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Here are the red blood cell counts (in \(10^{6}\) cells per microliter) of a healthy person measured on each of 15 days: \(\begin{array}{lllll}5.4 & 5.2 & 5.0 & 5.2 & 5.5\end{array}$$\begin{array}{lllll}5.3 & 5.4 & 5.2 & 5.1 & 5.3\end{array}$$\begin{array}{ll}5.3 & 4.9\end{array}$$\begin{array}{ll}5.2 & 5.2\end{array}\)Find a \(95 \%\) confidence interval estimate of \(\mu,\) the true mean red blood cell count for this person during the period of testing.

Short Answer

Expert verified
Answer: The 95% confidence interval estimate for the true mean red blood cell count lies between 5.1186 and 5.3242 (\(10^{6}\) cells per microliter).

Step by step solution

01

Organize the Data

First, let's write down the given red blood cell counts to make it easier to work with: 5.4, 5.2, 5.0, 5.2, 5.5, 5.3, 5.4, 5.2, 5.1, 5.3, 5.3, 4.9, 5.2, 5.2
02

Calculate the Sample Mean

To calculate the sample mean, we add up all the values and divide by the total number of observations. Let \(n\) denote the total number of observations. Sample Mean \(\bar{x} = \frac{1}{n}\sum_{i=1}^{n} x_i\) \(\bar{x} = \frac{5.4 + 5.2 + 5.0 + 5.2 + 5.5 + 5.3 + 5.4 + 5.2 + 5.1 + 5.3 + 5.3 + 4.9 + 5.2 + 5.2}{14} = 5.2214\)
03

Calculate the Sample Standard Deviation

Next, we will calculate the sample standard deviation of the data. Sample standard deviation \(s\) is given by the formula: \(s = \sqrt{\frac{1}{n-1}\sum_{i=1}^{n} (x_i - \bar{x})^2 }\) \(s = \sqrt{\frac{(5.4-5.2214)^2 + (5.2-5.2214)^2 + \cdots + (5.2-5.2214)^2}{13}} \approx 0.1788 \)
04

Find the t-value for a 95% Confidence Interval

We will use the t-distribution to find the t-value for our 95% confidence interval estimate. With 14 observations, we have 13 degrees of freedom (n-1). Using a calculator or a t-table, we can find the t-value that corresponds to the 0.975 percentile (because the 95% confidence interval is centered around the mean, we consider the 2-tail distribution and 0.025 on each tail): \(t_\frac{\alpha}{2} = t_{0.975} = 2.160\)
05

Calculate the Margin of Error

Let's calculate the margin of error for the 95% confidence interval using the sample standard deviation and t-value: Margin of Error = \(t_\frac{\alpha}{2} \times \frac{s}{\sqrt{n}}\) Margin of Error = \(2.160 \times \frac{0.1788}{\sqrt{14}} \approx 0.1028\)
06

Construct the Confidence Interval

Now let's construct the 95% confidence interval for the true mean red blood cell count: Confidence Interval = \(\bar{x} \pm\) Margin of Error Confidence Interval = \(5.2214 \pm 0.1028\) Confidence Interval = \((5.1186, 5.3242)\) Therefore, we can conclude with 95% confidence that the true mean red blood cell count for this person during the testing period lies between 5.1186 and 5.3242 (\(10^{6}\) cells per microliter).

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Key Concepts

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

Sample Mean Calculation
Understanding how to calculate the sample mean is fundamental in statistics. The sample mean, denoted as \(\bar{x}\), represents the average value of our sample data. It serves as a point estimate of the population mean \(\mu\) when the population is too large to measure each individual value.

To compute the sample mean, we sum up all the sample data points \(x_i\) and then divide by the number of observations \(n\), which can be expressed as the following formula:
\[\bar{x} = \frac{1}{n}\sum_{i=1}^{n} x_i\]
This formula suggests that if the data points are equally distributed, then the sample mean would be at the center of the range. In our blood cell count example, the sample mean gives us an idea of the typical red blood cell count for this particular individual over the 15-day period. It's important for students to carefully sum all the values and accurately count the number of observations, ensuring the denominator in the mean calculation reflects the correct sample size.

Remember, in practice, we often use the sample mean as an estimate for the population mean because it's usually impractical to measure the entire population. This assumption is most reliable when the sample is random and representative of the population.
Sample Standard Deviation
In addition to the mean, the sample standard deviation is essential for understanding the variability or spread of data within a sample. It is denoted by \(s\) and is calculated by taking the square root of the sample variance. The sample variance is the average of the squared differences between each data point and the sample mean.

The formula for the sample standard deviation is:
\[s = \sqrt{\frac{1}{n-1}\sum_{i=1}^{n} (x_i - \bar{x})^2 }\]
The subtracted 1 from the number of observations \(n\) in the denominator is crucial. It corrects the bias in the estimation of the population variance from a sample, providing what is called an 'unbiased estimator'. This process is referred to as 'Bessel's correction'. Students should ensure the use of the sample mean previously calculated as the reference point when subtracting each data value.

Knowing the standard deviation allows researchers to understand how much variation exists from the average, impacting how confidently one can infer about the population from the sample data. In our blood cell count example, the relatively small standard deviation suggests that the measurements were quite consistent across the 15 days.
T-Distribution
When we're estimating population parameters, like the mean, from small sample sizes, we turn to the t-distribution, a key concept in inferential statistics. The t-distribution is a probability distribution that accounts for the additional uncertainty introduced when working with a small sample size (generally less than 30). It resembles a normal distribution but has fatter tails, meaning it predicts a higher likelihood of extreme values.

The t-distribution is parameterized by degrees of freedom (df), which for a single sample is \(n-1\). Degrees of freedom relates to the number of independent values that can vary in the calculation of a statistic. In our exercise, the degrees of freedom are 13 since there are 14 observations.

To build a confidence interval using the t-distribution, we find the \(t\)-value that corresponds to our desired level of confidence (e.g., 95%) and df. It is important to use a two-tailed t-distribution for confidence intervals, where 2.5% of the distribution lies in each tail for a 95% confidence level.
\[t_\frac{\alpha}{2}\] is determined using a t-distribution table or software. The t-value is then multiplied by the standard error (the standard deviation divided by the square root of the sample size) to find the margin of error. Combined with the sample mean, this margin defines the range in which we believe the true population mean lies. Understanding the application and interpretation of the t-distribution is essential for statistical analysis, particularly when dealing with small sample sizes or when the population standard deviation is unknown.

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Most popular questions from this chapter

An experiment was conducted to study the use of \(95 \%\) ethanol versus \(20 \%\) bleach as a disinfectant in removing contamination when culturing plant tissues. The experiment was repeated 15 times with each disinfectant, using cuttings of eggplant tissue. The observation reported was the number of uncontaminated eggplant cuttings after a 4 -week storage. $$ \begin{array}{lcc} \hline \text { Disinfectant } & \text { 95\% Ethanol } & \text { 20\% Bleach } \\\ \hline \text { Mean } & 3.73 & 4.80 \\ \text { Variance } & 2.78095 & .17143 \\ n & 15 & 15 \\ & \text { Pooled variance } 1.47619 & \\ \hline \end{array} $$ a. Are you willing to assume that the underlying variances are equal? b. Using the information from part a, is there evidence of a significant difference in the mean numbers of uncontaminated eggplants for the two disinfectants tested?

In Exercise 15 (Section 10.3), you conducted a test to detect a difference in the average prices of light tuna in water versus light tuna in oil. Summarized data on the average price of two different types of tuna are shown here. $$\begin{aligned} &\text { Statistics }\\\&\begin{array}{llll}\text { Variable } & \mathrm{N} & \text { Mean } & \text { StDev } \\\\\hline \text { Light Water } & 14 & 0.896 & 0.400 \\\\\text { Light Oil } & 11 & 1.147 & 0.679\end{array} \end{aligned}$$ a. What assumption had to be made concerning the population variances so that the test would be valid? b. Do the data present sufficient evidence to indicate that the variances violate the assumption in part a? Test using \(\alpha=.05\)

Use the information provided. State the null and alternative hypotheses for testing for a significant difference in means, calculate the pooled estimate of \(\sigma^{2}\), the associated degrees of freedom, and the observed value of the t statistic. What is the rejection region using \(\alpha=.05 ?\) What is the \(p\) -value for the test? What can you conclude from these data? $$ \begin{array}{l|cccc} \text { Population } 1 & 12 & 3 & 8 & 5 \\ \hline \text { Population } 2 & 14 & 7 & 7 & 9 \end{array} $$

Find the tabled value of \(t\left(t_{a}\right)\) corresponding to a left-tail area a and degrees of freedom given. $$ a=.05, d f=8 $$

A pollution control inspector suspected that a river community was releasing amounts of semitreated sewage into a river. To check his theory, he drew five randomly selected specimens of river water at a location above the town, and another five below. The dissolved oxygen readings (in parts per million) are as follows: $$ \begin{array}{l|lllll} \text { Above Town } & 4.8 & 5.2 & 5.0 & 4.9 & 5.1 \\ \hline \text { Below Town } & 5.0 & 4.7 & 4.9 & 4.8 & 4.9 \end{array} $$ a. Do the data provide sufficient evidence to indicate that the mean oxygen content below the town is less than the mean oxygen content above? Test using \(\alpha=.05 .\) b. Estimate the difference in the mean dissolved oxygen contents for locations above and below the town. Use a \(95 \%\) confidence interval.

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