T/F?1.For a continuous distribution, the probability of an exact value is always zero.2. When constructing a confidence interval for a sample proportion, the t distribution is appropriate whenever the sample size is small3. The larger the p-value, the greater the chance of rejecting the null hypothesis4. The error term is the difference between the observed value of the dependent variable and the Mean value of the dependent variable5. The least squares simple linear regression line minimizes the sum of squares of the vertical deviations between the estimated line and the observed data points6. The sample mean, the sample proportion and the sample variance are all unbiased estimators of the corresponding population parameters7. You cannot make a Type II error when the null hypothesis is trueThe Ohio Department of Agriculture tested 203 fuel samples across the state in 1999 for accuracy of the reported octane level. For premium grade, 14 out of 105 samples failed (they didn’t meet ASTM specification and the FTC Octane posting rule). How many samples would be needed to create a 99% confidence interval that is within 0.02 of the true proportion of premium grade fuel-quality failures? 4148 2838 1913 744 54The least squares regression line minimizes the sum of the Differences between actual and predicted Y values Absolute deviations between actual and predicted Y values Absolute deviations between actual and predicted X values Squared differences between actual and predicted Y values Squared differences between actual and predicted X valuesGiven the following information about a hypothesis test of the difference between two means based on independent random samples, what is the calculated value of the test statistic? Assume that the samples are obtained from normally distributed populations having equal variances. HA: mu1>mu2, Xbar 1=12, Xbar 2=9, s1=5, s2=3, n1=13, n2=10. t = 1.674 t = 1.501 t = 2.823 t = 1.063In a study conducted by UCLA, it was found that 25% of college freshmen support increased military spending. If 10 college freshmen are randomly selected, find the probability that at less than 5 out of the selected 10 support increased military spending 0.0584 0.9219 0.1460 0.0781 0.9803 A salesperson goes door-to-door in a residential area to demonstrate the use of a new Household appliance to potential customers. She has found from her years of experience that after demonstration, the probability of purchase (long run average) is 0.40. To perform satisfactory on the job, the salesperson needs More than Four orders this week. If she performs 12 demonstrations this week, what is the probability of her being satisfactory? 0.5618 0.4382 0.2128 0.6652 0.7747A confidence interval increases in width as: The level of confidence increases Sample Size n decreases Sample Standard Deviation increases All of the aboveShow WorkAn insurance company estimates 45 percent of its claims have errors. The insurance company wants to estimate with 95 percent confidence the proportion of claims with errors. What sample size is needed if they wish to be within 4 percentage points of the actual?A small town has a population of 20,000 people. Among these 2,000 regularly visit a popular local bar. A sample of 225 people who visit the bar is surveyed for their annual expenditures in the bar. It is found that on average each person who regularly visits the bar spends about $2500 per year in the bar with a standard deviation of $450. Construct a 99 percent confidence interval around the mean annual expenditure in the barAt a recent meeting of educational researchers comparison were made between the type of college freshmen attend and the numbers who drop out. A random sample of freshmen show the following results: (keep two decimals in calculating expected frequencies) 4Yr public 4Yr private 2Yr public 2Yr privatedrop out 10 9 15 9don’t drop 26 28 18 27Use a significance level of .05 and determine if the type of school and the dropout rate are independentConsider the following partial computer output for a multiple regression model.Predictor Coefficient Standard DeviationConstant 41.225 6.380X1 1.081 1.353X2 -18.404 4.547Analysis of VarianceSource DF SSRegression 2 2270.11Error 26 3585.75Find Total Sum of Squares, Explained Variation, SSE, MSE, R-Squared, and Test the overall usefulness of the model at 1% level of significance calculating the F-Statistic.A set of final examination grades in a calculus course was found to be normally distributed with a mean of 77.2 and standard deviation of 5. Only 2.5% of the students taking the test scored higher than what grade?For a binomial probability experiment, with n=150 and p=.1, we can use the normal approximation to the binomial distribution even without continuity correctionFor a hypothesis test about a population proportion or mean, if the level of significance is less than the p-value, the null hypothesis is rejectedThe changing ecology of the swamps in Louisiana has been the subject of much environmental research. One water-quality parameter of concern is the total phosphorous level. Suppose that the EPA makes 15 measurements in one area of the swamp, yielding a mean level of total phosphorus of 12.3 parts per billion (ppb) and a standard deviation of 5.4 ppb. The EPA wants to test whether the data support the conclusion that the mean level is less than 15 ppb. Calculate the appropriate test statistic to test the hypotheses. 7.50 1.94 3.88 -1.94 -7.50 A state education agency designs and administers high school proficiency exams. Historically, time to complete the exam was an average of 120 minutes. Recently the format of the exam changed and the claim has been made that the time to complete the exam has changed. A sample of 50 new exam times yielded an average time of 118 minutes. The standard deviation is assumed to be 5 minutes. Calculate a 99% confidence interval. [117.61 120.09] [117.36 119.39] [116.18 119.82] [115.67 120.33] [115.82 120.18] A manufacturer of a chemical used in glue, attempting to control the amount of a hazardous chemical its workers are exposed to, has given instructions to halt production if the mean amount in the air exceeds 3.0ppm. A random sample of 50 air specimens produced the following statistics: sample mean=3.1 ppm , sample standard deviation=0.5 ppm. Calculate the appropriate test statistic to test the hypotheses. 0.20 -0.20 10.00 1.41 -1.41

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