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91Ó°ÊÓ

Construct a \(95 \%\) confidence interval estimate for the mean \(\mu\) using the sample information \(n=24, \bar{x}=16.7\) and \(s=2.6\).

Short Answer

Expert verified
The 95% confidence interval estimate for the population mean is approximately between 15.6 and 17.8.

Step by step solution

01

Understand the given parameters

We are given the sample size \(n = 24\), the sample mean \(\bar{x} = 16.7\), and the sample standard deviation \(s = 2.6\). We are asked to construct a 95% confidence interval for the mean, hence Confidence Level( \(CL = 0.95\).
02

Find the Degree of Freedom and the T-Score

Since it's a t-score scenario (the population standard deviation is unknown and the sample size is smaller than 30), we first find the degrees of freedom (df) using the formula \(df = n - 1\). From the given information we know \(n = 24\), so \(df = 24 - 1 = 23\). Next, we determine the t-score corresponding to a 0.95 confidence level from the student's t-distribution table, with 23 degrees of freedom. In this case, the t-score is approximately 2.069.
03

Calculate the Standard Error

Standard Error(SE) is calculated using the formula: \(SE = s/ \sqrt{n}\) Where: - \(s\) is the Sample Standard Deviation, which is 2.6 here- \(n\) is the number of samples, which is 24. When we substitute these values into the formula, we get an SE of approximately 0.53.
04

Calculate the Confidence Interval

Finally, to calculate the confidence interval, we use the formula: \(\bar{x} \pm T_{score} * SE\) Given that \(\bar{x}\) is 16.7, \(T_{score}\) is 2.069, and \(SE\) is approximately 0.53, substitution into the formula gets our confidence interval: \(CI = [16.7 - 2.069 * 0.53, 16.7 + 2.069 * 0.53]\), which is approximately [15.6, 17.8].

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Key Concepts

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

t-score
The t-score is a value derived from the Student's t-distribution, which is used when estimating population parameters when the sample size is small and the population standard deviation is unknown. It helps in determining how far the sample mean is from the population mean in terms of standard errors. When the sample size is less than 30, and especially when the population standard deviation is not available, we rely on the t-score rather than the z-score.
  • We calculate the t-score by looking up values in a t-distribution table, which is organized by levels of confidence and degrees of freedom.
  • The table gives us the critical value of t for our specific confidence level and degrees of freedom.
In our example, with a sample size of 24, the degrees of freedom is 23, and for a 95% confidence level, the t-score is approximately 2.069.
sample mean
The sample mean (\(ar{x}\)) represents the average value of a sample, which is a subset of the total population. It provides an estimate of the population mean (denoted as \(\mu\)). Measuring the sample mean is crucial because it is used in various statistical calculations, such as calculating the confidence interval, standard error, and more.
To compute the sample mean, you add up all the sample data points and divide by the number of data points: \[\bar{x} = \frac{(sum\ of\ sample\ values)}{(sample\ size)}\] In the exercise, the sample mean is given as 16.7. This value is central to constructing our confidence interval as it is the midpoint around which we calculate the range of possible values for the population mean.
standard deviation
Standard deviation (\(s\)) is a measure that describes the spread or variability of a dataset. In other words, it tells us how much the individual data points in a sample deviate from the sample mean. A low standard deviation means that most of the numbers are close to the average, while a high standard deviation indicates a wider spread of values.
  • To calculate the sample standard deviation, find the differences between each data point and the mean, square these differences, find the average of these squared differences, and take the square root of that average.
In the given problem, the sample standard deviation is provided as 2.6. This measure is crucial for calculating the standard error and, subsequently, the confidence interval, as it gives an indication of how much the sample mean might vary from the true population mean.
degrees of freedom
Degrees of freedom (\(df\)) are linked to the number of independent values or quantities that can vary in an analysis without breaking any constraints. In the context of confidence intervals and t-scores, degrees of freedom are primarily used to determine the appropriate critical t-score from the t-distribution table.
  • For a sample, the degrees of freedom are calculated as the sample size minus one: \(n - 1\).
  • The reason we subtract one is to account for the use of the sample mean in our calculations, which imposes a constraint on variability.
In our exercise, with 24 samples, the degrees of freedom is 23. This is an essential value, as it determines the specific t-score used to calculate the confidence interval, ensuring accurate statistical inference from the sample to the whole population.

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Most popular questions from this chapter

Even with a heightened awareness of beef quality, \(82 \%\) of Americans indicated their recent burger-eating behavior has remained the same, according to a recent T.G.I. Friday's restaurants random survey of 1027 Americans. In fact, half of Americans eat at least one beef burger each week. That's a minimum of 52 burgers each year. a. What is the point estimate for the proportion of all Americans who eat at least one beef burger per week? b. Find the \(98 \%\) confidence interval for the true proportion \(p\) in the binomial situation where \(n=1027\) and the observed proportion is one-half. c. Use the results of part b to estimate the percentage of all Americans who eat at least one beef burger per week.

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.

A random sample of 51 observations was selected from a normally distributed population. The sample mean was \(\bar{x}=98.2,\) and the sample variance was \(s^{2}=37.5 .\) Does this sample show sufficient reason to conclude that the population standard deviation is not equal to 8 at the 0.05 level of significance? a. Solve using the \(p\) -value approach. b. Solve using the classical approach.

In obtaining the sample size to estimate a proportion, the formula \(n=[z(\alpha / 2)]^{2} p q / E^{2}\) is used. If a reasonable estimate of \(p\) is not available, it is suggested that \(p=0.5\) be used because this will give the maximum value for \(n\). Calculate the value of \(p q=p(1-p)\) for \(p=0.1,0.2,0.3, \ldots, 0.8,0.9\) in order to obtain some idea about the behavior of the quantity \(p q\).

How important is the assumption "The sampled population is normally distributed" to the use of Student's \(t\) -distribution? Using a computer, simulate drawing 100 samples of size 10 from each of three different types of population distributions, namely, a normal, a uniform, and an exponential. First generate 1000 data values from the population and construct a histogram to see what the population looks like. Then generate 100 samples of size 10 from the same population; each row represents a sample. Calculate the mean and standard deviation for each of the 100 samples. Calculate \(t \neq\) for each of the 100 samples. Construct histograms of the 100 sample means and the 100 t \(\star\) values. (Additional details can be found in the Student Solutions Manual.) For the samples from the normal population: a. Does the \(\bar{x}\) distribution appear to be normal? Find percentages for intervals and compare them with the normal distribution. b. Does the distribution of \(t \star\) appear to have a \(t\) -distribution with df \(=9 ?\) Find percentages for intervals and compare them with the \(t\) -distribution. For the samples from the rectangular or uniform population: c. Does the \(\bar{x}\) distribution appear to be normal? Find percentages for intervals and compare them with the normal distribution. d. Does the distribution of \(t \star\) appear to have a \(t\) -distribution with df \(=9 ?\) Find percentages for intervals and compare them with the \(t\) -distribution. For the samples from the skewed (exponential) population: e. Does the \(\bar{x}\) distribution appear to be normal? Find percentages for intervals and compare them with the normal distribution. f. Does the distribution of \(t \star\) appear to have a \(t\) -distribution with df \(=9 ?\) Find percentages for intervals and compare them with the \(t\) -distribution. In summary: g. In each of the preceding three situations, the sampling distribution for \(\bar{x}\) appears to be slightly different from the distribution of \(t \star .\) Explain why. h. Does the normality condition appear to be necessary in order for the calculated test statistic \(t \star\) to have a Student's \(t\) -distribution? Explain.

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