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##### Project 6 Math 102

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Question;Project 6 Math 102Please write your name on the answer sheet. All Multiple Choice Questions.1 When a directional hypothesis is being tested and the rejection region is on the right of the mean, then the alternate hypothesis must be of the form:A Less thanB Greater thanC Not equalD Equal2 As the significance level decreases the chances of rejecting increases.A TrueB FalseC Cannot determineD Stays the same3 A directional alternate hypothesis results in a two tail test.A TrueB FalseC Cannot determineD Depends on the alpha4 If samples of size 36 are selected in a one sample hypothesis test using t-scores, then what would be the degrees of freedom?A 36B 37C 35D 345-13 In 1996 the mean monthly bill for cell phone users was \$45.70 with a standard deviation of \$22. In 1997 the mean monthly bill for cell phone use for a sample of 41 phone bills was \$42.40. At the 5% significance level, is there sufficient evidence to conclude that the cell phone use decreased from 1996 to 1997?)(Assume the population is a Normal Population.)5 What would be the Null Hypothesis?A ? = 42.40B ? 42.40E ? > 45.706 What is the Alternate Hypothesis?A ? = 42.40B ? 42.40E ? > 45.707 What is the given standard deviation?A 5B 42.40C 45.70D 22E 3.448 What is the sample mean outcome?A 5B 42.40C 45.70D 22E 3.449 What is the DSM Standard deviation?A 5B 42.40C 8.95D 22E 3.4410 What is the degrees of freedom?A 40B 21C 4D 8E There are no degrees of freedom needed for this example11 What is the value of the critical cut off score?A - 1.65B -2.33C -1.68D 1.68E 1.6512 What is the values of the test statistic?A -0.96B -1.65C -1.68D 0.96E None of the above13 What is the final decision?A Do Not Reject HoB Do Not Reject HaC Reject HoD Reject HaE Neither Reject or Do not Reject, since it is a tie.14-22 An automobile manufacturer claims that their leading compact car averages 32 mpg or more in the city. A local police department purchased 16 of these cars. The cars were driven exclusively in the city and averaged 28.2 mpg with a standard deviation of 5 mpg.Is the manufacturer?s claim too high? (Use? = 1%)(Assume the population is aNormal Population.)14 What would be the Null Hypothesis?A ? = 32B ? 28.2E ? = 515 What is the Alternate Hypothesis?A ? = 32B ? 3216 What is the given standard deviation?A 2.5B 5C 16D 14.1E 817 What is the sample mean outcome?A 35B 16C 5D 28.2E 3218 What is the DSM Standard deviation?A 2.5B 16C 5D 8E 1.2519 What is the degrees of freedom?A 16B 31C 4D 15E There are no degrees of freedom needed for this example20 What is the value of the critical cut off score?A - 1.75B -2.60C +1.75D -2.13 and 2.13E 2.6021 What is the values of the test statistic?A -3.04B -1.4C 5D 28.2E 1.422 What is the final decision?A Do Not Reject HoB Do Not Reject HaC Reject HoD Reject HaE Neither Reject nor Do not Reject, since it is a tie.23-31 According to the US Bureau of Labor Statistics, the average consumer spent \$1729 on apparel and services in 1997. That same year, 31 consumers in the Midwest spent a mean amount of \$1552 with a standard deviation of \$362. At the 5% level, do the data provide sufficient evidence to conclude that the 1997 expenditures for consumers in the Midwest were significantly different from the national mean of \$1729?)(Assume the population is a Normal Population.)23 What would be the Null Hypothesis?A ? = 1552B ? not equal 1552C ? = 28.2D ? not equal 1729E ? = 172924 What is the Alternate Hypothesis?A ? = 1552B ? not equal 1552C ? = 28.2D ? not equal 1729E ? = 172925 What is the given standard deviation?A 362B 65.02C 31D 1.63E 0.0526 What is the sample mean outcome?A 1729B 362C 1552D 31E 527 What is the DSM Standard deviation?A 65.02B 362C 1.63D 278.75E 9626.6628 What is the degrees of freedom?A 361B 30C 4D 31E There are no degrees of freedom needed for this example29 What is the value of the critical cut off score?A - 1.70 and 1.70B 2.04C -2.04 and 2.04D -1.70E -2.75 and 2.7530 What is the value of the test statistic?A -3.5B 2.72C 3.5D -2.04E -2.7231 What is the final decision?A Do Not Reject HoB Do Not Reject HaC Reject HoD Reject HaE Neither Reject or Do not Reject, since it is a tie.32-40 The Food and Nutrition Board of the National Academy of Sciences recommends a daily allowance of calcium for adults of 790 mg. A random sample of 38 people with incomes below the poverty level had a mean daily intake of 747.7 mg of calcium. At the 1% significance level, do the data provide evidence to conclude that the calcium intake of people with incomes below the poverty level is less than the recommended allowance of 790 mg? Assume that? = 210 mg.)(Assume the population is a Normal Population.)32 What would be the Null Hypothesis?A ? = 747.7B ? 790E ? = 79033 What is the Alternate Hypothesis?A ? = 747.7B ? 790E ? = 79034 What is the given standard deviation?A 1B 210C 790D 34.07E 2.6235 What is the sample mean outcome?A 210B 790C 747.7D 38E 121.2936 What is the DSM Standard deviation?A 1B 210C 790D 34.07E 2.6237 What is the degrees of freedom?A 37B 209C 38D 39E There are no degrees of freedom needed for this example38 What is the value of the critical cut off score?A - 1.68B -2.33C -2.42D -2.58E 1.6839 What is the values of the test statistic?A -3.8B -16.13C -.01D 16.13E -1.2440 What is the final decision?A Do Not Reject HoB Do Not Reject HaC Reject HoD Reject HaE Neither Reject or Do noeject, since it is a tie.41-49 The average retail price for bananas in 1998 was 49.0 cents per pound, as reported by the US Department of Agriculture. Recently, a random sample of 25 markets had the mean price for bananas of 52.4 cents per pound with a standard deviation of 3.8 cents. Can you conclude that the current mean retail price for bananas is significantly higher atthe 5% level of significance?)(Assume the population is a Normal Population.)41 What would be the Null Hypothesis?A ? = 52.4B ? 49E ? = 4942 What is the Alternate Hypothesis?A ? = 52.4B ? 49E ? = 4943 What is the given standard deviation?A 0.76B 3.8C 9.8D 10.48E 2544 What is the sample mean outcome?A 52.4B 25C 49D 3.8E 545 What is the DSM Standard deviation?A 10.48B 9.8C 3.8D 0.76E 2546 What is the degrees of freedom?A 24B 48C 4D 51E There are no degrees of freedom needed for this example47 What is the value of the critical cut off score?A - 1.71B 1.71C 1.65D 2.06E 2.4948 What is the values of the test statistic?A -4.47B 2.58C -2.58D 4.47E 0.8949 What is the final decision?A Do Not Reject HoB Do Not Reject HaC Reject HoD Reject HaE Neither Reject or Do not Reject, since it is a tie.50 Having a baby in a US hospital costs \$2628, on average, with a standard deviation of \$456. Active management of labor (AML) is a group of interventions designed to help reduce the length of labor and the rate of cesarean deliveries. Physicians were interested in determining whether AML would reduce the cost for delivery. So they researched the cost of 200 AML deliveries and found a mean cost of \$2498. At the 1% significance level, do the data provide sufficient evidence to conclude that, on average, AML reduces the cost of having a baby in the US hospitals?)(Assume the population is a Normal Population.)50 The above hypothesis test would make use of which statistic?a) Normal z scores b) T ? scores c) C ? scores d) r scores

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