assuming the conditions for the CLT are met, use the accompanying Minitab output to complete parts a and b below. N Event Sample p 95% CI for p 1026 664 0.647173 (0.617934, 0.676413) Question content area bottom b. Suppose a sports blogger wrote an article claiming that a majority of adults from this country believe that colleges and universities with big sports programs place too much emphasis on athletics over academics. Does this confidence interval support the blogger's claim? Explain your reasoning. A. No, it is not a plausible claim because the confidence interval contains 50%. B. No, it is not a plausible claim because the confidence interval does not contain only values above 50%. C. Yes, it is a…
The average annual miles driven per vehicle in the United States is 11.1 thousand miles, with σ ≈ 600 miles. Suppose that a random sample of 36 vehicles owned by residents of Chicago showed that the average mileage driven last year was 10.9 thousand miles. Does this indicate that the average miles driven per vehicle in Chicago is different from (higher or lower than) the national average? Use a 0.05 level of significance. State the null and alternate hypotheses. a) H0: p = 11.1; H1: p > 11.1 b) H0: p = 11.1; H1: p < 11.1 c) H0: μ = 11.1; H1: μ > 11.1 d) H0: μ = 11.1; H1: μ ≠ 11.1 e) H0: p = 11.1; H1: p ≠ 11.1 f) H0: μ = 11.1; H1: μ < 11.1 (b) What sampling distribution will you use? What assumptions are you making? a) The standard normal, since we assume that x has a normal distribution with known σ. b) The Student's t, since we assume that x has a normal distribution with unknown σ. c) The Student's t, since we assume that x has a normal distribution with known σ.…
Conduct a test at the alphaαequals=0.05 level of significance by determining (a) the null and alternative hypotheses, (b) the test statistic, and (c) the P-value. Assume the samples were obtained independently from a large population using simple random sampling. Test whether p1>p2. The sample data are x1=120, n1=248, x2=133, and n2=302. (a) Choose the correct null and alternative hypotheses below. A. Upper H 0 : p 1 equals p 2H0: p1=p2 versus Upper H 1 : p 1 not equals p 2H1: p1≠p2 B. Upper H 0 : p 1 equals p 2H0: p1=p2 versus Upper H 1 : p 1 less than p 2H1: p1<p2 C. Upper H 0 : p 1 equals 0H0: p1=0 versus Upper H 1 : p 1 not equals 0H1: p1≠0 D. Upper H 0 : p 1 equals p 2H0: p1=p2 versus Upper H 1 : p 1 greater than p 2H1: p1>p2 Your answer is correct. (b) Determine the test statistic. z0equals=nothing (Round to two decimal places as needed.)
A random sample of 54 students from Cabrillo College are selected from a normal population with mu equal to 20 and sigma equal to 5. After a treatment is administered to the students, the researcher calculated the sample mean to be 21. Is there enough evidence to conclude that the treatment had an effect at the 5% significance level? In step three, can we use the Critical Value Method? (a)Don't know (b)Cannot Determine (c)No (d)Yes
If all other values are held constant, what happens to the value of t when sample size increases ?
Use the standard error to construct a approximate prediction interval for Y using an alpha of 5%. (Round your answer to 3 decimal places.) Obs Assessed Floor Offices Entrances Age Freeway 1 1796 4790 4 2 8 0 2 1544 4720 3 2 12 0 3 2094 5940 4 2 2 0 4 1968 5720 4 2 34 1 5 1567 3660 3 2 38 1 6 1878 5000 4 2 31 1 7 949 2990 2 1 19 0 8 910 2610 2 1 48 0 9 1774 5650 4 2 42 0 10 1187 3570 2 1 4 1 11 1113 2930 3 2 15 1 12 671 1280 2 1 31 1 13 1678 4880 3 2 42 1 14 710 1620 1 2 35 1 15 678 1820 2 1 17 1 16 1585 4530 2 2 5 1 17 842 2570 2 1 13 0 18 1539 4690 2 2 45 0 19 433 1280 1 1 45 1 20 1268 4100 3 1 27 0 21 1251 3530 2 2 41 1 22 1094 3660 2 2 33 0 23 638 1110 1 2 50 1 24 999 2670 2 2 39 1 25 653 1100 1 1 20 1 26 1914 5810 4 3 17 0 27 772 2560 2 2 24 0 28 890 2340 3 1 5 0 29 1282 3690 2 2 15 1 30 1264 3580 3 2 27 0 31 1162 3610 2 1 8 1 32 1447 3960 3 2 17 0
A random sample of 1,175 people in a certain country were asked whether they thought climate change was aproblem. The sample proportion of those who think climate change is a problem was calculated, and a 95 percentconfidence interval was constructed as .Which of the following is a correct interpretation of the interval?(A) We are 95 percent confident that any sample of 1,175 people will produce a sample proportion between0.146 and 0.214.(B) We are 95 percent confident that the proportion of all people in the country who think climate change is aproblem is between 0.146 and 0.214.(C) We are 95 percent confident that the proportion of people in the sample who think climate change is aproblem is between 0.146 and 0.214.(D) The probability that 95 percent of all people in the country who think climate change is a problem is between0.146 and 0.214.(E) The probability is 0.95 that the proportion of all people in the country who think climate change is a problemis between 0.146 and 0.214

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