Find the degrees of freedom and t value of the following sample size and confidence levels. 1. n= 31 a. 90% confidence level b. 98% confidence level c. 99% confidence level
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- You are interested in finding a 95% confidence interval for the mean number of visits for physical therapy patients. The data below show the number of visits for 13 randomly selected physical therapy patients. Round answers to 3 decimal places where possible. 27 19 24 14 6 18 25 5 23 28 12 23 20 a. To compute the confidence interval use a ? t distribution. b. With 95% confidence the population mean number of visits per physical therapy patient is between___and ____visits. c. If many groups of 13 randomly selected physical therapy patients are studied, then a different confidence interval would be produced from each group. About____percent of these confidence intervals will contain the true population mean number of visits per patient and about_____percent will not contain the true population mean number of visits per patient.You are interested in finding a 90% confidence interval for the average number of days of class that college students miss each year. The data below show the number of missed days for 10 randomly selected college students. Round answers to 3 decimal places where possible. 1 3 10 10 12 11 10 6 10 3 a. To compute the confidence interval use a Correct distribution. b. With 90% confidence the population mean number of days of class that college students miss is between and days. c. If many groups of 10 randomly selected non-residential college students are surveyed, then a different confidence interval would be produced from each group. About percent of these confidence intervals will contain the true population mean number of missed class days and about percent will not contain the true population mean number of missed class days.You are interested in finding a 98% confidence interval for the average commute that non-residential students have to their college. The data below show the number of commute miles for 14 randomly selected non-residential college students. 8 6 20 26 11 21 23 10 9 23 16 24 23 17 a. To compute the confidence interval use a ? z t distribution. b. With 98% confidence the population mean commute for non-residential college students is between and miles. c. If many groups of 14 randomly selected non-residential college students are surveyed, then a different confidence interval would be produced from each group. About percent of these confidence intervals will contain the true population mean number of commute miles and about percent will not contain the true population mean number of commute miles.
- Sample Size: Calculate the sample size. Karen from Big Chip cookie company discovered that on average her cookies have 12.9 chocolate chips with a standard deviation of 3.4 chocolate chips. If she wants to reduce the error for a 99% confidence interval to 1 chocolate chip, how many cookies should she sample?Assuming that the population is normally distributed, construct a 95% confidence interval for the population mean for each of the samples below. Explain why these two samples produce different confidence intervals even though they have the same mean and range. Sample A: 1 2 4 4 5 5 7 8 Full data set Sample B: 1 2 3 4 5 6 7 8 Question content area bottom Part 1 Construct a 95% confidence interval for the population mean for sample A. enter your response here≤μ≤enter your response hereListed below are the amounts of net worth (in millions of dollars) of the ten wealthiest celebrities in a country. Construct a 99% confidence interval. 254 199 192 169 163 158 158 158 158 158 What is the confidence interval estimate of the population mean u?
- Use the given degree of confidence and sample data to construct a confidence interval for the population mean u. Assume that the population has a normal distribution. The principal randomly selected six students to take an aptitude test. Their scores were: 88.0 84.1 74.9 83.2 83.7 85.5 Determine a 90% confidence interval for the mean score for all students. Seleccione una: A. 79.49 < µ < 86.98 B. 86.98 < H < 79.49 C. 79.59 < u < 86.88 D. 86.88 < u <79.59You are interested in finding a 95% confidence interval for the average number of days of class that college students miss each year. The data below show the number of missed days for 13 randomly selected college students. Round answers to 3 decimal places where possible. 4 11 11 7 10 7 9 2 8 11 9 4 12 a. To compute the confidence interval use a distribution. b. With 95% confidence the population mean number of days of class that college students miss is between and days. c. If many groups of 13 randomly selected non-residential college students are surveyed, then a different confidence interval would be produced from each group. About percent of these confidence intervals will contain the true population mean number of missed class days and about percent will not contain the true population mean number of missed class days.You are interested in finding a 98% confidence interval for the average number of days of class that college students miss each year. The data below show the number of missed days for 14 randomly selected college students. Round answers to 3 decimal places where possible. 12 5 7 4 6 9 11 11 9 8 5 9 10 11 a. To compute the confidence interval use a distribution. b. With 98% confidence the population mean number of days of class that college students miss is between and days. c. If many groups of 14 randomly selected non-residential college students are surveyed, then a different confidence interval would be produced from each group. About percent of these confidence intervals will contain the true population mean number of missed class days and about percent will not contain the true population mean number of missed
- 9 The standard error of the mean is, a 0.881 b 0.735 c 0.612 d 0.510Use the given sample data to construct the indicated confidence interval for the population mean. The football coach randomly selected ten players and timed how long each player took to perform a certain drill. The times (in minutes) were: 12.7 14.6 11.7 12.6 11.5 11.2 7.2 10.3 7.2 9.1 Determine a 95% confidence interval for the mean time for all players. O A. (9.00, 12.62) O B. (9.57, 12.05) O C. (9.09, 12.53 ) O D. (9.29, 12.30 ) O E. (9.50, 12.12)10 The interval estimate is, a 12.2 13.4 b 11.8 13.8 c 11.1 14.5 d 10.0 15.6