Question

The calcium (Ca) content of a powdered mineral substance was analyzed 10 times with the following percent compositions recorded: $$ \begin{array}{lllll} .0271 & .0282 & .0279 & .0281 & .0268 \\ .0271 & .0281 & .0269 & .0275 & .0276 \end{array} $$ a. Find a \(99 \%\) confidence interval for the true calcium content of this substance. b. What does the phrase "99\% confident" mean? c. What assumptions must you make about the sampling procedure so that this confidence interval will be valid? What does this mean to the chemist who is performing the analysis?

Short answer

Answer: A 99% confidence interval for the true calcium content, such as (0.02638,0.02918), means that we are confident that the true population mean of calcium content will fall within this interval 99% of the time if we were to repeatedly draw samples of the same size from the population. The required assumptions for this confidence interval to be valid include that the sample is random and not biased, the sample data is independent, and the sample data is either normally distributed or the sample size is large enough (usually n > 30) for the Central Limit Theorem to apply. These assumptions help ensure that the confidence interval produced is valid and represents the true calcium content. If the chemist does not follow these assumptions, the confidence interval may not be reliable.

Step-by-step solution

1. Calculate the sample mean and sample standard deviation

First, we need to find the sample mean (\(\bar{x}\)) and the sample standard deviation (s) of the given data. The sample mean (\(\bar{x}\)) is calculated by summing up all the values and dividing by the number of values: $$\bar{x} = \frac{(.0271 + .0282 + .0279 +\dots+ .0275 + .0276)}{10} = 0.02778$$ Next, we calculate the sample standard deviation (s). Subtract the mean from each value, square each result, sum the squared values, divide the result by N-1 (degrees of freedom: 10-1 = 9), and finally take the square root. The equation is: $$s = \sqrt{\frac{(\bar{x}-x_1)^2 + (\bar{x}-x_2)^2 + \dots + (\bar{x}-x_{10})^2}{10-1}} = 0.00045$$

2. Determine the t-value

Now, we need to find the t-value for a 99% confidence interval with 9 degrees of freedom. We use a t-table or an online calculator to find the t-value. For a 99% confidence interval with 9 degrees of freedom, the t value approximately equals 3.250.

3. Calculate the confidence interval

We use our t-value, sample mean and sample standard deviation to find the confidence interval. The formula for the confidence interval is: $$\bar{x} \pm t \times \frac{s}{\sqrt{n}}$$ Plugging in our values, we get: $$0.02778 \pm 3.250 \times \frac{0.00045}{\sqrt{10}}$$ $$0.02778 \pm 0.00140$$ So, the 99% confidence interval for the true calcium content is: $$(0.02638,0.02918)$$ a. The 99% confidence interval for the true calcium content of this substance is \((0.02638,0.02918)\).

4. Explain the "99% confident" concept

When we say we are "99% confident" in the context of this problem, it means that we are confident that the true population mean of calcium content will fall within the interval \((0.02638,0.02918)\) 99% of the time if we were to repeatedly draw samples of the same size from the population. b. The phrase "99% confident" means that if we were to draw many samples and calculate the confidence interval for each, around 99% of those intervals would contain the true population mean for calcium content.

5. Assumptions and their significance

In order to create a valid confidence interval in this situation, the following assumptions must be made: 1. The sample is random and not biased. 2. The sample data is independent. 3. The sample data is normally distributed or the sample size is large enough (usually n > 30) for the Central Limit Theorem to apply. For the chemist performing the analysis, making these assumptions means that they can trust the confidence interval is reliable and representative of the true calcium content if the sampling procedure was done correctly. c. The assumptions we have to make are that the sampling procedure was random, independent, and either normally distributed or had a large enough size. These assumptions help ensure that the confidence interval produced is valid and represents the true calcium content. If the chemist does not follow these assumptions, the confidence interval may not be reliable.

Key concepts

Sample Mean

When analyzing the calcium content of a substance, one of the first statistical measures calculated is the sample mean. The sample mean, denoted by \( \bar{x} \), represents the average of a set of data points. It provides insight into the central tendency of the data. In this exercise, we have ten measurements of calcium content. To calculate the sample mean, you sum all the values recorded and divide by the total number of values. Here, the sample mean is calculated as follows:\[\bar{x} = \frac{(.0271 + .0282 + .0279 + .0281 + .0268 + .0271 + .0281 + .0269 + .0275 + .0276)}{10} = 0.02778\]This mean gives us a baseline understanding of the central value around which the individual recordings of calcium content are distributed. It acts as a benchmark to assess the variability in the data.

Sample Standard Deviation

The sample standard deviation is a measure of how much the values in a data set deviate from the sample mean. It provides an indication of the variability or spread of the data points. In simpler terms, it tells us how "spread out" the values are around the mean. For our calcium analysis, the sample standard deviation \( s \) is calculated through a multi-step process:

• Subtract the sample mean from each observed value.
• Square each of these differences.
• Sum all the squared differences.
• Divide this total by \( n-1 \), where \( n \) is the number of observations (in this case, 10).
• Finally, take the square root of the result.

Using this method, the sample standard deviation of the calcium content is calculated as approximately 0.00045. This result indicates that the individual calcium measurements do not deviate significantly from their mean, suggesting a relatively tight concentration around the mean value.

T-Distribution

The t-distribution is a type of probability distribution that is used when estimating population parameters for small sample sizes, or when the population standard deviation is not known. Unlike the normal distribution, the t-distribution has heavier tails, which means that it is more prone to producing values that fall farther from the mean.When calculating confidence intervals for the calcium content, the t-distribution is especially useful because our sample size is relatively small (n = 10). For our situation, with 9 degrees of freedom (\( n-1 \)), we use a t-table to find the critical t-value for a 99% confidence level, which is approximately 3.250. This t-value allows us to construct a confidence interval that reflects the uncertainty inherent in a small sample size.

Central Limit Theorem

The Central Limit Theorem (CLT) is a fundamental principle in statistics. It states that, regardless of the population's distribution, the distribution of the sample mean will approximate a normal distribution as the sample size grows sufficiently large, typically n > 30. In our analysis of calcium content, while our sample size is below the commonly suggested size for CLT to be directly applicable, the theorem provides assurance that as we continue to draw larger samples, the sample mean will tend to form a normal distribution. For small samples, the assumption of normal distribution is crucial if the CLT doesn't hold due to sample size. This underlying principle supports the use of the t-distribution when the sample size is small, providing a method to make inferences about the population mean even when exact distribution is unknown.