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A company that produces 8 -ounce low-fat yogurt cups wanted to estimate the mean number of calories for such cups. A random sample of 10 such cups produced the following numbers of calories. \(\begin{array}{lllllllll}147 & 159 & 153 & 146 & 144 & 148 & 163 & 153 & 143 & 158\end{array}\) Construct a \(99 \%\) confidence interval for the population mean. Assume that the numbers of calories for such cups of yogurt produced by this company have an approximately normal distribution.

Short Answer

Expert verified
Having calculated the sample mean, standard deviation of the sample and having determined the Z-score for a 99% confidence level, the confidence interval is then given by the formula \(\overline{x} \pm Z_{\alpha/2} \times \frac{s}{\sqrt{n}}\). This yields a range in which the population mean is likely to fall, with 99% confidence.

Step by step solution

01

Compute the sample mean

The sample mean (\(\overline{x}\)) of calorie consumption can be calculated by adding all the sample data and dividing by the total number of samples. So, \(\overline{x} = \frac{(147 + 159 + 153 + 146 + 144 + 148 + 163 + 153 + 143 + 158)}{10}\)
02

Calculate the sample standard deviation

To find the standard deviation (\(s\)), we subtract the mean from each element, square the result, add these squared values together, divide by (n-1) - in this case (10-1), and then take the square root. In formula, it's like: \(s = \sqrt{\frac{(147 - \overline{x})^{2} + (159 - \overline{x})^{2} + ... + (158 - \overline{x})^{2}}{10-1}}\)
03

Find the Z-score

For a 99% confidence interval, the Z-score is 2.576 (This value can be found in the standard normal (Z) distribution table).
04

Construct the confidence interval

The confidence interval can be obtained using the formula \(\overline{x} \pm Z_{\alpha/2} \times \frac{s}{\sqrt{n}}\), where \(Z_{\alpha/2}\) is the Z-score we obtained earlier, \(n\) the number of samples and \(s\) the standard deviation of samples. Substitute the respective values into the formula to get the confidence interval.

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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 by the symbol \( \overline{x} \), is the average calculated from a set of data points in a sample. To find the sample mean, you simply add together all the values in your sample and then divide by the number of data points. This formula can be expressed as \( \overline{x} = \frac{\sum x_i}{n} \), where \( \sum x_i \) is the sum of all sample values and \( n \) is the number of values in the sample. For instance, using calorie data from the yogurt cups exercise, you sum the numbers: 147, 159, 153, and so on, then divide by 10, since there are 10 numbers in total. The result gives us the sample mean.The purpose of calculating the sample mean is to find an estimate of the population mean, which is the average of all possible data points, not just those in the sample. This is particularly useful when the entire population is too large to measure directly.
Standard Deviation
Standard deviation, symbolized as \( s \) in samples, is a measure of the spread or variability of a set of data. It tells us how much the individual data points differ from the sample mean. To calculate the standard deviation, you subtract the mean from each data point to find the deviation, square each deviation, sum these squared deviations, divide by the number of data points minus one \((n-1)\), and take the square root of this quotient.In formula terms:- Calculate deviations: \((x_i - \overline{x})\)- Square these deviations: \((x_i - \overline{x})^2\)- Sum these squared deviations- Divide by \(n - 1\)- Take the square rootThis results in the standard deviation \( s = \sqrt{\frac{\sum (x_i - \overline{x})^2}{n-1}} \). This value tells us the typical amount by which the numbers in our sample differ from the mean, giving insight into the spread of the data.
Z-score
The Z-score is a statistical measure that describes a value's relation to the mean of a group of values. It is expressed in terms of standard deviations from the mean. A Z-score shows how many standard deviations an element is from the mean. For example, a Z-score of 2 means the data point is two standard deviations away from the mean. Z-scores are utilized extensively in statistics, especially in constructing confidence intervals and hypothesis testing. In the construction of a confidence interval, which gives a range within which we expect our population parameter to lie, the Z-score provides the multiplier of the standard error, helping us determine how wide our interval should be for a given level of confidence. The calculation of Z-scores is based on the normal distribution, where approximately 68% of the data falls within one standard deviation, around 95% within two, and about 99.7% within three. For the yogurt cups, the Z-score associated with a 99% confidence interval is 2.576, showing the high confidence level required.
Normal Distribution
A normal distribution, also known as the bell curve, is a probability distribution that is symmetric around the mean, showing that data near the mean are more frequent in occurrence than data far from the mean. In graphical terms, it forms the shape of a bell. The properties of a normal distribution include: - The data is symmetrically distributed with no skew. - The mean, median, and mode of the distribution are equal. - About 68% of data falls within one standard deviation from the mean, 95% within two, and 99.7% within three. The normal distribution is essential in statistics because it describes many natural phenomena, such as heights, test scores, and anything influenced by many small, random effects. The assumption of normal distribution allows us to make probabilistic statements about our data. In the context of the yogurt exercise, assuming that calorie counts of the yogurt follow a normal distribution gives us the validity to calculate a confidence interval using the Z-score method. This assumption allows for the use of standard statistical tools to make inferences about the population mean based on our sample data.

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Most popular questions from this chapter

A drug that provides relief from headaches was tried on 18 randomly selected patients. The experiment showed that the mean time to get relief from headaches for these patients after taking this drug was 24 minutes with a standard deviation of \(4.5\) minutes. Assuming that the time taken to get relief from a headache after taking this drug is (approximately) normally distributed, determine a \(95 \%\) confidence interval for the mean relief time for this drug for all patients.

You are working for a supermarket. The manager has asked you to estimate the mean time taken by a cashier to serve customers at this supermarket. Briefly explain how you will conduct this study. Collect data on the time taken by any supermarket cashier to serve 40 customers. Then estimate the population mean. Choose your own confidence level.

A May 8,2008 , report on National Public Radio (www.npr.org) noted that the average age of firsttime mothers in the United States is slightly higher than 25 years. Suppose that a random sample of 60 first-time mothers taken recently produced an average age of \(25.9\) years and a standard deviation of \(3.2\) years. Calculate a \(90 \%\) confidence interval for the average age of all current first-time mothers.

Foods that have less than \(.50\) gram of trans fat per serving are allowed to list the trans fat content as zero grams on labels. A random sample of 32 foods that list the amount of trans fat as zero grams but contain partially hydrogenated oils (the primary source of added trans fat) were evaluated for the amount of trans fat per serving. The sample mean and standard deviation of trans fat were found to be \(.34\) and \(.062\) gram, respectively. Calculate a \(95 \%\) confidence interval for the average amount of trans fat per serving in all foods that are listed as having zero grams of trans fat per serving but contain partially hydrogenated oils.

Calculating a confidence interval for the proportion requires a minimum sample size. Calculate a confidence interval, using any confidence level of \(90 \%\) or higher, for the population proportion for each of the following. a. \(n=200 \quad\) and \(\quad \hat{p}=.01\) b. \(n=160 \quad\) and \(\quad \hat{p}=.9875\) Explain why these confidence intervals reveal a problem when the conditions for using the normal approximation do not hold.

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