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91Ó°ÊÓ

Construct a \(90 \%\) confidence interval estimate for the mean \(\mu\) using the sample information \(n=53, \bar{x}=87.2\) and \(s=11.9\).

Short Answer

Expert verified
The 90% confidence interval for the mean is (83.512, 90.888).

Step by step solution

01

Identify the Z-score

First, determine the Z-score that corresponds to the desired confidence level. For a 90% confidence level, look up in the standard normal distribution table to find that the Z-score is approximately 1.645.
02

Compute the Standard Error

Next, compute the Standard Error of the mean, which is the standard deviation divided by the square root of the sample size, written as \(SE = \frac{s}{\sqrt{n}}\). Using the given values, \(SE = \frac{11.9}{\sqrt{53}}\). This calculates to approximately 1.634.
03

Compute the Confidence Interval

Then, compute the confidence interval by adding and subtracting the product of the Z-score and the Standard Error from the sample mean. The formula to calculate the confidence interval is \((\bar{X} - Z.{SE}, \bar{X} + Z.{SE})\) which is \((87.2 - 1.645 * 1.634, 87.2 + 1.645 * 1.634)\). The result is approximately (83.512, 90.888).

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Key Concepts

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

Standard Error
The standard error is an important concept when dealing with statistics, especially when calculating confidence intervals. It measures the precision of the sample mean as an estimate of the population mean. The formula for the standard error (SE) is the standard deviation of your sample divided by the square root of the sample size, represented as
  • \( SE = \frac{s}{\sqrt{n}} \)
Here, \(s\) is the sample standard deviation, and \(n\) is the sample size.
The smaller the standard error, the more precise your sample mean is as an estimate of the population mean. In simpler terms, a small standard error indicates that the sample mean is close to the actual population mean. To calculate it with our numbers, use \(s = 11.9\) and \(n = 53\). This gives:
  • \( SE = \frac{11.9}{\sqrt{53}} \approx 1.634 \)
Thus, our confidence interval's precision relies heavily on the standard error.
Z-score
The Z-score is crucial when you're constructing confidence intervals. It's a statistical measurement describing a value's relationship to the mean of a group of values in terms of standard deviations. To find a Z-score corresponding to the desired confidence level, like a 90% confidence interval, you can look it up in the standard normal distribution table. For a 90% confidence interval, the Z-score is usually around 1.645.
This means that the range spans 1.645 standard deviations away from the mean in both directions. It's this value that scales the standard error in the confidence interval formula. Just remember:
  • The higher the confidence level, the larger the Z-score.
  • A Z-score helps determine how extreme a value is.
By using a Z-score of 1.645, you can be 90% confident that the true population mean lies within the calculated interval.
Sample Mean
The sample mean \(\bar{x}\) is the arithmetic average of a set of sample values. It provides a point estimate of the population mean \(\mu\). The formula to calculate the sample mean is straightforward:
  • \( \bar{x} = \frac{\sum x_i}{n} \)
where \(x_i\) are the sample observations and \(n\) is the sample size.
In our exercise, the sample mean is given as \( \bar{x} = 87.2 \). This value acts as the midpoint for our confidence interval, where the interval extends above and below this mean. The sample mean is central because it's the best single estimate we have of the population mean from finite sample data.
Remember:
  • The sample mean is sensitive to extreme values or outliers.
  • Ensuring a representative sample bolsters the validity of your sample mean.
Understanding the sample mean is vital, as it heavily influences the final confidence interval estimation.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. A low standard deviation indicates that the values tend to be close to the mean of the set.
  • Conversely, a high standard deviation indicates a wider spread of data.
Standard deviation (\(s\)) is calculated as follows:
  • \( s = \sqrt{\frac{1}{n-1} \sum (x_i - \bar{x})^2} \)
Here, \(x_i\) are the individual data points, \(\bar{x}\) is the sample mean, and \(n\) is the number of observations.
In the exercise, the standard deviation is \(s = 11.9\), which reflects the variability of the data points from the mean. It's a crucial component when calculating the standard error and hence affects the width of the confidence interval.
Key points:
  • Smaller standard deviations lead to narrower confidence intervals.
  • Higher standard deviations result in broader intervals.
Standard deviation is key to understanding the reliability and precision of the calculated confidence interval.

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Most popular questions from this chapter

You hurry to the local emergency department in hopes of immediate, urgent care, only to find yourself waiting for what seems like hours. The manager of the large emergency department believes that his new procedures have substantially reduced the wait time for the average urgent care patient. He initiates a study to evaluate the wait time. The records of 18 randomly selected patients seen since the new procedures were put in place are checked, and the time between entering the emergency department and being seen by urgent care personnel was observed. The mean wait time was 17.82 minutes with a standard deviation of 5.68 minutes. Estimate the mean wait time using a \(99 \%\) confidence interval. Assume that wait times are normally distributed.

Calculate the test statistic \(z \star\) used in testing the following: a. \(H_{o}: p=0.70\) vs. \(H_{a}: p>0.70,\) with the sample \(n=300\) and \(x=224\) b. \(H_{o}: p=0.50\) vs. \(H_{a}: p<0.50,\) with the sample \(n=450\) and \(x=207\) c. \(H_{o}: p=0.35\) vs. \(H_{a}: p \neq 0.35,\) with the sample \(n=280\) and \(x=94\) d. \(H_{o}: p=0.90\) vs. \(H_{a}: p>0.90,\) with the sample \(n=550\) and \(x=508\)

It is claimed that the students at a certain university will score an average of 35 on a given test. Is the claim reasonable if a random sample of test scores from this university yields \(33,42,38,37,30,42 ?\) Complete a hypothesis test using \(\alpha=0.05 .\) Assume test results are normally distributed. a. Solve using the \(p\) -value approach. b. Solve using the classical approach.

a. Does it seem reasonable that the mean of the sampling distribution of observed values of \(p^{\prime}\) should be \(p,\) the true proportion? Explain. b. Explain why \(p^{\prime}\) is an unbiased estimator for the population \(p\).

Oranges are selected at random from a large shipment that just arrived. The sample is taken to estimate the size (circumference, in inches) of the oranges. The sample data are summarized as follows: \(n=100\) \(\Sigma x=878.2,\) and \(\Sigma(x-\bar{x})^{2}=49.91\). a. Determine the sample mean and standard deviation. b. What is the point estimate for \(\mu,\) the mean circumference of all oranges in the shipment? c. Find the \(95 \%\) confidence interval for \(\mu\)

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