91Ó°ÊÓ

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. $$ \begin{array}{lc} \text { Corn } & \text { Cauliflower } \\ \hline 5.6 & 15.9 \\ 7.1 & 13.4 \\ 4.5 & 17.6 \\ 6.0 & 16.8 \\ 7.9 & 15.8 \\ 4.8 & 16.3 \\ 5.7 & 17.1 \end{array} $$ a. Use these data to find a \(90 \%\) confidence interval for the difference between the mean amounts of oil required to produce these two crops. b. Based on the interval in part a, is there evidence of a difference in the average amount of oil required to produce these two crops? Explain.

Step by step solution

01

Calculate the mean oil required for each crop

To start, we need to calculate the mean amount of oil required for each crop. We can do this by adding up all the values for each crop and dividing by the number of samples (7 in this case). Mean for Corn: $$ \bar{X}_1 = \frac{5.6 + 7.1 + 4.5 + 6.0 + 7.9 + 4.8 + 5.7}{7} = 5.94 $$ Mean for Cauliflower: $$ \bar{X}_2 = \frac{15.9 + 13.4 + 17.6 + 16.8 + 15.8 + 16.3 + 17.1}{7} = 16.13 $$ So, the mean amount of oil required to produce an acre of corn is 5.94 barrels while the mean amount for cauliflower is 16.13 barrels.

02

Calculate the standard deviation and standard error for each crop

Next, we need to calculate the standard deviation and standard error for both crops. Standard Deviation for Corn (s1): $$ s_1 = \sqrt{\frac{(5.6-5.94)^2 + (7.1-5.94)^2 \dots + (5.7-5.94)^2}{7-1}} = 1.15 $$ Standard Deviation for Cauliflower (s2): $$ s_2 = \sqrt{\frac{(15.9-16.13)^2 + (13.4-16.13)^2 \dots + (17.1-16.13)^2}{7-1}} = 1.38 $$ Standard Error (SE): $$ SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{1.15^2}{7}+\frac{1.38^2}{7}} = 0.71 $$ The standard deviation for corn is 1.15 and for cauliflower is 1.38. The standard error is 0.71.

03

Calculate the t-score and confidence interval

For a 90% confidence interval, we need to find the t-score corresponding to the 0.05 level of significance in the t-distribution (because we use one tail each side). Since we have 7 samples for both groups and we are comparing two groups, we use the degrees of freedom (df) = n1 + n2 - 2 = 7 + 7 - 2 = 12. Using the t-table for 12 degrees of freedom and 0.05 significance level (one tail), the t-score is approximately 1.782. Now, calculate the margin of error (ME): $$ ME = t_{score} \times SE = 1.782 \times 0.71 = 1.27 $$ Finally, calculate the confidence interval for the difference of the means: $$ (\bar{X}_1 - \bar{X}_2) \pm ME = (5.94-16.13) \pm 1.27 = -10.19 \pm 1.27 $$ The confidence interval is between -11.46 and -8.92.

04

Interpret the results

Based on the 90% confidence interval that we calculated, it ranges from -11.46 to -8.92. This means that we are 90% confident that the difference in the mean amount of oil required to produce 1 acre of corn and 1 acre of cauliflower is in this range. Since the entire interval is negative, it indicates that the mean amount of oil required for corn is less than that of cauliflower. There is evidence of a significant difference in the average amount of oil required to produce these two crops.

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Key Concepts

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

Mean Difference

The mean difference is a central concept when comparing data sets. In our problem, we are looking at the means of two different crops, corn and cauliflower, and the amount of oil required for each. The mean difference essentially tells us how much more (or less) oil is needed for one crop compared to the other.

• The mean for corn was calculated to be 5.94 barrels.
• The mean for cauliflower was given as 16.13 barrels.

The mean difference is found by subtracting the mean of corn from the mean of cauliflower:
(5.94 - 16.13) = -10.19.

This result helps us understand that, on average, producing an acre of corn requires 10.19 barrels less oil than an acre of cauliflower. The sign of the mean difference (negative, in this case) also indicates which crop is more oil-intensive.

Standard Deviation

Standard deviation is a measure of how spread out the numbers in a data set are. When calculating standard deviation for the oil required by crops, it gives us an insight into the variability of oil consumption across different samples of corn and cauliflower fields.
For corn, the standard deviation was calculated to be 1.15 barrels, and for cauliflower, it was 1.38 barrels.

• This tells us that there's a moderate level of variation within each sample group, with cauliflower having slightly more variation.
• A lower standard deviation means the data points are closer to the mean, while a higher standard deviation indicates more spread.

Understanding standard deviation is crucial, as it impacts the overall precision and reliability of our calculated confidence interval.

T-Distribution

The t-distribution plays a key role when calculating confidence intervals for small sample sizes. It accounts for additional uncertainty due to limited data points. Here, our data consists of samples from 7 plots for each crop, which means we must consider this distribution when determining our confidence interval.

We used a t-score to find the critical value necessary for our 90% confidence interval. Given our degrees of freedom were 12 (7 samples per group, and 14 total observations minus 2), the corresponding t-score was found to be approximately 1.782.

• The t-distribution is crucial when sample sizes are too small to use the normal distribution, which requires larger sample sizes.
• The shape of the t-distribution is similar to the normal curve but with heavier tails, accommodating the additional variability from small samples.

Using this t-score allows us to accurately calculate the margin of error for our confidence interval.

Standard Error

The standard error measures the accuracy with which a sample mean represents the population mean. It’s derived from the standard deviations of the two data sets and the number of samples.
In our exercise, the standard error was calculated to be 0.71.

• Calculation involves the formula: \[SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}\]
• This means that each group's variability is divided by the number of samples before being combined.

A lower standard error indicates that the sample mean is a more accurate reflection of the true population mean. In our context, this tells us how precise the mean difference calculation is, ultimately affecting the width of our confidence interval. The standard error helps determine how much we can expect the sample mean to vary from the actual population mean, thus influencing our confidence of the results.