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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: \(6 \quad 17\) \(\begin{array}{lll}1 & 22 & 29\end{array}\) Assuming that these five students can be considered as representative of all students participating in the free checkup program, construct and interpret a \(95 \%\) confidence interval for the population 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 population mean number of months elapsed since the last visit to a dentist for the population of students participating in the program is approximately (5.87, 24.13) months. This means that we can be 95% confident that the true population mean falls within this interval.

Step by step solution

01

Calculate the sample mean

To calculate the sample mean, we sum up the number of months for each student and divide by the number of students (5). The sample mean is given by: \( \bar{x} = \frac{6 + 17 + 1 + 22 + 29}{5} = \frac{75}{5} = 15 \)
02

Calculate the sample standard deviation

The standard deviation is found by taking the square root of the average of the squared differences from the mean. s = \( \sqrt{\frac{\Sigma(x_i-\bar{x})^2}{n-1}} \) where \(x_i\) is each individual value, \(\bar{x}\) is the sample mean, and n is the number of observations. s = \( \sqrt{\frac{ (6-15)^2 + (17-15)^2 + (1-15)^2 + (22-15)^2 + (29-15)^2}{5-1}} \) s = \( \sqrt{\frac{ 81 + 4 + 196 + 49 + 196 }{4}} \) s = \( \sqrt{\frac{526}{4}} = \sqrt{131.5} \approx 11.47 \)
03

Determine degrees of freedom

The degrees of freedom for the t-distribution are given by: df = n - 1 = 5 - 1 = 4
04

Find the t-value

Look up the t-value corresponding to a 95% confidence level with 4 degrees of freedom using a t-distribution table. We can find that the t-value is approximately: t-value= 2.776
05

Calculate the margin of error

The margin of error is given by: Margin of Error = \( t-value \times \frac{s}{\sqrt{n}} \) Margin of Error = \( 2.776 \times \frac{11.47}{\sqrt{5}} \approx 9.13 \)
06

Construct the confidence interval

The confidence interval is given by: CI = \( \bar{x} \pm\) Margin of Error CI = \( 15 \pm 9.13 \) So, the 95% confidence interval for the population mean number of months elapsed since the last visit to a dentist for the population of students participating in the program is approximately (5.87, 24.13) months. This means that we can be 95% confident that the true population mean falls within this interval.

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Key Concepts

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

Population Mean Estimation
Population mean estimation involves predicting the average value of a variable for a larger group based on a smaller sample. This concept is crucial in statistics because it allows us to make educated guesses about a population's characteristics without surveying every individual. In the exercise, the mean time elapsed since the last dental visit for a group of students is estimated from the responses of just five participants. By calculating the sample mean, we get the best estimate of the group's mean. However, to understand how this sample mean relates to the entire population, we use a confidence interval.

A 95% confidence interval provides a range within which we can be reasonably sure (95% certain) that the population mean lies. The calculation of this range is based on the sample mean and accounts for variation within the sample as well as the sample size, ultimately giving us a more reliable estimate of the population mean.
Sample Standard Deviation Calculation
The sample standard deviation is a measure of how much each point in the sample varies from the mean of that sample. It effectively quantifies uncertainty and variability among the data points. In the context of our exercise, the sample standard deviation is calculated by taking the square root of the sum of squared differences between each student's response and the sample mean, divided by the degrees of freedom. This is a critical step in establishing the confidence interval because it's used to determine the margin of error, which adjusts our estimate to incorporate this variability, providing a more comprehensive and realistic range for our population mean estimation.

To ensure the students have an easier time understanding this concept, it would be beneficial to provide further examples of calculating standard deviation with varying data sets, and to show how even slight changes in data can affect the standard deviation.
Degrees of Freedom
Degrees of freedom refer to the number of values in a calculation that are free to vary. When estimating population parameters, like the mean, from sample data, the degrees of freedom are typically defined as the sample size minus one (- 1). This subtraction accounts for the fact that we've used the sample to estimate one parameter already, the mean, which constrains the variability of our data. In the exercise, because we have five students and we have already calculated the mean, we have four degrees of freedom.

Degrees of freedom are important because they directly impact the critical values we use in constructing confidence intervals, particularly when we refer to t-distribution, which tailors these values according to the sample size. Therefore, explaining to students the concept of degrees of freedom can help them better understand why it is subtracted and its relevance to their calculations.
T-distribution
The t-distribution is a type of probability distribution that is symmetric and bell-shaped, like the normal distribution, but has heavier tails. It's particularly useful when dealing with small sample sizes or when the population standard deviation is unknown, which is exactly our case in this exercise. As the degrees of freedom increase, the t-distribution approaches the normal distribution.

For constructing a confidence interval, the t-distribution is used to find a critical value (also known as the t-value) that, combined with the sample standard deviation and size, gives us the margin of error. It is imperative to understand that the t-distribution varies with degrees of freedom, which impacts the width of the confidence interval – with fewer degrees of freedom, the interval is wider, reflecting greater uncertainty. Students may benefit from interactive visuals or tools showing how the t-distribution changes with varying degrees of freedom to appreciate the impact on their confidence intervals.

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Most popular questions from this chapter

The formula used to calculate a confidence interval for the mean of a normal population is $$ \bar{x} \pm(t \text { critical value }) \frac{s}{\sqrt{n}} $$ What is the appropriate \(t\) critical value for each of the following confidence levels and sample sizes? a. \(90 \%\) confidence, \(n=12\) b. \(90 \%\) confidence, \(n=25\) c. \(95 \%\) confidence, \(n=10\)

Students in a representative sample of 65 first-year students selected from a large university in England participated in a study of academic procrastination ("Study Goals and Procrastination Tendencies at Different Stages of the Undergraduate Degree," Studies in Higher Education [2016]: 2028-2043). Each student in the sample completed the Tuckman Procrastination Scale, which measures procrastination tendencies. Scores on this scale can range from 16 to \(64,\) with scores over 40 indicating higher levels of procrastination. For the 65 first-year students in this study, the mean score on the procrastination scale was 37.02 and the standard deviation was 6.44 . a. Construct a \(95 \%\) confidence interval estimate of \(\mu,\) the mean procrastination scale for first-year students at this college. (Hint: See Example 12.7.) b. Based on your interval, is 40 a plausible value for the population mean score? What does this imply about the population of first-year students?

What percentage of the time will a variable that has a \(t\) distribution with the specified degrees of freedom fall in the indicated region? a. 10 df, between -2.23 and 2.23 b. 24 df, between -2.80 and 2.80 c. 24 df, to the right of 2.80

The dodo was a species of flightless bird that lived on the island of Mauritius in the Indian Ocean. The first record of human interaction with the dodo occurred in 1598 , and within 100 years the dodo was extinct due to hunting by humans and other newly introduced invasive species. After the extinction, the word "dodo" became synonymous with stupidity, implying that the birds lacked the intelligence to avoid or escape extinction. The closest existing relatives of the dodo are pigeons and doves. Researchers at the American Museum of Natural History used computed tomography (CT) scans to measure the brain size ("endocranial capacity") of one of the few existing preserved dodo birds, and to measure the brain sizes in samples of eight birds that are close relatives of dodos ("The First Endocast of the Extinct Dodo and an Anatomical Comparison Amongst Close Relatives," Zoological Journal of the Linnean Society [2016]: 950-953) The brain size for the dodo was \(4.17 \log \mathrm{mm}^{3}\). The following table contains the brain sizes for the sample of birds from related species (approximate values from a graph in the paper). a. Use the output at the bottom of the page from the Shiny App "Randomization Test for One Mean" to help you to carry out a randomization test of the hypothesis that the population mean brain size for birds that are relatives of the dodo differs from the established dodo brain size of 4.17 . b. What does the result of your test indicate about the brain size of the dodo?

Suppose that a random sample of size 64 is to be selected from a population with mean 40 and standard deviation 5. a. What are the mean and standard deviation of the sampling distribution of \(\bar{x}\) ? Describe the shape of the sampling distribution of \(\bar{x}\). b. What is the approximate probability that \(\bar{x}\) will be within 0.5 of the population mean \(\mu\) ? c. What is the approximate probability that \(\bar{x}\) will differ from \(\mu\) by more than \(0.7 ?\)

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