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Chapter 10: Problem 45

At a time when energy conservation is so important, some scientists think closer scrutiny should be given to the cost (in energy) of producing various forms of food. Suppose you wish to compare the mean amount of oil required to produce 1 acre of corn versus 1 acre of cauliflower. The readings (in barrels of oil per acre), based on 20-acre plots, seven for each crop, are shown in the table. Use these data to find a \(90 \%\) confidence interval for the difference between the mean amounts of oil required to produce these two crops. Corn Cauliflower \(\begin{array}{ll}5.6 & 15.9\end{array}\) 7.1 13.4 \(\begin{array}{ll}4.5 & 17.6\end{array}\) 6.0 16.8 7.9 15.8 \(\begin{array}{ll}4.8 & 16.3\end{array}\) 57 \(17 .\)

Short Answer

Expert verified
Answer: The 90% confidence interval for the difference between the mean amounts of oil required to produce 1 acre of corn and 1 acre of cauliflower is (-11.47, -8.53) barrels/acre.

Step by step solution

01

Calculate the sample means

First, calculate the mean amount of oil required to produce 1 acre of corn and 1 acre of cauliflower using the given data. Corn: \({(5.6 + 7.1 + 4.5 + 6.0 + 7.9 + 4.8 + 6.7)}/7 = 6 \; \mathrm{barrels/acre}\) Cauliflower: \({(15.9 + 13.4 + 17.6 + 16.8 + 15.8 + 16.3 + 17)}/7 = 16 \; \mathrm{barrels/acre}\)
02

Calculate the sample standard deviations

Next, we will calculate the standard deviations for both samples using the formula \(\displaystyle \sqrt{\frac{\sum{(x_{i}-\bar{x})^2}}{(n-1)}}\). For corn, the standard deviation is \(s_{1} = \displaystyle \sqrt{\frac{(5.6 - 6)^2 + (7.1 - 6)^2 + \cdots + (6.7 - 6)^2}{6}} = 1.36 \; \mathrm{barrels/acre}\) For cauliflower, the standard deviation is \(s_{2} = \displaystyle \sqrt{\frac{(15.9 - 16)^2 + (13.4 - 16)^2 + \cdots + (17 - 16)^2}{6}} = 1.69 \; \mathrm{barrels/acre}\)
03

Calculate the t-score and margin of error

To find the 90% confidence interval for the difference between the means, we must determine the critical t-value and margin of error. The degrees of freedom are \(n_{1} + n_{2} - 2 = 7 + 7 - 2 = 12\). Using a t-table or calculator, the critical t-value for a 90% confidence interval with 12 degrees of freedom is approximately \(t_{crit} = 1.782\). Then we need to find the margin of error, which is given by the formula: \(ME = t_{crit} \sqrt{\frac{s_{1}^2}{n_{1}} + \frac{s_{2}^2}{n_{2}}} = 1.782\sqrt{\frac{1.36^2}{7}+\frac{1.69^2}{7}} = 1.47\)
04

Calculate the confidence interval

Now we can calculate the confidence interval by finding the difference between the sample means and adding/subtracting the margin of error. \((\bar{x}_{1} - \bar{x}_{2}) \pm ME = (6 - 16) \pm 1.47 = -10 \pm 1.47\) So, the 90% confidence interval for the difference between the mean amounts of oil required to produce 1 acre of corn and 1 acre of cauliflower is \((-11.47, -8.53)\) barrels/acre.
05

Interpret the result

The 90% confidence interval tells us that we can be 90% certain that the true difference in the mean amounts of oil required to produce 1 acre of corn and 1 acre of cauliflower lies between -11.47 and -8.53 barrels/acre. This means that it takes, on average, between 8.53 and 11.47 barrels more oil to produce 1 acre of cauliflower than 1 acre of corn.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a fundamental concept in statistics. It helps us determine if there's enough evidence to support a specific claim about a population parameter, based on sample data. In this exercise, we are looking at the difference in oil consumption between corn and cauliflower.

We begin by forming two hypotheses:
  • Null Hypothesis (\( H_0 \)): There is no difference in oil usage between the crops.
  • Alternative Hypothesis (\( H_1 \)): There is a significant difference in oil usage.
The goal is to use data to decide whether to reject the null hypothesis in favor of the alternative. This is where our confidence interval helps. If the interval does not include zero, it suggests a significant difference exists.
Sample Mean
The sample mean is a measure of central tendency that gives us an average value from a sample. In this context, the sample mean helps us understand the average oil required per acre for each crop.

To find the sample mean, we sum up all the values for each crop and divide by the number of observations:
  • For corn, the mean is calculated as: \( \frac{5.6 + 7.1 + 4.5 + 6.0 + 7.9 + 4.8 + 6.7}{7} = 6 \; \text{barrels/acre} \).
  • For cauliflower, the mean is: \( \frac{15.9 + 13.4 + 17.6 + 16.8 + 15.8 + 16.3 + 17}{7} = 16 \; \text{barrels/acre} \).
These means are used to compare the two samples and play a crucial role in calculating the confidence interval.
Standard Deviation
Standard deviation measures how much the data varies from the mean. It gives us an idea of the spread or dispersion of our data set. In this problem, understanding the variability in oil usage is key to confidence interval calculations.

The formula for standard deviation is:\[ s = \sqrt{\frac{\sum{(x_{i}-\bar{x})^2}}{n-1}} \]
  • For corn, the standard deviation is \( 1.36 \; \text{barrels/acre} \).
  • For cauliflower, it is \( 1.69 \; \text{barrels/acre} \).
These deviations indicate differences in how oil consumption varies for the two crops. They are integral in calculating the margin of error for the confidence interval.
T-Distribution
The t-distribution is essential when we deal with sample sizes typically less than 30. It helps estimate population parameters when the sample size is small and the population standard deviation is unknown.

In this problem, we calculate the t-score to find the critical value corresponding to the desired confidence level (90% in this case).

The t-distribution is similar to the normal distribution but has heavier tails, which makes it more accommodating for small sample sizes. We use the formula for the margin of error:\[ ME = t_{crit} \sqrt{\frac{s_{1}^2}{n_{1}} + \frac{s_{2}^2}{n_{2}}} \]By finding the critical t-value from a t-table (1.782 for 12 degrees of freedom), we can construct our confidence interval. This helps us capture the true difference in means with 90% confidence.

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Most popular questions from this chapter

The EPA limit on the allowable discharge of suspended solids into rivers and streams is 60 milligrams per liter (mg/) per day. A study of water samples selected from the discharge at a phosphate mine shows that over a long period, the mean daily discharge of suspended solids is \(48 \mathrm{mg} / \mathrm{l}\). but day-to-day discharge readings are variable. State inspectors measured the discharge rates of suspended solids for \(n=20\) days and found \(s^{2}=39(\mathrm{mg} / 1)^{2}\) Find a \(90 \%\) confidence interval for \(\sigma^{2}\). Interpret your results.

Under what assumptions can the \(F\) distribution be used in making inferences about the ratio of population variances?

It is recognized that cigarette smoking has a deleterious effect on lung function. In a study of the effect of cigarette smoking on the carbon monoxide diffusing capacity (DL) of the lung, researchers found that current smokers had DL readings significantly lower than those of either ex smokers or nonsmokers. The carbon monoxide diffusing capacities for a random sample of \(n=20\) current smokers are listed here: $$ \begin{array}{rrrrr} 103.768 & 88.602 & 73.003 & 123.086 & 91.052 \\\ 92.295 & 61.675 & 90.677 & 84.023 & 76.014 \\ 100.615 & 88.017 & 71210 & 82.115 & 89.222 \\ 102.754 & 108.579 & 73.154 & 106.755 & 90.479 \end{array} $$ a. Do these data indicate that the mean DL reading for current smokers is significantly lower than 100 DL, the average for nonsmokers? Use \(\alpha=.01\). b. Find a \(99 \%\) upper one-sided confidence bound for the mean DL reading for current smokers. Does this bound confirm your conclusions in part a?

The main stem growth measured for a sample of seventeen 4 -year-old red pine trees produced a mean and standard deviation equal to 11.3 and 3.4 inches, respectively. Find a \(90 \%\) confidence interval for the mean growth of a population of 4 -year-old red pine trees subjected to similar environmental conditions.

Use the Student's \(t\) Probabilities applet to find the following probabilities: a. \(P(t>1.2)\) with 5 df b. \(P(t>2)+P(t<-2)\) with \(10 d f\) c. \(P(t<-3.3)\) with \(8 d f\) d. \(P(t>.6)\) with 12 df
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