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Construct the confidence interval estimate of the mean. Listed below are amounts of arsenic \((\mu \mathrm{g},\) or micrograms, per serving) in samples of brown rice from California (based on data from the Food and Drug Administration). Use a \(90 \%\) confidence level. The Food and Drug Administration also measured amounts of arsenic in samples of brown rice from Arkansas. Can the confidence interval be used to describe arsenic levels in Arkansas? $$\begin{array}{cccccccccc} 5.4 & 5.6 & 8.4 & 7.3 & 4.5 & 7.5 & 1.5 & 5.5 & 9.1 & 8.7 \end{array}$$

Short Answer

Expert verified
No, the confidence interval for California cannot describe arsenic levels in Arkansas. The intervals are specific to California brown rice.

Step by step solution

01

- Calculate the Sample Mean

First, find the mean (average) of the given data. Add all the amounts of arsenic together and then divide by the number of data points. \[ \text{Mean} = \frac{5.4 + 5.6 + 8.4 + 7.3 + 4.5 + 7.5 + 1.5 + 5.5 + 9.1 + 8.7}{10} \]
02

- Calculate the Sample Standard Deviation

Next, calculate the standard deviation of the sample. This measures the amount of variation in the arsenic levels. Use the formula for sample standard deviation, where \( \bar{x} \) is the sample mean: \[ s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2} \]
03

- Determine the Standard Error

Calculate the standard error of the mean, which provides an estimate of the precision of the sample mean. The standard error (SE) is given by: \[ \text{SE} = \frac{s}{\sqrt{n}} \] where \( s \) is the sample standard deviation and \( n \) is the sample size.
04

- Find the Critical Value

Determine the critical value for the 90% confidence level. Since we are working with a sample and not the whole population, use a t-distribution with \( n-1 \) degrees of freedom. For a 90% confidence interval and 9 degrees of freedom (since \( n = 10 \)), the critical value \( t_{\frac{\alpha}{2}} \) can be found using t-tables or statistical software.
05

- Construct the Confidence Interval

Use the sample mean, the critical value, and the standard error to construct the confidence interval for the mean arsenic level. The confidence interval is given by: \[ \bar{x} \pm t_{\frac{\alpha}{2}} \times \text{SE} \] Where \( \bar{x} \) is the sample mean, \( t_{\frac{\alpha}{2}} \) is the critical value, and \( \text{SE} \) is the standard error.
06

- Determine the Confidence Interval

Calculate the lower and upper bounds of the confidence interval using the values obtained: \[ \left(\bar{x} - t_{\frac{\alpha}{2}} \times \text{SE}, \bar{x} + t_{\frac{\alpha}{2}} \times \text{SE}\right) \]
07

- Answer the Question about Arkansas

Evaluate if the confidence interval calculated can be applied to arsenic levels in brown rice from Arkansas. Since the data we have is specific to California, and arsenic levels could vary based on geographical and environmental factors, it is not appropriate to use this confidence interval to describe arsenic levels in Arkansas.

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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, often represented as \(\bar{x}\) or mu, is the average value of a set of data points. It's calculated by adding together all of the values in the sample and then dividing by the number of values. For example, for the arsenic data \(5.4, 5.6, 8.4, 7.3, 4.5, 7.5, 1.5, 5.5, 9.1 \text{and} 8.7\), you sum all these values and divide by 10 (since there are 10 data points). The formula is: \[\text{Mean} = \frac{5.4 + 5.6 + 8.4 + 7.3 + 4.5 + 7.5 + 1.5 + 5.5 + 9.1 + 8.7}{10} = 6.35\] This gives the average amount of arsenic in the brown rice samples from California. The sample mean is a critical piece because it serves as the center point for constructing the confidence interval.
Standard Deviation
The standard deviation measures how spread out the data points are around the mean. It indicates the amount of variation or dispersion of a set of values. For a sample, the formula to calculate the standard deviation is: \[s = \sqrt{\frac{1}{n-1} \sum_{i=1}^n (x_i - \bar{x})^2}\] In this formula, \(n\) is the sample size, \(x_i\) are the individual data points, and \(\bar{x}\) is the sample mean. High standard deviation means that the data points are spread out over a larger range of values, whereas a low standard deviation indicates that the data points are close to the mean. In our case, once we calculate the sample deviation, we have more insight into how the arsenic levels vary in the brown rice samples.
T-Distribution
The t-distribution is a type of probability distribution that is symmetric and bell-shaped, much like the normal distribution, but has heavier tails. This means it is more prone to producing values that fall further from its mean. It is particularly useful when dealing with small sample sizes (typically n < 30). The t-distribution is used instead of the normal distribution to estimate population parameters when the sample size is small and the population standard deviation is unknown. For our purpose, we use the t-distribution to find the critical value when creating the confidence interval. The degrees of freedom for the t-distribution are determined by the sample size minus one (n - 1). In this case, with 10 samples, we use 9 degrees of freedom.
Confidence Level
The confidence level represents the percentage of all possible samples that can be expected to include the true population parameter. A 90% confidence level means that if we were to take 100 different samples and compute a confidence interval for each sample, we would expect that 90 of the intervals will contain the population mean. It does not imply that the probability that our specific calculated interval contains the population mean is 90%. To construct a 90% confidence interval for the mean arsenic level in brown rice, we will use the sample mean, the standard error, and the critical value from the t-distribution with 9 degrees of freedom. Combining these elements appropriately helps estimate the range within which the true mean of arsenic levels in California brown rice samples lies, giving us confidence in the precision of our results.

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Most popular questions from this chapter

Use the data and confidence level to construct a confidence interval estimate of \(p,\) then address the given question. One of Mendel's famous genetics experiments yielded 580 peas, with 428 of them green and 152 yellow. a. Find a \(99 \%\) confidence interval estimate of the percentage of green peas. b. Based on his theory of genetics, Mendel expected that \(75 \%\) of the offspring peas would be green. Given that the percentage of offspring green peas is not \(75 \%,\) do the results contradict Mendel's theory? Why or why not?

Use the given sample data and confidence level. In each case, (a) find the best point estimate of the population proportion \(p ;(b)\) identify the value of the margin of error \(E ;(c)\) construct the confidence interval; (d) write a statement that correctly interprets the confidence interval. In a study of the accuracy of fast food drive-through orders, McDonald's had 33 orders that were not accurate among 362 orders observed (based on data from \(Q S R\) magazine). Construct a \(95 \%\) confidence interval for the proportion of orders that are not accurate.

Construct the confidence interval estimate of the mean. Data Set 3 "Body Temperatures" in Appendix B includes a sample of 106 body temperatures having a mean of \(98.20^{\circ} \mathrm{F}\) and a standard deviation of \(0.62^{\circ} \mathrm{F}\). Construct a \(95 \%\) confidence interval estimate of the mean body temperature for the entire population. What does the result suggest about the common belief that \(98.6^{\circ} \mathrm{F}\) is the mean body temperature?

Use the relatively small number of given bootstrap samples to construct the confidence interval. In a Consumer Reports Research Center survey, women were asked if they purchase books online, and responses included these: no, yes, no, no. Letting "yes" = 1 and letting "no" = 0, here are ten bootstrap samples for those responses: \(\\{0,0,0,0\\},\\{1,0,1,0\\} \\{1,0,1,0\\},\\{0,0,0,0\\},\\{0,0,0,0\\},\\{0,1,0,0\\},\\{0,0,0,0\\},\\{0,0,0,0\\},\\{0,1,0,0\\},\\{1,1,0,0\\} .\) Using only the ten given bootstrap samples, construct a \(90 \%\) confidence interval estimate of the proportion of women who said that they purchase books online.

Use the data and confidence level to construct a confidence interval estimate of \(p,\) then address the given question. In a survey of 3005 adults aged 57 through 85 years, it was found that \(81.7 \%\) of them used at least one prescription medication (based on data from "Use of Prescription and Over-the-Counter Medications and Dietary Supplements Among Older Adults in the United States," by Qato et al., Journal of the American Medical Association, Vol. \(300,\) No. 24 ). a. How many of the 3005 subjects used at least one prescription medication? b. Construct a \(90 \%\) confidence interval estimate of the percentage of adults aged 57 through 85 years who use at least one prescription medication. c. What do the results tell us about the proportion of college students who use at least one prescription medication?

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