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91Ó°ÊÓ

The attendance at the Savannah Colts minor league baseball game last night was \(400 .\) A random sample of 50 of those in attendance revealed that the mean number of soft drinks consumed per person was 1.86 with a standard deviation of \(0.50 .\) Develop a 99 percent confidence interval for the mean number of soft drinks consumed per person.

Short Answer

Expert verified
The 99% confidence interval is between 1.6706 and 2.0494 soft drinks.

Step by step solution

01

Identify the Problem

We need to find a 99% confidence interval for the mean number of soft drinks consumed per person based on a sample of size 50 with a sample mean of 1.86 and a standard deviation of 0.50.
02

Select the Appropriate Formula

Since the population standard deviation is unknown and our sample size is less than 30, we will use the t-distribution for calculating the confidence interval. The formula to use is: \[ \bar{x} \pm t_{\alpha/2} \left( \frac{s}{\sqrt{n}} \right) \]where \(\bar{x}\) is the sample mean, \(s\) is the sample standard deviation, \(n\) is the sample size, and \(t_{\alpha/2}\) is the t-value for a 99% confidence interval with \(n-1\) degrees of freedom.
03

Calculate the Standard Error

Calculate the standard error (SE) using the formula: \[ SE = \frac{s}{\sqrt{n}} = \frac{0.50}{\sqrt{50}} = 0.0707 \]
04

Find the t-Value

For a 99% confidence interval and \(49\) degrees of freedom, use a t-table to find that \(t_{\alpha/2} \approx 2.678\) (value from t-distribution table).
05

Calculate the Margin of Error

Calculate the margin of error (ME) using:\[ ME = t_{\alpha/2} \times SE = 2.678 \times 0.0707 = 0.1894 \]
06

Determine the Confidence Interval

Using the margin of error, determine the confidence interval:\[ \text{Confidence Interval} = \bar{x} \pm ME = 1.86 \pm 0.1894 \]This results in an interval of \([1.6706, 2.0494]\).

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Key Concepts

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

t-distribution
In statistics, the t-distribution is essentially a family of curves. It is used when the sample size is small, which is often the case with real-world data. Its shape is similar to the normal distribution but has thicker tails. This means that it can better accommodate the variability expected in smaller samples.

The t-distribution is particularly useful when the population standard deviation is not known, and you only have a small sample size. In such cases, the t-distribution will provide a more reliable estimate of the confidence interval than the normal distribution. The critical value from the t-distribution, denoted as \( t_{\alpha/2} \), depends on the desired level of confidence and the degrees of freedom (sample size minus one).
  • The degrees of freedom adjust the shape of the t-distribution curve; fewer degrees of freedom make the tails thicker.
  • As the sample size grows larger, the t-distribution gradually approaches the shape of the normal distribution.
Understanding the t-distribution is crucial for correctly estimating confidence intervals for small samples.
standard error
The standard error (SE) provides a measure of how much the sample mean might vary from the true population mean. It is a crucial concept in statistical inference because it quantifies the uncertainty in the estimate of the sample mean. Specifically, the standard error is the standard deviation of the sampling distribution of the sample mean.

The formula to calculate the standard error is given by:\[ SE = \frac{s}{\sqrt{n}} \]where \( s \) is the sample standard deviation, and \( n \) is the sample size. Importantly, the standard error decreases as the sample size increases, which means with larger samples, the estimate of the sample mean is more precise.
  • The smaller the standard error, the closer the sample mean is likely to be to the population mean.
  • A high standard error indicates more variability within the sample and thus less confidence that the sample mean is an accurate reflection of the population mean.
Grasping the concept of standard error is essential for interpreting confidence intervals.
sample mean
The sample mean is a basic concept in statistics. It represents the average value of a sample, which is a subset of the total population under study. Calculating the sample mean is typically one of the first steps in statistical analysis because it gives a quick snapshot of the data characteristics.

For instance, in the baseball game exercise, the sample mean of soft drinks consumed is calculated by adding up all the soft drinks consumed by the 50 sampled attendees and then dividing by 50. Mathematically, it can be expressed as:\[ \bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i \]where \( n \) is the sample size, and \( x_i \) are the sample observations.
  • The sample mean provides a single value that summarizes the dataset as a whole.
  • It is the central point of the dataset and a critical estimate of the population mean when a full dataset is not available.
Understanding the sample mean is a fundamental stepping stone to grasp more advanced statistical concepts, such as hypothesis testing and confidence intervals.
margin of error
The margin of error (ME) reflects the range within which we expect the true population mean to lie with a certain level of confidence. It is calculated by multiplying the critical value from the t-distribution by the standard error. In formula terms, this is represented as:\[ ME = t_{\alpha/2} \times SE \]where \( t_{\alpha/2} \) is the t-distribution critical value, and \( SE \) is the standard error.

The margin of error provides a buffer around the sample mean, allowing researchers to express the confidence interval. For the baseball game example, the margin of error tells us how much the sample mean of 1.86 drinks per person might differ from the true population mean.
  • A larger margin of error suggests more variability, meaning less certainty about where the true mean lies.
  • A smaller margin of error indicates that the sample mean is a more precise estimate of the population mean.
Understanding the margin of error is critical for making informed judgments based on the data, as it conveys both the precision and reliability of the sample estimates.

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Most popular questions from this chapter

Families USA, a monthly magazine that discusses issues related to health and health costs, surveyed 20 of its subscribers. It found that the mean annual health insurance premium for a family with coverage through an employer was \(\$ 10,979 .\) The standard deviation of the sample was \(\$ 1,000 .\) a. Based on this sample information, develop a 90 percent confidence interval for the population mean yearly premium. b. How large a sample is needed to find the population mean within \(\$ 250\) at 99 percent confidence?

14\. The Greater Pittsburgh Area Chamber of Commerce wants to estimate the mean time workers who are employed in the downtown area spend getting to work. A sample of 15 workers reveals the following number of minutes spent traveling. \begin{tabular}{|llllllll|} \hline 29 & 38 & 38 & 33 & 38 & 21 & 45 & 34 \\ 40 & 37 & 37 & 42 & 30 & 29 & 35 & \\ \hline \end{tabular} Develop a 98 percent confidence interval for the population mean. Interpret the result.

It is estimated that 60 percent of U.S. households subscribe to cable TV. You would like to verify this statement for your class in mass communications. If you want your estimate to be within 5 percentage points. with a 95 percent level of confidence, how large of a sample is required?

Marty Rowatti recently assumed the position of director of the YMCA of South Jersey. He would like some current data on how long current members of the YMCA have been members. To investigate, suppose he selects a random sample of 40 current members. The mean length of membership of those included in the sample is 8.32 years and the standard deviation is 3.07 years. a. What is the mean of the population? b. Develop a 90 percent confidence interval for the population mean. c. The previous director, in the summary report she prepared as she retired, indicated the mean length of membership was now "almost 10 years." Does the sample information substantiate this claim? Cite evidence.

The National Collegiate Athletic Association (NCAA) reported that the mean number of hours spent per week on coaching and recruiting by college football assistant coaches during the season was \(70 .\) A random sample of 50 assistant coaches showed the sample mean to be 68.6 hours, with a standard deviation of 8.2 hours. a. Using the sample data, construct a 99 percent confidence interval for the population mean. b. Does the 99 percent confidence interval include the value suggested by the NCAA? Interpret this result. c. Suppose you decided to switch from a 99 to a 95 percent confidence interval. Without performing any calculations, will the interval increase, decrease, or stay the same? Which of the values in the formula will change?

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