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Chapter 8: Problem 70

Several years ago, the U.S. Agency for International Development provided 238,300 metric tons of corn-soy blend (CSB) for emergency relief in countries throughout the world. \(\mathrm{CSB}\) is a highly nutritious, low-cost fortified food. As part of a study to evaluate appropriate vitamin C levels in this food, measurements were taken on samples of CSB produced in a factory. \({ }^{28}\) The following data are the amounts of vitamin C, measured in milligrams per 100 grams \((\mathrm{mg} / 100 \mathrm{~g})\) of blend, for a random sample of size 8 from one production run: $$\begin{array}{llllllll}\hline 26 & 31 & 23 & 22 & 11 & 22 & 14 & 31 \\\\\hline\end{array}$$ Construct and interpret a \(95 \%\) confidence interval for the mean amount of vitamin \(\mathrm{C} \mu\) in the \(\mathrm{CSB}\) from this production run.

Short Answer

Expert verified
The 95% confidence interval is (16.5, 28.5) mg/100g.

Step by step solution

01

Calculate Sample Mean

First, compute the mean of the given data set. Add all the values together and divide by the number of data points. Formula:\[ \bar{x} = \frac{\text{sum of all values}}{\text{number of values}} = \frac{26 + 31 + 23 + 22 + 11 + 22 + 14 + 31}{8} = \frac{180}{8} = 22.5 \]The sample mean \( \bar{x} \) is 22.5 mg/100g.
02

Calculate the Sample Standard Deviation

To calculate the sample standard deviation, first find the variance. Use the formula:\[s^2 = \frac{\sum (x_i - \bar{x})^2}{n-1}\]Substitute the values, where \( x_i \) are the data points and \( n \) is the sample size:\[s^2 = \frac{(26-22.5)^2 + (31-22.5)^2 + (23-22.5)^2 + (22-22.5)^2 + (11-22.5)^2 + (22-22.5)^2 + (14-22.5)^2 + (31-22.5)^2}{7}\]\[s^2 = \frac{12.25 + 72.25 + 0.25 + 0.25 + 132.25 + 0.25 + 72.25 + 72.25}{7} = \frac{362}{7} = 51.71 \]Now, find the standard deviation \( s \):\[s = \sqrt{51.71} \approx 7.19 \]
03

Determine the t-score for Confidence Interval

Use the t-distribution table to find the critical t-value for a 95% confidence level and 7 degrees of freedom (\( n-1 \)), where \( n = 8 \). For most t-tables, \( t_{0.025,7} \approx 2.364 \).
04

Calculate Margin of Error

The margin of error (ME) is calculated as:\[ME = t \times \frac{s}{\sqrt{n}}\]Substitute the known values:\[ME = 2.364 \times \frac{7.19}{\sqrt{8}} \approx 2.364 \times 2.54 \approx 6.00 \]
05

Construct the Confidence Interval

Now form the 95% confidence interval for the mean vitamin C content. The formula is:\[(\bar{x} - ME, \bar{x} + ME) = (22.5 - 6.00, 22.5 + 6.00) = (16.5, 28.5) \]
06

Interpret the Confidence Interval

The 95% confidence interval for the mean amount of vitamin C is from 16.5 to 28.5 mg/100g. This means we are 95% confident that the true mean amount of vitamin C in the CSB from this production run falls within this interval.

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Key Concepts

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

Sample Standard Deviation
The sample standard deviation is a measure used to quantify the amount of variation or dispersion in a set of data values. In the context of our exercise, it tells us how spread out the vitamin C measurements are from the sample mean. To compute it, we first determine the variance by averaging the squared differences of each data point from the mean.

For our dataset, we calculate the variance as follows:
  • Subtract the sample mean (22.5) from each value in the dataset.
  • Square each of these differences.
  • Sum all the squared differences.
  • Divide by one less than the sample size (7 for our sample of 8).
The standard deviation is simply the square root of the variance. In this exercise, it's approximately 7.19 mg/100g, showing us how much individual vitamin C measurements deviate, on average, from the sample mean.
T-Distribution
The t-distribution is a statistical distribution used primarily when dealing with small sample sizes. It is similar to the normal distribution but has thicker tails, which provides a more conservative estimate for uncertainty when sample sizes are small.

In the context of the confidence interval calculation for the vitamin C exercise, the t-distribution helps account for the variability in estimating the population mean from our small sample. With our sample of size 8, the degrees of freedom are 7 (n-1). Using a t-table, we found the t-score for a 95% confidence level, which is crucial for determining how much the mean estimate might vary if we took another sample. This t-value was approximately 2.364, reflecting a more conservative estimate than the standard normal distribution would yield. This conservativeness is essential in maintaining our confidence level's integrity, ensuring that our interval indeed covers the true mean 95% of the time.
Margin of Error
The margin of error quantifies the extent of uncertainty or potential error in our estimate of the population mean. It expresses how much above and below our sample mean the true population mean might lie, considering the sample data variability.

To calculate it, we use the formula:
  • Multiply the sample standard deviation by the t-score.
  • Divide by the square root of the sample size.
For this exercise, we multiplied the t-score (2.364) by the sample standard deviation (7.19) and divided by the square root of 8 to arrive at a margin of error of approximately 6 mg/100g. This value tells us that the true mean vitamin C level could reasonably be about 6 mg/100g higher or lower than our sample mean of 22.5 mg/100g.
Sample Mean
The sample mean is a measure of central tendency, representing an average of the data points in the sample. It provides a best estimate of the population mean in the absence of full population data.

For the given exercise, we calculated the sample mean by summing all the vitamin C measurements from our sample and dividing by the number of observations:
  • Sum of the data points: 26 + 31 + 23 + 22 + 11 + 22 + 14 + 31 = 180
  • Number of observations: 8
The result is a sample mean of 22.5 mg/100g, which acts as an approximation for the average vitamin C content we expect in the population of CSB. This measure, along with the margin of error, allows us to construct the confidence interval and gain a better understanding of where the true population mean may lie.

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Most popular questions from this chapter

Young people have a better chance of full-time employment and good wages if they are good with numbers. How strong are the quantitative skills of young Americans of working age? One source of data is the National Assessment of Educational Progress (NAEP) Young Adult Literacy Assessment Survey, which is based on a nationwide probability sample of households. The NAEP survey includes a short test of quantitative skills, coveringmainly basic arithmetic and the ability to apply it to realistic problems. Scores on the test range from 0 to \(500 .\) For example, a person who scores \(233 \mathrm{can}\) add the amounts of two checks appearing on a bank deposit slip; someone scoring 325 can determine the price of a meal from a menu; a person scoring 375 can transform a price in cents per ounce into dollars per pound. Suppose that you give the NAEP test to an SRS of 840 people from a large population in which the scores have mean 280 and standard deviation \(\sigma=60\). The mean \(\bar{x}\) of the 840 scores will vary if you take repeated samples. (a) Describe the shape, center, and spread of the sampling distribution of \(\bar{x}\). (b) Sketch the sampling distribution of \(\bar{x}\). Mark its mean and the values \(1,2,\) and 3 standard deviations on either side of the mean. (c) According to the \(68-95-99.7\) rule, about \(95 \%\) of all values of \(\bar{x}\) lie within a distance \(m\) of the mean of the sampling distribution. What is \(m ?\) Shade the region on the axis of your sketch that is within \(m\) of the mean. (d) Whenever \(\bar{x}\) falls in the region you shaded, the population mean \(\mu\) lies in the confidence interval \(\bar{x} \pm m\). For what percent of all possible samples does the interval capture \(\mu ?\)

One reason for using a \(t\) distribution instead of the standard Normal curve to find critical values when calculating a level \(C\) confidence interval for a population mean is that (a) \(z\) can be used only for large samples. (b) \(z\) requires that you know the population standard deviation \(\sigma\). (c) \(z\) requires that you can regard your data as an \(\mathrm{SRS}\) from the population. (d) \(z\) requires that the sample size is at most \(10 \%\) of the population size. (e) a \(z\) critical value will lead to a wider interval than a \(t\) critical value.

Determine the point estimator you would use and calculate the value of the point estimate. What proportion of students are willing to report cheating by other students? A student project put this question to an SRS of 172 undergraduates at a large university: "You witness two students cheating on a quiz. Do you go to the professor?" Only 19 answered "Yes."

A Pew Intemet and American Life Project survey found that 392 of 799 randomly selected teens reported texting with their friends every day. (a) Calculate and interpret a \(95 \%\) confidence interval for the population proportion \(p\) that would report texting with their friends every day. (b) Is it plausible that the true proportion of American teens who text with their friends every day is \(0.45 ?\) Use your result from part (a) to support your answer.

The admissions director from Big City University found that (107.8,116.2) is a \(95 \%\) confidence interval for the mean IQ score of all freshmen. Discuss whether each of the following explanations is correct. (a) There is a \(95 \%\) probability that the interval from 107.8 to 116.2 contains \(\mu\) (b) There is a \(95 \%\) chance that the interval (107.8, 116.2 ) contains \(\bar{x}\) (c) This interval was constructed using a method that produces intervals that capture the true mean in \(95 \%\) of all possible samples. (d) If we take many samples, about \(95 \%\) of them will contain the interval (107.8,116.2) (e) The probability that the interval (107.8,116.2) captures \(\mu\) is either 0 or 1 , but we don't know which.
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