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- Consider the following data: 5.3, 5.5, 7.5,7.7,7.7, 7.7,7.7, 7.9, 7.9,7.9,8.1,8.0,8.2,8.4,8.6,8.6,8.8, 8.8, 8.9. The First Quartile value, Q1, of this data set is: The Second Quartile value, Q2, of this data set is: The Third Quartile value, Q3, of this data set is: The Inter Quartile Range, IQR value, of this data set is: The interval for outlier detection using the Quartile infromation for this data set is:Which of the following is a biased estimator? That is, which of the following does not target the population parameter? Choose the correct answer below. Variance Proportion Median MeanA random sample of male college baseball players and a random sample of male college soccer players were obtained independently and weighed. The accompanying table shows the unstacked weights (in pounds). The distributions of both data sets suggest that the population distributions are roughly Normal. Determine whether the difference in means is significant, using a significance level of 0.05. Click the icon to view the data table. Let µBaseball be the population mean weight (in pounds) of male college baseball players and let uSoccer be the population mean weight (in pounds) of male college soccer players. Determine the hypotheses for this test. Ho: HBaseball - HSoccer %3D Ha: HBaseball- HSoccer + 0 Find the test statistic for this test. (Round to two decimal places as needed.)
- 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 you decide to compare the average monthly costs of similar insurance policies from the two companies. For a random sample of 13 people who buy insurance from Company A, the mean cost is $150 per month with a standard deviation of $19. For 9 randomly selected customers of Company B, you find that they pay a mean of $157 per month with a standard deviation of $16. Assume that both populations are approximately normal and that the population variances are equal to test Company A’s claim at the 0.05 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. Step 3 of 3: Draw a conclusion and interpret the decision.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 you decide to compare the average monthly costs of similar insurance policies from the two companies. For a random sample of 11 people who buy insurance from Company A, the mean cost is $150 per month with a standard deviation of $14. For 14 randomly selected customers of Company B, you find that they pay a mean of $158 per month with a standard deviation of $12. Assume that both populations are approximately normal and that the population variances are equal to test Company A’s claim at 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.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 you decide to compare the average monthly costs of similar insurance policies from the two companies. For a random sample of 13 people who buy insurance from Company A, the mean cost is $150 per month with a standard deviation of $19. For 9 randomly selected customers of Company B, you find that they pay a mean of $157 per month with a standard deviation of $16. Assume that both populations are approximately normal and that the population variances are equal to test Company A's claim at the 0.05 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.
- Why is it important to report both a measure of central tendency and variability to describe a set of scores? Group of answer choices Sets of data with different central tendencies may have different patterns of variability. Sets of data with the same central tendency may have the same patterns of variability. Sets of data with the same variability may have different patterns of central tendency. Sets of data with the same central tendency may have different patterns of variability.Grand Strand Family Medical Center treats minor medical emergencies for visitors to the Myrtle Beach area. There are two facilities, one in the Little River Area and the other in Murrells Inlet. The Quality Assurance Department wishes to compare the mean waiting time for patients at the two locations. Samples of the waiting times for each location, reported in minutes, follow: Location Waiting Time Little River 31.73 28.77 29.53 22.08 29.47 18.60 32.94 25.18 29.82 26.49 Murrells Inlet 22.93 23.92 26.92 27.20 26.44 25.62 30.61 29.44 23.09 23.10 26.69 22.31 Assume the population standard deviations are not the same. At the 0.05 significance level, is there a difference in the mean waiting time? Click here for the Excel Data File Find the degrees of freedom for unequal variance test. (Round down your answer to the nearest whole number.) State the decision rule for 0.05 significance level: H0: μLitttle River = μMurrells Inlet; H1: μLitttle River ≠…In a study of birth order and intelligence, IQ tests were given to 18 - and 19 -year-old men to estimate the size of the difference, if any, between the mean IQs of firstborn sons and secondborn sons. The following data for 10 firstborn sons and 10 secondborn sons are consistent with the means and standard deviations reported in the article. It is reasonable to assume that the samples come from populations that are approximately normal. Can you conclude that the mean IQ of firstborn sons differs from the mean IQ of secondborn sons? Let μ1 denote the mean IQ of firstborn sons and μ2 denote the mean IQ of secondborn sons. Use the =α0.05 level and the P -value method with the TI-84 Plus calculator. reject or not is there evidence or not Firstborn 101 84 122 84 88 97 130 109 83 93 Secondborn 115 125 84 112 113 113 129 83 97 126
- 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 you decide to compare the average monthly costs of similar insurance policies from the two companies. For a random sample of 13 people who buy insurance from Company A, the mean cost is $150 per month with a standard deviation of $19. For 9 randomly selected customers of Company B, you find that they pay a mean of $157 per month with a standard deviation of $16. Assume that both populations are approximately normal and that the population variances are equal to test Company A's claim at the 0.05 level of significance. Let customers of Company A be Population 1 and let customers of Company B be Population 2. Step 1 of 3: State the null and alternative hypotheses for the test. Fill in the blank below. Ho: M₁ = μ₂ Ha:M₁ •H₂Fuel efficiency of Prius: Fueleconomy.gov, the official US government source for fuel economy information, allows users to share gas mileage information on their vehicles. The histogram below shows the distribution of gas mileage in miles per gallon (MPG) from 14 users who drive a 2012 Toyota Prius. The sample mean is 53.3 MPG and the standard deviation is 5.2 MPG. Note that these data are user estimates and since the source data cannot be verified, the accuracy of these estimates are not guaranteed. (a) The EPA claims that a 2012 Prius gets 50 MPG (city and highway mileage combined). Do these data provide strong evidence against this estimate for drivers who participate on fueleconomy.gov? The test statistic is: __________________(please round to two decimal places)The p-value:___________________________ for this hypothesis test is: (please round to four decimal places)(c) Please comment on the primary thing that is wrong about the following statement: There is sufficient evidence…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 you decide to compare the average monthly costs of similar insurance policies from the two companies. For a random sample of 15 people who buy insurance from Company A, the mean cost is $154 per month with a standard deviation of $13. For 11 randomly selected customers of Company B, you find that they pay a mean of $159 per month with a standard deviation of $16. Assume that both populations are approximately normal and that the population variances are equal to test Company A’s claim at the 0.02 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.