Here,
population mean = sample mean
population SS =sample SS
population variance = population SS / 16 = 2083 /16 = 130.188
population standard deviation = sqrt(population variance) = sqrt(130.188) = 11.41
sample variance = sample SS / 16 = 2083 /16 = 130.188
sample standard deviation = sqrt(sample variance) = sqrt(138.867) = 11.78
Prof. records the time in minutes does it take 16 students to complete an exam. Compute...
A professor records the time (in minutes) that it takes 16 students to complete an exam. Compute the SS, the variance, and the standard deviation assuming the 16 students constitute a population and assuming the 16 students constitute a sample. (Round your answers for variance and standard deviation to two decimal places.) 23 44 42 29 41 20 25 43 40 52 39 36 50 25 14 33
A professor records the time (in minutes) that it takes 16 students to complete an exam. Compute the SS, the variance, and the standard deviation assuming the 16 students constitute a population and assuming the 16 students constitute a sample. (Round your answers for variance and standard deviation to two decimal places.) 50 51 39 41 29 33 48 21 14 25 24 38 27 40 44 32
the time (in minutes) that it takes 16 students to complete an exam. Compute the SS, the variance, and the standard deviation g the 16 students constitute a population and assuming the 16 students constitute a sample. (Round your answers for variance and standard deviation to two decimal places 37 31 27 18 45 30 48 38 12 21 43 46 36 23 2346 (a) the 16 students constitute a population 9 standard deviation min (b) the 16 students constitute...
A random sample of 60 business students required an average of 50.7 minutes to complete a statistics exam. Assume that the population standard deviation to complete the exam is 10.4 minutes. 1.Calculate the margin of error if the confidence level is 98%.
The time required for Dr. B's students to complete the Statistics Exam is approximately normally distributed with a mean of 40.4 minutes and a standard deviation of 2.2 minutes. Let X be the random variable "the time required for Dr. B's students to complete the Statistics Exam." 6. With the above setting what time marks the 90th percentile? A. 37.562 minutes B. 37.584 minutes C. 43.238 minutes D. 43.216 minutes E. None of the above 7. Which of the following...
2. In a large class, an in-person exam, where the lazy students couldn't do cheating, in a random sample of 8 lazy students, average exam score was 42.6 with sample standard deviation 4.4, while in a random sample of 10 non-lazy students, the average exam score was 49.8 with sample standard deviation 3.1. Assume that the two populations (Exam scores for the students who do not cheat) are normally distributed. Exam was out of 60points. (a) Assume that the two...
1. The following data represent the travel time (in minutes) to school for seven students enrolled in Chaffey College. Treat the seven students as a sample. Student Travel time John 32 Jeff 24 Kevin 15 Ron 40 Noel 32 Mary 23 Christy 27 (a) Find standard deviation of the sample. (b) Find variance of the sample.
A professor finds that for 16 randomly selected students the time needed to complete an assignment had a sample mean of 66 minutes and a sample standard deviation of 40 minutes. If the time needed to complete the assignment is normally distributed, then the standard error of the mean is a. 5 b. 20 c. 16 d. 10
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Test: Exam#1 This Question: 1 pt 16 of 17 (17 complete) The ages (in years) of a random sample of shoppers at a gaming store are shown. Determine the range, mean, variance, and standard deviation of the sample data set 12, 21, 23, 14, 13, 17, 22.16, 15, 16 The range is 11 (Simplify your answer.) The mean is 16.9 (Simplify your answer. Round to the nearest tenth as needed.) The variance is (Simplify your answer. Round to the nearest...