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Fortunately, we aren't really interested in the number of seeds velvetleaf plants produce (see Exercise \(20.41\) ). The velvetleaf seed beetle feeds on the seeds and might be a natural weed control. Here are the total seeds, seeds infected by the beetle, and percent of seeds infected for 28 velvetleaf plants: $$ \begin{array}{l|rrrrrrrrrr} \hline \text { Seeds } & 2450 & 2504 & 2114 & 1110 & 2137 & 8015 & 1623 & 1531 & 2008 & 1716 \\ \text { Infected } & 135 & 101 & 76 & 24 & 121 & 189 & 31 & 44 & 73 & 12 \\ \text { Percent } & 5.5 & 4.0 & 3.6 & 2.2 & 5.7 & 2.4 & 1.9 & 2.9 & 3.6 & 0.7 \\\ \hline \text { Seeds } & 721 & 863 & 1136 & 2819 & 1911 & 2101 & 1051 & 218 & 1711 & 164 \\ \text { Infected } & 27 & 40 & 41 & 79 & 82 & 85 & 42 & 0 & 64 & 7 \\ \text { Percent } & 3.7 & 4.6 & 3.6 & 2.8 & 4.3 & 4.0 & 4.0 & 0.0 & 3.7 & 4.3 \\\ \hline \text { Seeds } & 2228 & 363 & 5973 & 1050 & 1961 & 1809 & 130 & 880 & & \\ \text { Infected } & 156 & 31 & 240 & 91 & 137 & 92 & 5 & 23 & & \\ \text { Percent } & 7.0 & 8.5 & 4.0 & 8.7 & 7.0 & 5.1 & 3.8 & 2.6 & & \\ \hline \end{array} $$ Do a complete analysis of the percent of seeds infected by the beetle. Include a \(90 \%\) confidence interval for the mean percent infected in the population of all velvetleaf plants. Do you think that the beetle is very helpful in controlling the weed? Why is analyzing percent of seeds infected more useful than analyzing number of seeds infected?

Short Answer

Expert verified
The 90% confidence interval helps determine beetle effectiveness. Percent infected is better as it adjusts for plant size differences.

Step by step solution

01

Gather Data

Begin by listing all the percent values of seeds infected by the beetle from the given data: 5.5, 4.0, 3.6, 2.2, 5.7, 2.4, 1.9, 2.9, 3.6, 0.7, 3.7, 4.6, 3.6, 2.8, 4.3, 4.0, 4.0, 0.0, 3.7, 4.3, 7.0, 8.5, 4.0, 8.7, 7.0, 5.1, 3.8, 2.6.
02

Calculate Mean

Compute the mean of these percentage values. Sum up all the percentages and divide by the total number of observations (28). Calculate as follows: \(\text{Mean} = \frac{5.5 + 4.0 + ... + 2.6}{28}\).
03

Calculate Standard Deviation

Determine the standard deviation of the percent values. First, find the difference between each value and the mean, square these differences, and then compute the average of these squared differences to find the variance. Finally, take the square root of the variance to get the standard deviation.
04

Determine Standard Error

Calculate the standard error of the mean percent infected using the formula: \(\text{Standard Error} = \frac{\text{Standard Deviation}}{\sqrt{n}}\), where \(n=28\).
05

Find Critical Value

For a 90% confidence interval and 27 degrees of freedom (n-1), find the t-score using a t-distribution table. This will likely be around 1.703.
06

Calculate Confidence Interval

Calculate the 90% confidence interval for the mean percent using the formula: \(\text{Confidence Interval} = \text{Mean} \pm \text{t-score} \times \text{Standard Error}\).
07

Interpret the Confidence Interval

Analyze the interval to understand the range within which the true mean percent infected falls with 90% confidence. Discuss whether this indicates effectiveness of the beetles in controlling the population.
08

Compare Percent vs Number Infected

Discuss why analyzing the percentage of seeds infected is more useful than the number. It accounts for variability in total seeds and provides a proportional representation of infection rate.

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Key Concepts

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

Confidence Interval
In statistics, a confidence interval provides an estimated range of values which is likely to include an unknown population parameter, such as the mean. This gives us a margin of error where we're reasonably confident the true average falls within a particular range. In our exercise, a 90% confidence interval was calculated for the mean percentage of seeds infected by the beetle.

The key components in calculating a confidence interval include:
  • Mean (\(\bar{x}\)): The average value of your data set.
  • Standard Error (SE): A measure of the statistical accuracy of an estimate, calculated by dividing the standard deviation by the square root of the number of observations \((SE = \frac{SD}{\sqrt{n}})\).
  • t-score: A value obtained from a t-distribution table which accounts for the desired confidence level and the degrees of freedom in your sample size. For a 90% confidence level, this is typically around 1.703 when the sample size is 28.
The formula for the confidence interval is: \(\text{Confidence Interval} = \bar{x} \pm (t \cdot SE)\).

This calculated interval helps assess whether the beetles are indeed effective in infecting a significant number of seeds across the entire population.
Mean Calculation
The mean, often referred to as the average, is a central value of a finite set of numbers calculated by summing all values and dividing by the count of numbers. Calculating the mean percentage of seeds infected by beetles gives us a basic understanding of their effectiveness.

To calculate the mean:
  • Add together all the percentage values you collected.
  • Divide the sum by the number of observations. In this case, there are 28 plants.
For example, if your sum of percent infected is 100.0, then the mean percent infected would be \(\frac{100.0}{28} = 3.57\%.\)

Finding the mean helps set a baseline to understand the overall infection trend by beetles across a range of velvetleaf plants.
Standard Deviation
Standard deviation (SD) is a measure of the amount of variation or dispersion in a set of values. Understanding the SD of the percentage of seeds infected allows us to assess how much variation there is in seed infection across different plants.

To find the standard deviation:
  • First, compute the variance by:
    • Calculating the mean.
    • Finding the difference between each percentage value and the mean, then squaring those differences.
    • Averaging these squared differences to get the variance.
  • The standard deviation is simply the square root of this variance.
For example, if the variance is 2.89, then the standard deviation would be \(\sqrt{2.89} = 1.7\).

In summary, smaller standard deviation values indicate that the data points tend to be close to the mean, whereas larger values signify more spread out data.
Analyzing the standard deviation gives us context on how consistent the beetles are in infecting similar proportions of seeds.
t-distribution
The t-distribution is a type of probability distribution that is symmetrical and bell-shaped, like the normal distribution, but it has heavier tails. It emerges when estimating the mean of a normally distributed population in situations where the sample size is small and the population standard deviation is unknown.

In the context of calculating the confidence interval for the mean percent of seeds infected, the t-distribution is used to find the appropriate t-score. This score is crucial for creating a confidence interval with a small sample size, as it accounts for additional variability due to the limited data.

The t-score can be found in a t-distribution table based on:
  • The desired confidence level (in this case, 90%).
  • The degrees of freedom, which is the sample size minus one (\(n-1\)). For 28 samples, that's 27 degrees of freedom.
Due to its adaptability with smaller samples and compensating for fewer data points, the t-distribution offers a more accurate estimation when compared to the standard normal distribution.

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Most popular questions from this chapter

The Trial Urban District Assessment (TUDA) is a government-sponsored study of student achievement in large urban school districts. TUDA gives a mathematics test scored from 0 to 500 . A score of 262 is a "basic" mathematics level and a score of 299 is "proficient." Scores for a random sample of 1100 eighth- graders in Dallas had \(x \bar{x}=271\) with standard error 1.3. 16 (a) We don't have the 1100 individual scores, but use of the \(t\) procedures is surely safe. Why? (b) Give a \(99 \%\) confidence interval for the mean score of all Dallas eighthgraders. (Be careful: the report gives the standard error of \(x^{-} \bar{x}\), not the standard deviation s.) (c) Urban children often perform below the basic level. Is there good evidence that the mean for all Dallas eighth-graders is more than the basic level?

In a randomized comparative experiment on the effect of color on the performance of a cognitive task, researchers randomly divided 69 subjects ( 27 males and 42 females ranging in age from 17-25 years) into three groups. Participants were asked to solve a series of six anagrams. One group was presented with the anagrams on a blue screen, one group saw them on a red screen, and one group had a neutral screen. The time, in seconds, taken to solve the anagrams was recorded. The paper reporting the study gives \(x-\bar{x}=11.58\) and \(s=4.37\) for the times of the 23 members of the neutral group. \({ }^{17}\) (a) Give a 95\% confidence interval for the mean time in the population from which the subjects were recruited. (b) What conditions for the population and the study design are required by the procedure you used in part (a)? Which of these conditions are important for the validity of the procedure in this case?

Use software or Table \(C\) to find (a) the critical value for a one-sided test with level \(\alpha=0.01\) based on the \(t_{2}\) distribution. (b) the critical value for a \(90 \%\) confidence interval based on the \(t_{28}\) distribution.

Velvetleaf is a particularly annoying weed in corn fields. It produces lots of seeds, and the seeds wait in the soil for years until conditions are right. How many seeds do velvetleaf plants produce? Here are counts from 28 plants that came up in a corn field when no herbicide was used: 28 \(\begin{array}{ll}245025042114111021378015 & 1623\end{array}\) \(\begin{array}{lllllllllll}721 & 863 & 1136 & 2819 & 1911 & 2101 & 1051 & 218 & 1711 & 164\end{array}\) \(22283635973 \quad 105019611809 \quad 130 \quad 880\) We would like to give a confidence interval for the mean number of seeds produced by velvetleaf plants. Alas, the \(t\) interval can't be safely used for these data. Why not?

Researchers at Texas A\&M studied the effect of using standing height desks in an elementary school on the energy expended by nine students. In their paper about the study, they reported descriptive statistics for the nine students. These descriptive statistics were expressed as a mean plus or minus a standard deviation. \({ }^{3}\) One such descriptive statistic was the weight of students before using the standing desks, which was reported as \(27.0 \pm 7.9\) kilograms. What are \(\mathrm{x}^{-} \bar{x}\) and the standard error of the mean for these students? (This exercise is also a warning to read carefully: that \(27.0 \pm\) \(7.9\) is not a confidence interval, yet summaries in this form are common in scientific reports.)

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