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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 as follows: \(\begin{array}{lllll}6 & 17 & 11 & 22 & 29\end{array}\) 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 solution will involve applying statistical concepts related to calculating the sample mean, the sample standard deviation, the standard error, finding corresponding critical value and finally calculating the 95% confidence interval. The final result will be an interval that estimates the average number of months since the last dentist visit for the population of students participating in the free checkup program.

Step by step solution

01

Calculating the sample mean

First, calculate the mean (average) of the provided data. This can be done by adding up the months recorded for all five students and then dividing by the number of students. In this case, the calculation is as follows: (6+17+11+22+29)/5.
02

Calculating the standard error

Next, calculate the standard error of the mean. This is the standard deviation of the sample divided by the square root of the sample size. Since this is a small sample and we don't know the population standard deviation, we must first calculate the sample standard deviation using the formula for the sample standard deviation. Then proceed with the calculation of the standard error.
03

Finding the critical value

Then, find the critical value that corresponds to a 95% confidence level. Since the sample size is small and the population standard deviation is unknown, use the t-distribution. For a 95% confidence level and 4 degrees of freedom (since n-1=5-1), the critical value (also called t-star) is commonly 2.776.
04

Constructing the confidence interval

The formula for a confidence interval is: Confidence Interval = sample mean ± (critical value * standard error). Substitute the values previously calculated into this formula to obtain the 95% confidence interval.

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Most popular questions from this chapter

Suppose that a random sample of 50 bottles of a par. ticular brand of cough medicine is selected and the alcohol content of each bottle is determined. Let \(\mu\) denote the average alcohol content for the population of all bottles of the brand under study. Suppose that the sample of 50 results in a \(95 \%\) confidence interval for \(\mu\) of \((7.8,9.4)\). a. Would a \(90 \%\) confidence interval have been narrower or wider than the given interval? Explain your answer. b. Consider the following statement: There is a \(95 \%\) chance that \(\mu\) is between \(7.8\) and \(9.4\). Is this statement correct? Why or why not? c. Consider the following statement: If the process of selecting a sample of size 50 and then computing the corresponding \(95 \%\) confidence interval is repeated 100 times, 95 of the resulting intervals will include \(\mu\). Is this statement correct? Why or why not?

The eating habits of 12 bats were examined in the article "Foraging Behavior of the Indian False Vampire Bat" (Biotropica \([1991]: 63-67) .\) These bats consume insects and frogs. For these 12 bats, the mean time to consume a frog was \(\bar{x}=21.9 \mathrm{~min}\). Suppose that the standard deviation was \(s=7.7 \mathrm{~min} .\) Construct and interpret a \(90 \%\) confidence interval for the mean suppertime of a vampire bat whose meal consists of a frog. What assumptions must be reasonable for the one-sample \(t\) interval to be appropriate?

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. No Fast Food \(\begin{array}{llllllll}2331 & 1918 & 1009 & 1730 & 1469 & 2053 & 2143 & 1981 \\ 1852 & 1777 & 1765 & 1827 & 1648 & 1506 & 2669 & \\ \text { Fast Food } & & & & & & \\ 2523 & 1758 & 934 & 2328 & 2434 & 2267 & 2526 & 1195 \\ 890 & 1511 & 875 & 2207 & 1811 & 1250 & 2117 & \end{array}\) 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 estimate the 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.

A random sample of \(n=12\) four-year-old red pine trees was selected, and the diameter (in inches) of each tree's main stem was measured. The resulting observations are as follows: \(\begin{array}{llllllll}11.3 & 10.7 & 12.4 & 15.2 & 10.1 & 12.1 & 16.2 & 10.5\end{array}\) \(\begin{array}{llll}11.4 & 11.0 & 10.7 & 12.0\end{array}\) a. Compute a point estimate of \(\sigma\), the population standard deviation of main stem diameter. What statistic did you use to obtain your estimate? b. Making no assumptions about the shape of the population distribution of diameters, give a point estimate for the population median diameter. What statistic did you use to obtain the estimate? c. Suppose that the population distribution of diameter is symmetric but with heavier tails than the normal distribution. Give a point estimate of the population mean diameter based on a statistic that gives some protection against the presence of outliers in the sample. What statistic did you use? d. Suppose that the diameter distribution is normal. Then the 90 th percentile of the diameter distribution is \(\mu\) t \(1.28 \sigma\) (so \(90 \%\) of all trees have diameters less than this value). Compute a point estimate for this percentile. (Hint: First compute an estimate of \(\mu\) in this case; then use it along with your estimate of \(\sigma\) from Part (a).)

An article in the Chicago Tribune (August 29,1999 ) reported that in a poll of residents of the Chicago suburbs, \(43 \%\) felt that their financial situation had improved during the past year. The following statement is from the article: "The findings of this Tribune poll are based on interviews with 930 randomly selected suburban residents. The sample included suburban Cook County plus DuPage, Kane, Lake, McHenry, and Will Counties. In a sample of this size, one can say with \(95 \%\) certainty that results will differ by no more than 3 percent from results obtained if all residents had been included in the poll." Comment on this statement. Give a statistical argument to justify the claim that the estimate of \(43 \%\) is within \(3 \%\) of the true proportion of residents who feel that their financial situation has improved.

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