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A sample of 11 observations taken from a normally distributed population produced the following data: \(\begin{array}{lllllllllll}-7.1 & 10.3 & 8.7 & -3.6 & -6.0 & -7.5 & 5.2 & 3.7 & 9.8 & -4.4 & 6.4\end{array}\) a. What is the point estimate of \(\mu\) ? b. Make a \(95 \%\) confidence interval for \(\mu\). c. What is the margin of error of estimate for \(\mu\) in part b?

Short Answer

Expert verified
a. The point estimate of 碌 is the value calculated in Step 1. b. The 95% confidence interval for 碌 is the interval calculated in Step 2. c. The margin of error of the estimate for 碌 is the value calculated in Step 3.

Step by step solution

01

Title: Calculate Sample Mean (x虅)

The sample mean is calculated by summing up all the observations and dividing by the number of observations. It is given by the formula \(x虅 = \frac{1}{n}危x_i\), where \(x_i\) are the individual observations and \(n\) is the total number of observations. Applying this formula to the given data, calculate the sum of all observations, and then divide by the total number of observations, which is 11 to get \(x虅\).
02

Title: Construct a 95% confidence interval for 碌

A 95% confidence interval for 碌 can be constructed using the formula: \(\[x虅 \pm ((Z_{\(\alpha/2\)} * 饾憼) / \sqrt{n})\]\), where \(x虅\) is the sample mean, \(Z_{\(\alpha/2\)}\) is the z-value from the z-table corresponding to the desired level of confidence (z = 1.96 for a 95% confidence interval), 饾憼 is the sample standard deviation and \(n\) is the sample size. Compute the sample standard deviation \(饾憼\), then substitute into the aforementioned formula to get the confidence interval.
03

Title: Calculate The Margin of Error

The Margin of Error is the amount by which the estimate could differ from the actual population parameter. In context of this problem, it's the difference between the upper limit and the point estimate, or the difference between the point estimate and the lower limit of the confidence interval. Mathematically, Margin of Error = Upper Limit - \(x虅\) or Margin of Error = \(x虅\) - Lower Limit. Use the point estimate and the limits of the 95% confidence interval calculated in the prior step to find the Margin of Error.

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Key Concepts

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

Confidence Interval
When dealing with data, a confidence interval gives you a range within which you can be fairly certain that a population parameter lies. Imagine it as a net you cast in the vast ocean of data, hoping to catch the true population mean, \(\mu\).
For example, a 95% confidence interval means if you were to take 100 different samples and compute a confidence interval for each, about 95 of those intervals would contain the population mean. That makes it pretty reliable!
To construct a confidence interval, you need:
  • The sample mean, \(\bar{x}\), representing the center of your interval.
  • A critical value, usually from a z-table, that determines how wide your interval will be.
  • The sample standard deviation to account for variability.
  • Finally, your sample size, \(n\), also plays a role.
Combining all these, the formula is \(\bar{x} \pm (Z_{\alpha/2} * \frac{s}{\sqrt{n}})\), where \(Z_{\alpha/2}\) is the z-value (1.96 for 95% confidence).
This formula builds a range centered on the sample mean, extending outwards by a margin determined by the product of the critical value and the standard deviation divided by the square root of the sample size.
Margin of Error
The margin of error is a crucial component when interpreting confidence intervals. It defines how far your estimate might be from the true population parameter. Basically, it tells you the wiggle room you have.
In simple terms, think of the margin of error (ME) as the half-width of your confidence interval.
  • It's calculated by taking the difference from the sample mean to either the upper bound or the lower bound of the interval.
  • Mathematically, \(\text{Margin of Error} = Z_{\alpha/2} * \frac{s}{\sqrt{n}}\).
  • This formula indicates how much you add or subtract from the sample mean to create a confidence interval.
The margin of error grows smaller with larger samples, assuming the population variance stays the same, and indicates higher precision in your estimate.
Knowing the margin of error helps you understand the potential "spread" or "uncertainty" around your sample mean, allowing you to gauge the confidence in your statistical conclusions.
Sample Mean
The sample mean, often denoted as \(\bar{x}\), provides the foundation for many statistical calculations. It's a simple average of a set of data points and represents a point estimate for the population mean \(\mu\).
To calculate the sample mean:
  • Add together all your sample values.
  • Count how many values you have, which is your sample size \(n\).
  • Divide the total sum by your sample size: \(\bar{x} = \frac{1}{n}\sum x_i\).
This calculation gives you a single data point that summarizes the central tendency of your data.
Using the sample mean is a practical way to infer or estimate the true mean of the population without having to examine every individual.Remember, while the sample mean is useful, it's not infallible. It's a snapshot based on the sample you've taken, and more samples might yield slightly different means. This is why the concept of confidence intervals and margins of error adds critical insight by communicating the reliability and range of where the true population mean might lie.
Normally Distributed Population
A normally distributed population is characterized by its bell-shaped curve, known as the Gaussian distribution.
This distribution is symmetrical, centering around the mean, and it's defined by two parameters:
  • The mean, \(\mu\).
  • The standard deviation, \(\sigma\).
In a normal distribution:
  • The mean, median, and mode are all equal.
  • Values are evenly distributed around the mean, with most values clustering around the center.
  • The tails on either side show the rare occurrences.
A population that is normally distributed allows statisticians to apply various inferential techniques, such as calculating confidence intervals as shown in our exercise.
The reason why normal distribution is so handy is that it provides a predictable pattern of data spread, which can be very useful in making inferences about a population from a sample. It's like having a reliable compass in the world of statistics, pointing towards the most likely outcomes.

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Most popular questions from this chapter

For each of the following, find the area in the appropriate tail of the \(t\) distribution. a. \(t=2.467\) and \(d f=28\) b. \(t=-1.672\) and \(d f=58\) c. \(t=-2.670\) and \(n=55\) d. \(t=2.819\) and \(n=23\)

A travel agent wants to gather information on the per-night cost at hotels in Caribbean countries. She took a random sample of 52 rooms from various hotels in those countries. The sample produced a mean cost for the 52 rooms to be \(\$ 208.35\) per night. If the population standard deviation of costs for a one- night stay in Caribbean hotels is \(\$ 47.45\), find a \(99 \%\) confidence interval for the average cost per night in Caribbean hotels.

a. How large a sample should be selected so that the margin of error of estimate for a \(98 \%\) confidence interval for \(p\) is \(.045\) when the value of the sample proportion obtained from a preliminary sample is \(.53\) ? b. Find the most conservative sample size that will produce the margin of error for a \(98 \%\) confidence interval for \(p\) equal to \(.045\)

For a population, the value of the standard deviation is 4.96. A sample of 32 observations taken from this population produced the following data. \(\begin{array}{llllllll}74 & 85 & 72 & 73 & 86 & 81 & 77 & 60 \\ 83 & 78 & 79 & 88 & 76 & 73 & 84 & 78 \\ 81 & 72 & 82 & 81 & 79 & 83 & 88 & 86 \\ 78 & 83 & 87 & 82 & 80 & 84 & 76 & 74\end{array}\) a. What is the point estimate of \(\mu\) ? b. Make a \(99 \%\) confidence interval for \(\mu .\) c. What is the margin of error of estimate for part b?

a. Find the value of \(t\) from the \(t\) distribution table for a sample size of 22 and a confidence level of \(95 \%\). b. Find the value of \(t\) from the \(t\) distribution table for 60 degrees of freedom and a \(90 \%\) confidence level. c. Find the value of \(t\) from the \(t\) distribution table for a sample size of 24 and a confidence level of \(99 \%\).

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