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Assume that the population of \(x\) values has an approximately normal distribution. Wildlife: Mountain Lions How much do wild mountain lions weigh? The 77 th Annual Report of the New Mexico Department of Game and Fish, edited by Bill Montoya, gave the following information. Adult wild mountain lions 18 months or older) captured and released for the first time in the San Andres Mountains gave the following weights (pounds): \(\begin{array}{llllll}68 & 104 & 128 & 122 & 60 & 64\end{array}\) (a) Use a calculator with mean and sample standard deviation keys to verify that \(\bar{x}=91.0\) pounds and \(s \approx 30.7\) pounds. (b) Find a \(75 \%\) confidence interval for the population average weight \(\mu\) of all adult mountain lions in the specified region. (c) Interpretation What does the confidence interval mean in the context of this problem?

Step by step solution

01

Calculate the Mean

First, we need to verify the given mean \(\bar{x}\). The mean is calculated by adding all observed values and dividing by the number of values: \[ \bar{x} = \frac{68 + 104 + 128 + 122 + 60 + 64}{6} = \frac{546}{6} = 91.0 \text{ pounds} \] Thus, \(\bar{x} = 91.0\) pounds. This matches the problem statement.

02

Calculate the Sample Standard Deviation

Next, verify the sample standard deviation \(s\). The formula for the sample standard deviation is: \[ s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2} \] Calculate each squared deviation, sum them up, and divide by \(n-1\), then take the square root:- \( (68-91)^2 = 529 \)- \( (104-91)^2 = 169 \)- \( (128-91)^2 = 1369 \)- \( (122-91)^2 = 961 \)- \( (60-91)^2 = 961 \)- \( (64-91)^2 = 729 \)Sum: \( 529 + 169 + 1369 + 961 + 961 + 729 = 4718 \)\[ s = \sqrt{\frac{4718}{5}} \approx 30.7 \text{ pounds} \]This confirms \(s \approx 30.7\) pounds.

03

Determine the Confidence Interval Components

For a \(75\%\) confidence interval, we need the critical value \(t\) from the t-distribution. Since \(n=6\), \(df = n-1 = 5\). Find \(t\) for \(75\%\) CI:From a t-table, \(t_{0.25,5} \approx 1.475\).

04

Calculate the Margin of Error

The margin of error (ME) is determined by the formula:\[ ME = t \cdot \frac{s}{\sqrt{n}} \] Substitute the known values:\[ ME = 1.475 \cdot \frac{30.7}{\sqrt{6}} \approx 18.47 \text{ pounds} \]

05

Calculate the Confidence Interval

Now use the margin of error to calculate the confidence interval:\[ CI = \bar{x} \pm ME \] \[ CI = 91.0 \pm 18.47 \] \[ CI = (72.53, 109.47) \] The \(75\%\) confidence interval for the mountain lion weight is \((72.53, 109.47)\) pounds.

06

Interpretation of the Confidence Interval

The confidence interval \((72.53, 109.47)\) pounds means that we are \(75\%\) confident the true average weight \(\mu\) of all adult mountain lions in that region is between \(72.53\) and \(109.47\) pounds.

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Key Concepts

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

Confidence Interval

A confidence interval is a range of values, derived from a sample, used to estimate an unknown population parameter. In simpler terms, it helps us understand how close our sample data might be to the true value of the population parameter we are interested in.
In our exercise, the goal is to determine the average weight of adult mountain lions in a specified region. We calculated a confidence interval for the average weight of these mountain lions based on the sample data we have.

• Confidence Level: This tells us the degree of certainty we have that the true parameter lies within our confidence interval. Here, it is 75% confidence level.
• Interpretation: The range (72.53 pounds, 109.47 pounds) from our calculation means we are 75% certain that the true average weight of all mountain lions in that region lies within this range.

The confidence interval is essentially about balancing between precision and the level of confidence. A higher confidence level typically means a wider interval.

Sample Standard Deviation

The sample standard deviation is a key measure in statistics that tells us how much the values in our data set vary from the mean of that sample. It's particularly useful for understanding the spread or dispersion in a set of data points.
To compute the sample standard deviation (denoted as \(s\)), we use the following formula:

• The deviations of each data point from the sample mean are squared to remove any negative signs.
• We sum up all those squared deviations.
• This sum is divided by \((n-1)\), where \(n\) is the number of observations. This adjustment is because we are working with a sample, not the whole population.
• The square root of this quotient provides the sample standard deviation.

In our scenario, after computing, we find that the sample standard deviation is approximately 30.7 pounds. This value offers insight into how much individual mountain lion weights deviate from the sample mean of 91.0 pounds.

Mean Calculation

Calculating the mean, or average, of a sample is one of the foundational tasks in statistics. The mean gives us the central value of a data set and serves as a reference point to understand other statistical measures.
In simple terms, the mean is calculated by:

• Adding up all individual data values in a set.
• Dividing this total by the number of values (often denoted as \( n \)).

For the mountain lions' data set, the weights provided were 68, 104, 128, 122, 60, and 64 pounds. We add these values to get a total of 546 pounds. Since there are six mountain lions, we divide 546 by 6 to find that the mean weight is 91.0 pounds.
Understanding the mean calculation helps in identifying the typical weight in our data set and also serves as a crucial part of calculating standard deviation and confidence intervals.