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 2.85 2.66 3 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 3.17 3.37 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 95% confidence interval for the decrease in penetrability of compressed soil relative to intermediate soil. Compute degrees of freedom using the conservative method. Interval: to

Answer 1

Answer:

Step-by-step explanation:

Given that;

Compressed Soil 2.85 2.66 3 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 3.17 3.37 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

                            Compressed soil           intermediate soil

Mean                      2.907                              3.286

SD                          0.1414                              0.2377

Here we have,

$\bar x_1 =3.286\\ \bar x_2 =2.907\\s_1 =0.2377\\ s_2 =0.1414\\n_1=19\\n_2=20$

The pooled Standard deviation

$s_p=\sqrt{\frac{(n_1-1)s_1^2(n_2-1)s_2^2}{n_1+n_2-2} } \\\\=0.1943$

So standard error for difference in population mean is

$s_{\bar x_1 - \bar x_2} = \sqrt{\frac{s_p^2}{n_1}+\frac{s_p^2}{n_2} } \\\\=0.0623$

by inputting the values we get 0.0623

Degree of freedom for t is

df = 19 + 20 - 2 = 37,

so t-critical value is 2.715.

So required confidence interval for

$\mu_{1} - \mu_{2}$ will be

$( \bar x_1 - \bar x_2) \pm t_{critical}*s_{\bar x_1 - \bar x_2}$

$=0.3793 \pm2.715*0.063\\\\=0.3793\pm0.1690$

So required confidence interval is (0.2103 , 0.5483).