A certain number of university students have been asked how many hours of homework they do...
Based on all student records at Camford University, students spend an average of 5.5 hours per week playing organized sports. The population's standard deviation is 2.2 hours per week. Based on a sample of 121 students, Healthy Lifestyles Incorporated (HLI) would like to apply the central limit theorem to make various estimates. a. Compute the standard error of the sample mean. (Round your answer to 1 decimal place.) Standard error C. Calculate the probability that the sample mean will be...
significance level is .05
10. Towson University wants to estimate how many hours per week students work at paying jobs. How many students do they need to survey, in order to have an estimate, to within a half hour, the mean number of hours a TU student works per week? Assume that in a survey done five years ago, the sample standard deviation was.42 hours.
10. Towson University wants to estimate how many hours per week students work at paying...
A sample of 200 college freshmen was asked how many hours per week they spent playing video games. The following frequency distribution presents the results. Number of Hours Frequency 1.0- 3.9 20 4.0- 6.9 32 7.0- 9.9 47 10.0- 12.9 28 13.0- 15.9 24 16.0- 18.9 18 19.0- 21.9 12 22.0- 24.9 7 25.0- 27.9 4 28.0- 30.9 8 1) What is the class width? 2) What percentage of students play video games for less than 19 hours per week?...
A survey among freshmen at a certain university revealed that the number of hours spent studying the week before final exams was normally distributed with mean 26 and standard deviation 5. Use the TI-84 PLUS calculator to answer the following. Round the answer to at least four decimal places. (a) What proportion of students studied more than 36 hours? (b) What is the probability that a randomly selected student spent between 13 and 32 hours studying? (c) What proportion of...
A survey among freshmen at a certain university revealed that the number of hours spent studying the week before final exam was normally distributed with mean 29 and standard deviation 5. Find percentile corresponding to P equals 59% of the number of hours studying. Round to 4 decimal places.
A professor wants to estimate how many hours per week her students study. A simple random sample of 56 students had a mean of 19 hours of studying per week. Construct a 98% confidence interval for the mean number of hours a student studies per week. Assume that the population standard deviation is known to be 2.4 hours per week. Round to two decimal places.
2. Auniversity asked 10 graduate students how many hours of homework they were planning to do that week. Here are their responses: 14 13 15 21 19 24 25 28 25 31 Answer the following questions about this data: a. What's the mean number of hours of homework done by these students per week? What's the median? Why are these figures different? b. What's the range of the data? c. Is the data skewed? If so, in what direction? How...
In the Department of Education at UR University, student records suggest that the population of students spends an average of 5.60 hours per week playing organized sports. The population's standard deviation is 2.00 hours per week. Based on a sample of 49 students, Healthy Lifestyles Incorporated (HLI) would like to apply the central limit theorem to make various estimates. a. Compute the standard error of the sample mean. (Round your answer to 2 decimal places.) Standard error b....
A survey among freshmen at a certain university revealed that the number of hours spent studying the week before final exams was normally distributed with mean 35 and standard deviation 7. Find percentile corresponding to p = 36% of the number of hours studying. Write only a number as your answer. Round to two decimal places (for example: 20.81).
A professor at a university wants to estimate the average number of hours of sleep students get during exam week. On the first day of exams, she asked 27 students how many hours they had slept the night before. The average of the sample was 3.79 with a standard deviation of 1.224. When estimating the average amount of sleep with a 99% confidence interval, what is the margin of error? 1) 0.3549 2) 0.6546 3) 0.5839 4) 0.2356 5) 0.6527