1. Five students visiting the student health center for a free dental examination during National Dental Hygiene Month were asked how many months had passed since their last visit to a dentist. Their responses were as follows.7 19 12 20 27 Assuming that these five students can be considered a random sample of all students participating in the free checkup program, construct a 95% confidence interval for the mean number of months elapsed since the last visit to a dentist for the population of students participating in the program. (Round your answers to three decimal places.) , 2. The authors of the paper "Length of Stay, Wait Time to Surgery and 30-Day Mortality for Patients with Hip Fractures After Opening of a Dedicated Orthopedic Weekend Trauma Room"† were interested in estimating the mean time that patients who broke a hip had to wait for surgery after the opening of a new hospital facility. They reported that for a representative sample of 204 people with a fractured hip, the sample mean time between arriving at the hospital and surgery to repair the hip was 28.5 hours and that the sample standard deviation of the wait times was 16.8 hours.If we had access to the raw data (the 204 individual wait-time observations), we might begin by looking at a boxplot. The authors of the paper commented that there were several outliers in the data set, which might cause us to question the normality of the wait-time distribution, but because the sample size is large, it is still appropriate to use the t confidence interval.We can use the confidence interval of this section to estimate the actual mean wait time for surgery. We know the following.sample size = n = 204 sample mean wait time = x = 28.5 hours sample standard deviation = s = 16.8 hoursThe sample was thought to be representative of the population of patients with a fractured hip. So, with ? denoting the mean wait time for surgery for patients with a fractured hip, we can estimate ? using a 90% confidence interval.From Appendix Table 3, we use t critical value = 1.645 (from the z critical value row because df = n − 1 = 203 > 120, the largest number of degrees of freedom in the table). The 90% confidence interval for ? isx ± (t critical value) s n = 28.5 ± (1.645) 16.8 204 = 28.5 ± 1.94 = (26.56, 30.44).Based on this sample, we are 90% confident that ? is between 26.56 hours and 30.44 hours. This interval is fairly narrow indicating that our information about the value of ? is relatively precise.The paper also gave data on surgery wait times for a representative sample of 405 patients with fractured hips who were seen at this hospital before the new facility was opened. The mean wait time for the patients in this sample was 31.5 hours and the standard deviation of wait times was 27.0 hours. Because the sample was a representative sample and the sample size was large, it is appropriate to use the one-sample t confidence interval to estimate the mean wait time for surgery before the new facility was opened.A graphing calculator or statistical software can produce a one-sample t confidence interval. Using a 90% confidence level, output from Minitab for the sample of patients who had surgery before the new facility was opened is shown below.One-Sample TNMeanStDevSE Mean90% CI40531.5027.001.34(29.29, 33.71)The 90% confidence interval for the mean wait time before the new facility extends from 29.29 hours to 33.71 hours.Calculate the 99% confidence interval for the mean wait time before the new facility was opened. (Use technology to calculate the critical value. Round your answers to three decimal places.) ,

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MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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1. Five students visiting the student health center for a free dental examination during National Dental Hygiene Month were asked how many months had passed since their last visit to a dentist. Their responses were as follows.

7 19 12 20 27

Assuming that these five students can be considered a random sample of all students participating in the free checkup program, construct a 95% confidence interval for the mean number of months elapsed since the last visit to a dentist for the population of students participating in the program. (Round your answers to three decimal places.)

The authors of the paper "Length of Stay, Wait Time to Surgery and 30-Day Mortality for Patients with Hip Fractures After Opening of a Dedicated Orthopedic Weekend Trauma Room"† were interested in estimating the mean time that patients who broke a hip had to wait for surgery after the opening of a new hospital facility. They reported that for a representative sample of 204 people with a fractured hip, the sample mean time between arriving at the hospital and surgery to repair the hip was 28.5 hours and that the sample standard deviation of the wait times was 16.8 hours.

If we had access to the raw data (the 204 individual wait-time observations), we might begin by looking at a boxplot. The authors of the paper commented that there were several outliers in the data set, which might cause us to question the normality of the wait-time distribution, but because the sample size is large, it is still appropriate to use the t confidence interval.

We can use the confidence interval of this section to estimate the actual mean wait time for surgery. We know the following.

sample size	=	n = 204

sample mean wait time	=	x = 28.5 hours

sample standard deviation	=	s = 16.8 hours

The sample was thought to be representative of the population of patients with a fractured hip. So, with ? denoting the mean wait time for surgery for patients with a fractured hip, we can estimate ? using a 90% confidence interval.

From Appendix Table 3, we use t critical value = 1.645 (from the z critical value row because

df = n − 1 = 203 > 120,

the largest number of degrees of freedom in the table). The 90% confidence interval for ? is

x ± (t critical value)

28.5 ± (1.645)

16.8

204

28.5 ± 1.94

(26.56, 30.44).

Based on this sample, we are 90% confident that ? is between 26.56 hours and 30.44 hours. This interval is fairly narrow indicating that our information about the value of ? is relatively precise.

The paper also gave data on surgery wait times for a representative sample of 405 patients with fractured hips who were seen at this hospital before the new facility was opened. The mean wait time for the patients in this sample was 31.5 hours and the standard deviation of wait times was 27.0 hours. Because the sample was a representative sample and the sample size was large, it is appropriate to use the one-sample t confidence interval to estimate the mean wait time for surgery before the new facility was opened.

A graphing calculator or statistical software can produce a one-sample t confidence interval. Using a 90% confidence level, output from Minitab for the sample of patients who had surgery before the new facility was opened is shown below.

One-Sample T

N	Mean	StDev	SE Mean	90% CI
405	31.50	27.00	1.34	(29.29, 33.71)

The 90% confidence interval for the mean wait time before the new facility extends from 29.29 hours to 33.71 hours.

Calculate the 99% confidence interval for the mean wait time before the new facility was opened. (Use technology to calculate the critical value. Round your answers to three decimal places.)

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