91Ó°ÊÓ

The pathogen Phytophthora capsici causes bell peppers to wilt and die. Because bell peppers are an important commercial crop, this disease has undergone a great deal of agricultural research. It is thought that too much water aids the spread of the pathogen. Two fields are under study. The first step in the research project is to compare the mean soil water content for the two fields (Source: Journal of Agricultural, Biological, and Environmental Statistics, Vol. 2, No. 2). Units are percent water by volume of soil. \(\begin{aligned} &\text { Field A samples, } x_{1} \text { : }\\\ &\begin{array}{llllrll} 10.2 & 10.7 & 15.5 & 10.4 & 9.9 & 10.0 & 16.6 \\ 15.1 & 15.2 & 13.8 & 14.1 & 11.4 & 11.5 & 11.0 \end{array} \end{aligned}\) \(\begin{aligned} &\text { Field B samples, } x_{2} \text { : }\\\ &\begin{array}{rrrrrrrr} 8.1 & 8.5 & 8.4 & 7.3 & 8.0 & 7.1 & 13.9 & 12.2 \\ 13.4 & 11.3 & 12.6 & 12.6 & 12.7 & 12.4 & 11.3 & 12.5 \end{array} \end{aligned}\) i. Use a calculator with mean and standard deviation keys to verify that \(\bar{x}_{1} \approx\) \(12.53, s_{1} \approx 2.39, \bar{x}_{2} \approx 10.77\), and \(s_{2} \approx 2.40\) ii. Assuming the distribution of soil water content in each field is moundshaped and symmetric, use a \(5 \%\) level of significance to test the claim that field \(\mathrm{A}\) has, on average, a higher soil water content than field \(\mathrm{B}\).

Step by step solution

01

Calculate the Sample Means

To verify the sample mean for Field A, sum all the data points for Field A: \(10.2 + 10.7 + 15.5 + 10.4 + 9.9 + 10.0 + 16.6 + 15.1 + 15.2 + 13.8 + 14.1 + 11.4 + 11.5 + 11.0 = 175.5\). Divide this sum by the number of samples, which is 14. Therefore, \(\bar{x}_1 = \frac{175.5}{14} \approx 12.53\). Repeat the process for Field B.

02

Calculate the Sample Standard Deviations

To verify the standard deviation for Field A, use the formula: \(s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}\). Substituting the data points and the mean \(\bar{x}_1\), we find \(s_1 \approx 2.39\). Repeat this calculation for Field B to confirm \(s_2 \approx 2.40\).

03

State the Null and Alternative Hypotheses

The null hypothesis \(H_0\) is \(\mu_A = \mu_B\) (Field A and Field B have the same mean soil water content). The alternative hypothesis \(H_1\) is \(\mu_A > \mu_B\) (Field A has a higher mean soil water content).

04

Calculate the Test Statistic

Assuming equal variances, use the formula for the t-statistic: \(t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\left(\frac{s_1^2}{n_1}\right) + \left(\frac{s_2^2}{n_2}\right)}}\), where \(n_1 = 14\) and \(n_2 = 16\). Substituting the values gives \(t \approx 2.19\).

05

Determine the Critical Value and Conclusion

Look up the critical value for \(t\) with \(df \approx 28\) at \(\alpha = 0.05\) for a one-tailed test, which is approximately \(1.701\). Since \(t_{calc} = 2.19 > t_{critical} = 1.701\), reject the null hypothesis.

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Key Concepts

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

Sample Mean Calculation

A sample mean is a statistical measure that provides an average of a set of data points. It's crucial for summarizing data and making comparisons between different groups. To find the sample mean, simply sum up all the data values and then divide by the number of samples. Let's dive into how we calculate it.
For Field A, the sample mean is calculated by adding all soil water content readings: \[175.5 = 10.2 + 10.7 + 15.5 + 10.4 + 9.9 + 10.0 + 16.6 + 15.1 + 15.2 + 13.8 + 14.1 + 11.4 + 11.5 + 11.0 \].
Then, divide this sum by the number of samples, which is 14. This yields a mean \[\bar{x}_1 = \frac{175.5}{14} \approx 12.53\].
Repeat the same process for Field B to find \[\bar{x}_2\approx 10.77\].
This tells us that on average, soil from Field A has higher water content than Field B.

Standard Deviation Calculation

Standard deviation is a measure of the amount of variation or dispersion in a set of values. A lower standard deviation suggests that the values tend to be close to the mean, while a higher standard deviation indicates they are spread out over a wider range.
To calculate the standard deviation, you use the formula: \[s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}\].
For Field A, substitute each data point and the calculated mean, \(\bar{x}_1 = 12.53\), to find the deviations from the mean. After squaring each deviation and summing them, divide by \(n-1\), where \(n\) is the number of measurements.
For Field A, \(s_1\approx 2.39\). The same process applies to Field B, confirming \(s_2\approx 2.40\).
This calculation is essential to understanding how much individual water contents in the fields vary from their averages.

Null and Alternative Hypothesis

Statistical hypothesis testing begins with formulating two different statements: the null and alternative hypotheses. These are crucial for any kind of statistical testing.
The **null hypothesis (\(H_0\))** posits no effect or no difference. In this scenario, it suggests that the mean soil water content of Field A is equal to that of Field B (\(\mu_A = \mu_B\)).
On the other hand, the **alternative hypothesis (\(H_1\))** suggests a specific effect or difference. Here, it claims that Field A has a higher mean soil water content than Field B (\(\mu_A > \mu_B\)).
Formulating these hypotheses is the foundation for establishing the presence or absence of an effect or difference in statistical analysis.

t-Test

The t-test is a statistical test used to determine if there is a significant difference between the means of two groups. It is particularly useful when dealing with small sample sizes or unknown variances.
For our scenario, we use the t-test to compare soil water content means between the two fields. The formula for the t-statistic, assuming equal variances, is:
\[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\left(\frac{s_1^2}{n_1}\right) + \left(\frac{s_2^2}{n_2}\right)}} \]
Here \(\bar{x}_1\) and \(\bar{x}_2\) are the sample means, \(s_1\) and \(s_2\) are the sample standard deviations, and \(n_1\) and \(n_2\) are the sample sizes for fields A and B, respectively.
Upon calculation, this results in \(t \approx 2.19\). This figure becomes critical when determining the significance of our results in the next step.

Critical Value Determination

The critical value in hypothesis testing helps decide whether to reject the null hypothesis. It's determined from the t-distribution table based on the desired significance level (\(\alpha\)) and degrees of freedom (df).
For our example, we use a significance level of \(0.05\), and approximate degrees of freedom (often calculated using a conservative approach for two independent samples).
Looking up the critical value for a one-tailed test with \(df \approx 28\), we find approximately \(1.701\).
When comparing the calculated t-value (\(t_{calc}\approx 2.19\)) to this critical value, we see that \(t_{calc} > t_{critical}\).
This leads us to reject the null hypothesis, supporting the claim that Field A does indeed have a higher mean soil water content than Field B.