Week Three Homework
Week Three Homework
33. The National Collegiate Athletic Association (NCAA) reported that the mean number of hours spent per week coaching and recruiting by college football assistant coaches during the season was 70. A random sample of 50 assistant coaches showed the sample mean to be 68.6 hours, with a standard deviation of 8.2 hours. A. Using the sample data, construct a 99% confidence interval for the population mean. The confidence interval is as follows: 68.6- 2.58*8.2/ã50, 686+ 2.58*8.2/ã50 CI(65.6-71.5) B. Does the 99% confidence interval include the value suggested by the NCAA? Interpret the result. The result is interpreted by the result because 70 is inside the interval. C. Suppose you decide to switch from a 99% to a 95% confidence interval. Without performing any calculations, will the interval increase, decrease, or stay the same? Which of the values in the formula will change? If we switched to a 95 percent interval then the interval will decrease. We will have a smaller confidence interval then part a. Only the Z value will change. We would use Z (0.025)=1.96 instead of Z(0.005)=2.58.
42. As a condition of employment, Fashion industries applicants must pass a drug test. Of the last 220 applicants, 14 failed the test. Develop a 99% confidence interval for the proportion of applicants that fail the test. Would it be reasonable to conclude that more than 10% of the applicants are now failing the test? 1. The sample proportion, p = 14/220 = 0.063636 and q = 1 - p = 0.936364 The standard error: SE = Sqrt(pq/n) = Sqrt(0.063636 * 0.936364/220) = Sqrt(0.000271) = 0.016457 For 99% confidence, z = 2.576 A 99 percent confidence interval for the proportion of applicants that fail the test is given by: (p – 2.576 * SE, p + 2.576 * SE)= (0.063636 - 2.576 * 0.016457, 0.063636 + 2.576 * 0.016457) = (0.063636 – 0.042394, 0.063636 + 0.042394) = (0.021242, 0.106031) The 99 percent
33. The National Collegiate Athletic Association (NCAA) reported that the mean number of hours spent per week coaching and recruiting by college football assistant coaches during the season was 70. A random sample of 50 assistant coaches showed the sample mean to be 68.6 hours, with a standard deviation of 8.2 hours. A. Using the sample data, construct a 99% confidence interval for the population mean. The confidence interval is as follows: 68.6- 2.58*8.2/ã50, 686+ 2.58*8.2/ã50 CI(65.6-71.5) B. Does the 99% confidence interval include the value suggested by the NCAA? Interpret the result. The result is interpreted by the result because 70 is inside the interval. C. Suppose you decide to switch from a 99% to a 95% confidence interval. Without performing any calculations, will the interval increase, decrease, or stay the same? Which of the values in the formula will change? If we switched to a 95 percent interval then the interval will decrease. We will have a smaller confidence interval then part a. Only the Z value will change. We would use Z (0.025)=1.96 instead of Z(0.005)=2.58.
42. As a condition of employment, Fashion industries applicants must pass a drug test. Of the last 220 applicants, 14 failed the test. Develop a 99% confidence interval for the proportion of applicants that fail the test. Would it be reasonable to conclude that more than 10% of the applicants are now failing the test? 1. The sample proportion, p = 14/220 = 0.063636 and q = 1 - p = 0.936364 The standard error: SE = Sqrt(pq/n) = Sqrt(0.063636 * 0.936364/220) = Sqrt(0.000271) = 0.016457 For 99% confidence, z = 2.576 A 99 percent confidence interval for the proportion of applicants that fail the test is given by: (p – 2.576 * SE, p + 2.576 * SE)= (0.063636 - 2.576 * 0.016457, 0.063636 + 2.576 * 0.016457) = (0.063636 – 0.042394, 0.063636 + 0.042394) = (0.021242, 0.106031) The 99 percent