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A random sample of 20 acres gave a mean yield of wheat equal to \(41.2\) bushels per acre with a standard deviation of 3 bushels. Assuming that the yield of wheat per acre is normally distributed, construct a \(90 \%\) confidence interval for the population mean \(\mu\).

Short Answer

Expert verified
The 90% confidence interval for the population mean µ of wheat yield per acre is calculated with the method shown above using the given sample mean and standard deviation.

Step by step solution

01

Calculate Sample Standard Error

Calculate the standard error of the sample using the formula \(SE = \frac{s}{\sqrt{n}}\) where \(s\) is the standard deviation (3 bushels) and \(n\) is the sample size (20).
02

Find t-value

Find the t-value for the 90% confidence level. Since the sample size is 20, the degrees of freedom will be \(n - 1 = 19\). Using a t-distribution table, or statistical software, find the t-value that corresponds to the 90% confidence level and 19 degrees of freedom.
03

Construct Confidence Interval

Construct the 90% confidence interval using the formula \((\bar{x} - t*SE, \bar{x} + t*SE)\), where \(\bar{x}\) is the sample mean (41.2 bushels), \(t\) is the t-value from Step 2 and \(SE\) is the standard error from Step 1.

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Key Concepts

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

Sample Standard Error
When working with samples, it's important to estimate how much the sample mean might differ from the actual population mean. This is where the sample standard error comes into play. It measures the variability of the sample mean and helps us understand how close our sample mean is likely to be to the population mean.

The formula to calculate the sample standard error (SE) is given by:
  • \( SE = \frac{s}{\sqrt{n}} \)
where:
  • \(s\) is the sample standard deviation.
  • \(n\) is the sample size.
For example, in our exercise, the standard deviation \(s\) is 3 bushels, and the sample size \(n\) is 20. Using the formula, you can calculate the standard error to see how much the average yield per acre might vary in different samples of 20 acres.
t-Distribution
The t-distribution is a vital concept when dealing with small sample sizes. Unlike the normal distribution, which assumes precise knowledge of the population standard deviation, the t-distribution accounts for the added uncertainty when only sample data is available. This makes it a perfect tool for creating confidence intervals with smaller samples.

It has a shape similar to the normal distribution, especially the bell curve, but with thicker tails. These thicker tails reflect the greater variability and uncertainty associated with small sample sizes. As the sample size increases, the t-distribution approaches a normal distribution. In our exercise, the t-distribution helps in determining the t-value that reflects our desired level of confidence.
Degrees of Freedom
Degrees of Freedom (df) are a key concept when working with statistical calculations and especially when looking at the t-distribution. They generally represent the number of values that are free to vary in a calculation.

In the context of a sample, the degrees of freedom are usually the sample size minus one: \( df = n - 1 \). This adjustment is necessary because we're using the sample to estimate population parameters, and one number is used to calculate the sample's average.

In the wheat yield problem, the sample size is 20, which gives us \( df = 20 - 1 = 19 \). It's crucial to use the correct degree of freedom as it influences the accuracy of the t-value used in our confidence interval.
Population Mean
The population mean, often denoted as \( \mu \), is a central concept in statistics. It represents the average of all possible observations in the entire population. In contrast, a sample mean \( \bar{x} \) is the average of observations in a sample.

The true population mean is usually unknown, especially when dealing with small sample sizes, hence the importance of estimating it through confidence intervals.

In the problem, we aim to estimate the population mean yield of wheat per acre using our sample statistics. By constructing a confidence interval, we determine a range within which we believe the true population mean \( \mu \) lies, given a certain level of confidence (90% in this case). This range gives us a statistical assurance about where the true average wheat yield is likely located, based on the sample mean of 41.2 bushels.

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Most popular questions from this chapter

At the end of Section \(8.2\), we noted that we always round up when calculating the minimum sample size for a confidence interval for \(\mu\) with a specified margin of error and confidence level. Using the formula for the margin of error, explain why we must always round up in this situation.

In a Time/Money Magazine poll of Americans of age 18 years and older, \(65 \%\) agreed with the statement, "We are less sure our children will achieve the American Dream" (Time, October 10,2011 ). Assume that this poll was based on a random sample of 1600 Americans. a. Construct a \(95 \%\) confidence interval for the proportion of all Americans of age 18 years and older who will agree with the aforementioned statement. b. Explain why we need to construct a confidence interval. Why can we not simply say that \(65 \%\) of all Americans of age 18 years and older agree with the aforementioned statement?

The following data give the number of pitches thrown by both teams in each of a random sample of 24 Major League Baseball games played between the beginning of the 2012 season and May 16, \(2012 .\) \(\begin{array}{llllllll}234 & 281 & 264 & 251 & 284 & 266 & 337 & 291 \\ 309 & 245 & 331 & 284 & 239 & 282 & 226 & 286 \\ 361 & 278 & 317 & 306 & 325 & 256 & 295 & 276\end{array}\) a. Create a histogram of these data using the class intervals 210 to less than 230,230 to less than 250,250 to less than 270 , and so on. Based on the histogram, does it seem reasonable to assume that these data are approximately normally distributed? b. Calculate the value of the point estimate of the corresponding population mean. c. Assuming that the distribution of total number of pitches thrown by both teams in Major League Baseball games is approximately normal, construct a \(99 \%\) confidence interval for the average number of pitches thrown by both teams in a Major League Baseball game.

A city planner wants to estimate the average monthly residential water usage in the city. He selected a random sample of 40 households from the city, which gave the mean water usage to be \(3415.70\) gallons over a 1 -month period. Based on earlier data, the population standard deviation of the monthly residential water usage in this city is \(389.60\) gallons. Make a \(95 \%\) confidence interval for the average monthly residential water usage for all households in this city.

When one is attempting to determine the required sample size for estimating a population mean, and the information on the population standard deviation is not available, it may be feasible to take a small preliminary sample and use the sample standard deviation to estimate the required sample size, \(n\). Suppose that we want to estimate \(\mu\), the mean commuting distance for students at a community college, to a margin of error within 1 mile with a confidence level of \(95 \%\). A random sample of 20 students yields a standard deviation of \(4.1\) miles. Use this value of the sample standard deviation, \(s\), to estimate the required sample size, \(n\). Assume that the corresponding population has a normal distribution.

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