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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

The interval from \(-2.33\) to \(1.75\) captures an area of \(.95\) under the \(z\) curve. This implies that another large-sample \(95 \%\) confidence interval for \(\mu\) has lower limit \(\bar{x}=2.33 \frac{\sigma}{\sqrt{n}}\) and upper limit \(\bar{x}\) + \(1.75 \frac{\sigma}{\sqrt{n}}\). Would you recommend using this \(95 \%\) interval over the \(95 \%\) interval \(\bar{x} \pm 1.96 \frac{\sigma}{\sqrt{n}}\) discussed in the text? Explain. (Hint: Look at the width of each interval.)

USA Today (October 14, 2002) reported that \(36 \%\) of adult drivers admit that they often or sometimes talk on a cell phone when driving. This estimate was based on data from a sample of 1004 adult drivers, and a bound on the error of estimation of \(3.1 \%\) was reported. Assuming a \(95 \%\) confidence level, do you agree with the reported bound on the error? Explain.

The article "Most Canadians Plan to Buy Treats. Many Will Buy Pumpkins, Decorations and/or Costumes" (Ipsos-Reid, October 24, 2005) summarized results from a survey of 1000 randomly selected Canadian residents. Each individual in the sample was asked how much he or she anticipated spending on Halloween during \(2005 .\) The resulting sample mean and standard deviation were \(\$ 46.65\) and \(\$ 83.70\), respectively. a. Explain how it could be possible for the standard deviation of the anticipated Halloween expense to be larger than the mean anticipated expense. b. Is it reasonable to think that the distribution of the variable anticipated Halloween expense is approximately normal? Explain why or why not. c. Is it appropriate to use the \(t\) confidence interval to estimate the mean anticipated Halloween expense for Canadian residents? Explain why or why not. d. If appropriate, construct and interpret a \(99 \%\) confidence interval for the mean anticipated Halloween expense for Canadian residents.

One thousand randomly selected adult Americans participated in a survey conducted by the Assodated Press (June 2006 ). When asked "Do you think it is sometimes justified to lie or do you think lying is never justified?" \(52 \%\) responded that lying was never justified. When asked about lying to avoid hurting someone's feelings, 650 responded that this was often or sometimes okay. a. Construct a \(90 \%\) confidence interval for the proportion of adult Americans who think lying is never justified. b. Construct a \(90 \%\) confidence interval for the proportion of adult American who think that it is often or sometimes okay to lie to avoid hurting someone's feelings. c. Comment on the apparent inconsistency in the responses given by the individuals in this sample.

How much money do people spend on graduation gifts? In 2007, the National Retail Federation (www.nrf.com) surveyed 2815 consumers who reported that they bought one or more graduation gifts that year. The sample was selected in a way designed to produce a sample representative of adult Americans who purchased graduation gifts in \(2007 .\) For this sample, the mean amount spent per gift was \(\$ 55.05 .\) Suppose that the sample standard deviation was \(\$ 20 .\) Construct and interpret a \(98 \%\) confidence interval for the mean amount of money spent per graduation gift in 2007 .
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