Question

A biologist wants to test whether chemically modifying a particular plant family will reduce their ability to retain water and moisture. In order to test this, a fixed volume of water is given to plants that have been chemically modified, and plants in a control group that have not been modified. The volume of moisture retained by each plant some time after is then recorded.

The statistics are given in the table below:

Volume of moisture retained
Group Sample Size Sample Mean (ml) Sample Standard Deviation (ml)
Control Group 24 146 114
Modified Group 23 106 115
At level of alpha = 0.05, the null hypothesis that chemical modification does not reduce the plant's ability to retain water and moisture is ________.

Answer 1

Answer:

the answer is 4.56

Step-by-step explanation:

because it is i use my mommy calculator

Answer 2

There would be enough information to establish that chemical alteration has no effect on the plant's capacity to absorb moisture from the air. A further solution is below.

According to the question,

• $n_1 = 23, n_2 = 24$
• $\bar{x_1} = 140, \bar{x_2} = 152$
• $s_1 = 94, s_2 = 60$

The degree of freedom are:

→ $d.f = n_1+n_2-2$

        $= 23+24-2$

        $= 45$

The null and alternative hypothesis are:

• $H_0: \mu_1 \leq \mu_2$
• $H_1: \mu_1 \geq \mu_2$

The value of test statistics will be:

→ $t = \frac{\bar{x_1}- \bar{x_2}}{\sqrt{\frac{s_1^2}{n_1} +\frac{s_2^2}{n_2} } }$

By substituting the values, we get

     $= \frac{140-152}{\sqrt{\frac{94^2}{23} +\frac{60^2}{24} } }$

     $= -1.288028$

The P value for the given test statistics will be:

→ $P-value = P(t \leq -|t_{cal}|)$

                   $= P(t \leq -|-1.288028|)$

                   $= P(t \leq - 1.288028)$

                   $= 0.1066014$

or,

                   $= 0.1066$

Thus the above response is appropriate.

                 

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