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Chapter 9: Problem 50

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: \(\begin{array}{ccccc}6 & 17 & 11 & 22 & 29\end{array}\) 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.

Short Answer

Expert verified
The 95% confidence interval for the average number of months since the last dentist visit among students participating in this program, based on this sample, is approximately (5.54, 28.46) months.

Step by step solution

01

Calculate the Sample Mean

To find the sample mean, all the sample values must be added together and divided by the sample size. For this set: \( \frac{(6 + 17 + 11 + 22 + 29)}{5} = 17 \) months.
02

Calculate the Sample Standard Deviation

The standard deviation of the sample data can be calculated using the formula: \( \sqrt{\frac{\sum(x_{i} - \mu)^2}{n - 1}} \), where \( x_{i} \) are the sample data, \( \mu \) is the sample mean and \( n \) is the sample size. In this case, the sample standard deviation is approximately \( \sqrt{\frac{(17 - 6)^2 + (17 - 17)^2 + (17 - 11)^2 + (17 - 22)^2 + (17 - 29)^2}}{5 - 1}} \) = 9.17 months.
03

Determine the Degrees of Freedom

The degrees of freedom for the t-statistic is equal to the sample size minus 1. In this case, it's 5 - 1 = 4.
04

Determine the Critical Value (t*)

The critical value for a 95% confidence interval can be found from the t-distribution table given degrees of freedom and the desired level of confidence. With 4 degree of freedom the t* is approximately 2.776.
05

Calculate the Confidence Interval

Now, the 95% confidence interval for the population mean can be computed using the following formula: \( \mu = \bar{x} \pm t^* \frac{s}{\sqrt{n}} \), where \( \bar{x} \) is the sample mean, \( t^* \) is the critical value from the t-distribution, \( s \) is the sample standard deviation, and \( n \) is the sample size. By substituting the given values, the interval is \( 17 \pm 2.776 \frac{9.17}{\sqrt{5}} = 17 \pm 11.46 \). Hence, the 95% confidence interval for the population mean is approximately (5.54, 28.46) months.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sample Mean
The sample mean is a fundamental concept in statistics that helps us understand the average value of a data set. It is calculated by summing all the individual data points and then dividing by the total number of data points. In this exercise, we have data from five students about how long it has been since they last visited a dentist, with the values being 6, 17, 11, 22, and 29 months. To compute the sample mean, we sum these values to get 85 and divide by the number of observations, which is 5. This results in a sample mean of 17 months.
The sample mean gives us a single value that represents the center of the data, helping us make comparisons and predictions about larger groups or populations.
Standard Deviation
Standard deviation measures how spread out the numbers in a data set are from the mean. It provides an idea of the variability or diversity of the data. To compute the standard deviation, we start by calculating the variance. This involves finding the difference between each data point and the sample mean, squaring those differences, summing them up, and dividing by the number of observations minus one, which is the degrees of freedom.
For our data set, the calculated standard deviation is approximately 9.17 months. A larger standard deviation indicates more spread in the data, while a smaller one signifies that data points are close to the mean.
  • Steps:
  • Find the difference from the mean for each data point.
  • Square these differences.
  • Sum them up.
  • Divide by 4 (since 5-1 = 4).
  • Take the square root of the result.
Degrees of Freedom
Degrees of freedom is a statistical concept that refers to the number of independent values or quantities that can vary in an analysis without being constrained by other values. It's often used in the context of statistical tests and calculations, such as standard deviation and t-distribution. In the given exercise, the degrees of freedom for calculating the standard deviation and later for the confidence interval is the sample size minus one.
So, for the five students' data, the degrees of freedom is 5 - 1 = 4. This adjustment is necessary because when we estimate a parameter from a sample, we lose some degree of freedom as it's being used to calculate the sample mean itself.
t-Distribution
The t-distribution is a probability distribution that is used in statistics, particularly when dealing with small sample sizes, or when the population standard deviation is unknown. It is similar to the normal distribution but has heavier tails, meaning it is more prone to producing values far away from its mean.
In the context of confidence intervals, the t-distribution allows us to account for the additional uncertainty inherent in estimating a population parameter from a small sample. For a 95% confidence interval and 4 degrees of freedom, we refer to a t-distribution table to find the critical value, often denoted as \(t^*\).
In this scenario, \(t^*\) is approximately 2.776. This value plays a crucial role in adjusting the confidence interval width. The t-distribution becomes more like the normal distribution as the sample size grows, resulting in narrower confidence intervals.

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Most popular questions from this chapter

In a survey of 1000 randomly selected adults in the United States, participants were asked what their most favorite and what their least favorite subject was when they were in school (Associated Press, August 17, 2005 ). In what might seem like a contradiction, math was chosen more often than any other subject in both categories! Math was chosen by 230 of the 1000 as the favorite subject, and it was also chosen by 370 of the 1000 as the least favorite subject. a. Construct a \(95 \%\) confidence interval for the proportion of U.S. adults for whom math was the favorite subject in school. b. Construct a \(95 \%\) confidence interval for the proportion of U.S. adults for whom math was the least favorite subject.

Data consistent with summary quantities in the article referenced in Exercise \(9.3\) on total calorie consumption on a particular day are given for a sample of children who did not eat fast food on that day and for a sample of children who did eat fast food on that day. Assume that it is reasonable to regard these samples as representative of the population of children in the United States. $$\begin{aligned} &\text { No Fast Food }\\\ &\begin{array}{llllllll} 2331 & 1918 & 1009 & 1730 & 1469 & 2053 & 2143 & 1981 \\ 1852 & 1777 & 1765 & 1827 & 1648 & 1506 & 2669 & \end{array} \end{aligned}$$ $$\begin{aligned} &\text { Fast Food } \\ &\begin{array}{rrrrrrrr} 2523 & 1758 & 934 & 2328 & 2434 & 2267 & 2526 & 1195 \\ 890 & 1511 & 875 & 2207 & 1811 & 1250 & 2117 & \end{array} \end{aligned}$$ a. Use the given information to estimate the mean calorie intake for children in the United States on a day when no fast food is consumed. b. Use the given information to estimate the mean calorie intake for children in the United States on a day when fast food is consumed. c. Use the given information to produce estimates of the standard deviations of calorie intake for days when no fast food is consumed and for days when fast food is consumed.

The article "Career Expert Provides DOs and DON'Ts for Job Seekers on Social Networking" (CareerBuilder.com, August 19, 2009) included data from a survey of 2667 hiring managers and human resource professionals. The article noted that many employers are using social networks to screen job applicants and that this practice is becoming more common. Of the 2667 people who participated in the survey, 1200 indicated that they use social networking sites (such as Facebook, MySpace, and LinkedIn) to research job applicants. For the purposes of this exercise, assume that the sample is representative of hiring managers and human resource professionals. Construct and interpret a \(95 \%\) confidence interval for the proportion of hiring managers and human resource professionals who use social networking sites to research job applicants.

The Associated Press (December 16,1991 ) reported that in a random sample of 507 people, only 142 correctly described the Bill of Rights as the first 10 amendments to the U.S. Constitution. Calculate a \(95 \%\) confidence interval for the proportion of the entire population that could give a correct description.

The formula used to compute a large-sample confidence interval for \(p\) is $$\hat{p} \pm(z \text { critical value }) \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$ What is the appropriate \(z\) critical value for each of the following confidence levels? a. \(95 \%\) d. \(80 \%\) b. \(90 \%\) e. \(85 \%\) c. \(99 \%\)
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