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Question
In a simple linear regression problem, suppose that
= 12.48 and 
= 124.8. Then the percentage of variation explained 
must be 0.90.Answer
This answer is hidden. It contains 1 characters.
Related questions
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The significance level also determines the rejection region.
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NARRBEGIN: SA_108_112The following values have been calculated using the TDIST and TINV functions in Excel. These values come from a t- distribution with 15 degrees of freedom.These values represent the probability to the right of the given positive values.Valuet -probability1.000.16661.200.12441.400.0909These values represent the positive t- value for a given probability in both tails (sum of both tails).Probabilityt -value0.201.34060.101.75310.052.1315NARRENDWhat would be the t-values where 0.95 of the values would fall within this interval?
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NARRBEGIN: SA_108_112The following values have been calculated using the TDIST and TINV functions in Excel. These values come from a t- distribution with 15 degrees of freedom.These values represent the probability to the right of the given positive values.
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NARRBEGIN: SA_108_112The following values have been calculated using the TDIST and TINV functions in Excel. These values come from a t- distribution with 15 degrees of freedom.These values represent the probability to the right of the given positive values.Valuet -probability1.000.16661.200.12441.400.0909These values represent the positive t- value for a given probability in both tails (sum of both tails).Probabilityt -value0.201.34060.101.75310.052.1315NARRENDWhat is the probability of a t-value between "1.40 and +1.40?
Q:
NARRBEGIN: SA_101_107The personnel department of a large corporation wants to estimate the family dental expenses of its employees to determine the feasibility of providing a dental insurance plan. A random sample of 12 employees in 2004 reveals the following family dental expenses (in dollars): 115, 370, 250, 93, 540, 225, 177, 425, 318, 182, 275, and 228. Use StatTools for your calculations.NARREND(A) Construct a 90% confidence interval estimate of the mean family dental expenses for all employees of this corporation.(B) What assumption about the population distribution must be made to answer (A)?(C) Interpret the 90% confidence interval constructed in (A).(D) Suppose you used a 95% confidence interval in (A). What would be your answer?(E) Suppose the fourth value were 593 instead of 93. What would be your answer to (A)? What effect does this change have on the confidence interval?(F) Construct a 90% confidence interval estimate for the standard deviation of family dental expenses for all employees of this corporation.(G) Interpret the 90% confidence interval constructed in (E).
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NARRBEGIN: SA_95_96An automobile dealer wants to estimate the proportion of customers who still own the cars they purchased six years ago. A random sample of 200 customers selected from the automobile dealer's records indicates that 88 still own cars that were purchased six years earlier.NARREND(A) Construct a 95% confidence interval estimate of the population proportion of all customers who still own the cars they purchased six years ago(B) How can the result in (A) be used by the automobile dealer to study satisfaction with cars purchased at the dealership?
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NARRBEGIN: SA_79_80
Auditors of Independent Bank are interested in comparing the reported value of all 1775 customer savings account balances with their own findings regarding the actual value of such assets. Rather than reviewing the records of each savings account at the bank, the auditors randomly selected a sample of 100 savings account balances from the frame. The sample mean and sample standard deviations were $505.75 and 360.95, respectively.
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(A) Construct a 90% confidence interval for the total value of all savings account balances within this bank. Assume that the population consists of all savings account balances in the frame.
(B) Interpret the 90% confidence interval constructed in (A).
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NARRBEGIN: SA_75_78The percent defective for parts produced by a manufacturing process is targeted at 4%. The process is monitored daily by taking samples of sizes n = 160 units. Suppose that today's sample contains 14 defectives.NARREND(A) Determine a 95% confidence interval for the proportion defective for the process today.(B) Based on your answer to (A), is it still reasonable to think the overall proportion defective produced by today's process is actually the targeted 4%? Explain your reasoning.(C) The confidence interval in (A) is based on the assumption of a large sample size. Is this sample size sufficiently large in this example? Explain how you arrived at your answer.(D) How many units would have to be sampled to be 95% confident that you can estimate the fraction of defective parts within 2% (using the information from today's sample)?
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NARRBEGIN: SA_67_68A sample of 9 production managers with over 15 years of experience has an average salary of $71,000 and a sample standard deviation of $18,000.NARREND(A) You can be 95% confident that the mean salary for all production managers with at least 15 years of experience is between what two numbers (the t-multiple with 8 degrees of freedom is 2.306)? What assumption are you making about the distribution of salaries?(B) What sample size would be needed to ensure that we could estimate the true mean salary of all production managers with more than 15 years of experience and have only 5 chances in 100 of being off by more than $4200?
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Samples of exam scores for employees before and after a training class would be examples of paired data
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The approximate standard error of the point estimate of the population total is .
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The upper limit of the 90% confidence interval for the population proportionp, given that n = 100; and = 0.20 is 0.2658.
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A 90% confidence interval estimate for a population mean is determined to be 72.8 to 79.6. If the confidence level is reduced to 80%, the confidence interval for becomes narrower.
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In order to construct a confidence interval estimate of the population mean, the value of must be given.
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A confidence interval is an interval estimate for which there is a specified degree of certainty that the actual true value of the population parameter will fall within the interval.
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If two random samples of size 40 each are selected independently from two populations whose variances are 35 and 45, then the standard error of the sampling distribution of the sample mean difference, , equals 1.4142.
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Two independent samples of sizes 20 and 25 are randomly selected from two normal populations with equal variances. In order to test the difference between the population means, the test statistic is:
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A parameter such as is sometimes referred to as a ________ parameter, because many times we need its value even though it is not the parameter of primary interest.
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NARRBEGIN: SA_55_62A landowner in Texas is offered $200,000 for the exploration rights to oil on her land, along with a 25% royalty on the future profits if oil is discovered. The landowner is also tempted to develop the field herself, believing that the interest in her land is a good indication that oil is present. In that case, she will have to contract a local drilling company to drill an exploratory well on her own. The cost for such a well is $750,000, which is lost forever if no oil is found. If oil is discovered, however, the landowner expects to earn future profits of $7,500,000. Finally, the landowner estimates (with the help of her geologist friend) the probability of finding oil on this site to be 60%.NARREND(A) Construct a decision tree to help the landowner make her decision. Make sure to label all decision and chance nodes and include appropriate costs, payoffs and probabilities.(B) What should the landowner do? Why?(C) Suppose the landowner is uncertain about the reliability of her geologist friend's estimate of the probability that oil will be found on her land. If she thinks the probability could be anywhere between 40% and 80%, would that change her decision?(D) Suppose that, in addition to the uncertainty about the probability of finding oil, the landowner is also uncertain about the cost of the exploratory well (could vary +/- 25%) and the future profits (could vary +/- 50%). To which of these variables is the expected value most sensitive?(E) What does the risk profile show about the relative risk levels for the landowner's two options?(F) Suppose the landowner suspects that she may be a somewhat risk-averse decision maker, because the she doesn"t feel there is as much of a difference between the two options as their expected values would indicate. She consults with a decision analysis expert who asks her to decide between two hypothetical alternatives: 1) a gamble with equal probabilities of winning an amount $X and losing an amount "$X/2, and 2) doing nothing, with a payoff of $0. The point at which she cannot decide between 1) and 2) is when X=$1,500,000. What is her risk tolerance if she uses an exponential utility function to model her preferences?(G) Apply the risk tolerance given in your answer to the previous question to the landowner's decision tree in (A). What is the optimal decision in this case? What is the resulting certainty equivalent?(H) If the landowner could hire an expert geologist prepare a report to help her make her decision, what is the most that information could be worth? Assume the geologist's information is perfectly reliable.
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A spider chart shows both the range (as a percentage) of the variability of the input variables as well as the resulting changes in the expected value
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The expected monetary value (EMV) criterion is sometimes referred to as "playing the averages" and for that reason should only be used for recurring decisions.
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For each possible decision and each possible outcome, the payoff table lists the monetary value earned by an organization, where a positive value represents a profit and a negative value represents a loss.
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The expected value of perfect information (EVPI) is equal to:
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b. EMV with free perfect information " EMV with information
c. EMV with free perfect information " EMV with no information
d. EMV with perfect information " EMV with less than perfect information