Question

Farmers know that driving heavy equipment on wet soil compresses the soil and injures future crops. Here are data on the "penetrability" of the same type of soil at two levels of compression. Penetrability is a measure of how much resistance plant roots will meet when they try to grow through the soil. Compressed Soil 2.82 2.66 2.98 2.82 2.76 2.81 2.78 3.08 2.94 2.86 3.08 2.82 2.78 2.98 3.00 2.78 2.96 2.90 3.18 3.16 Intermediate Soil 3.18 3.38 3.1 3.40 3.38 3.14 3.18 3.26 2.96 3.02 3.54 3.36 3.18 3.12 3.86 2.92 3.46 3.44 3.62 4.26 Use the data, omitting the high outlier, to give a 99% confidence interval for the decrease in penetrability of compressed soil relative to intermediate soil. Compute degrees of freedom using the conservative method.

Answer 1

Answer:

Step-by-step explanation:

Hello!

To see if driving heavy equipment on wet soil compresses it causing harm to future crops, the penetrability of two types of soil were measured:

Sample 1: Compressed soil

X₁: penetrability of a plot with compressed soil.

n₁= 20 plots

X[bar]₁= 2.90

S₁= 0.14

Sample 2: Intermediate soil

X₂: penetrability of a plot with intermediate soil.

n₂= 20 (with outlier) n₂= 19 plots (without outlier)

X[bar]₂= 3.34 (with outlier) X[bar]₂= 2.29 (without outlier)

S₂= 0.32 (with outlier) S₂= 0.24 (without outlier)

Outlier: 4.26

Assuming all conditions are met and ignoring the outlier in the second sample, you have to construct a 99% CI for the difference between the average penetration in the compressed soil and the intermediate soil. To do so, you have to use a t-statistic for two independent samples:

Parámeter of interest: μ₁-μ₂

Interval:

[(X[bar]₁-X[bar]₂)±$t_{n_1+n_2-2;1-\alpha/2}$*Sa$\sqrt{\frac{1}{n_1} +\frac{1}{n_2} }$]

$t_{n_1+n_2-2;1-\alpha/2}= t_{20+19-2;1-(0.01/2)}= t_{37; 0.995}= 2.715$

$Sa= \sqrt{\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2} } = \sqrt{\frac{19*0.0196+18*0.0576}{20+19-2} }= 0.195= 0.20$

[(2.90-2.29)±2.715*0.20$\sqrt{\frac{1}{20} +\frac{1}{19} }$]

[0.436; 0.784]

I hope this helps!

Answer 2

Using the t-distribution, it is found that the 99% confidence interval for the decrease in penetrability of compressed soil relative to intermediate soil is (-0.418, 1.182).

Using a calculator, we find that for the compressed soil, the mean and the standard deviation are given by:

$\mu_C = 2.9075, s_C = 0.1427$

For the intermediate soil, omitting the high outlier of 4.26, we have that:

$\mu_I = 3.2895, s_I = 0.2385$

The distribution of the decrease has mean and standard deviation given by:

$\overline{x} = \mu_I - \mu_C = 3.2895 - 2.9075 = 0.382$

$s = \sqrt{s_C^2 + s_I^2} = \sqrt{0.1427^2 + 0.2385^2} = 0.2779$

The interval is given by:

$\overline{x} \pm ts$

By the conservative method, the amount of degrees of freedom is one less than the smaller sample size, which in this case is 19, for the intermediate soil.

• Then, critical value, using a t-distribution calculator, for a two-tailed 99% confidence interval, with 19 - 1 = 18 df, is t = 2.8784.

Then:

$\overline{x} - ts = 0.382 - 2.8784(0.2779) = -0.418$

$\overline{x} + ts = 0.382 + 2.8784(0.2779) = 1.182$

The 99% confidence interval for the decrease in penetrability of compressed soil relative to intermediate soil is (-0.418, 1.182).

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