Question

A random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specimen, resulting in the accompanying data. 1.10 5.09 1.20 3.50 0.69 4.47 0.97 1.59 4.60 5.02 4.67 5.22 3.31 1.17 0.76 0.32 0.55 1.45 0.14 4.47 2.69 3.98 3.17 3.03 2.21 1.17 1.57 2.62 1.66 2.05 The values of the sample mean and sample standard deviation are 2.481 and 1.616, respectively. Suppose we know the population distribution is normal, does this data suggest that the true average percentage of organic matter in such soil is something other than 3%? Carry out a test of the appropriate hypotheses at significance level 0.10 using the P-value method.

Answer 1

Answer:

We conclude that the true average percentage of organic matter in such soil is different from 3%.

Step-by-step explanation:

We are given that the values of the sample mean and sample standard deviation are 2.481 and 1.616, respectively.

Suppose we know the population distribution is normal, we have to test the hypothesis that does this data suggest that the true average percentage of organic matter in such soil is something other than 3%.

Let $\mu$ = true average percentage of organic matter in such soil

SO, Null Hypothesis, $H_0$ : $\mu$ = 3%   {means that the true average percentage of organic matter in such soil is equal to 3%}

Alternate Hypothesis, $H_A$ : $\mu \neq$ 3%   {means that the true average percentage of organic matter in such soil is different than 3%}

The test statistics that will be used here is One-sample t test statistics because we don't know about the population standard deviation;

                           T.S.  = $\frac{\bar X -\mu}{{\frac{s}{\sqrt{n} } } }$  ~ $t_n_-_1$

where,  $\bar X$ = sample mean amount of organic matter = 2.481%

              s = sample standard deviation = 1.616%

              n = sample of soil specimens = 30

So, test statistics  =  $\frac{0.02481-0.03}{{\frac{0.01616}{\sqrt{30} } } }$  ~ $t_2_9$

                               =  -1.759

Now, P-value of the test statistics is given by;

       P-value = P( $t_2_9$ > -1.759) = 0.046 or 4.6%

• If the P-value of test statistics is more than the level of significance, then we will not reject our null hypothesis as it will not fall in the rejection region.

• If the P-value of test statistics is less than the level of significance, then we will reject our null hypothesis as it will fall in the rejection region.

Now, here the P-value is 0.046 which is clearly smaller than the level of significance of 0.05 (for two-tailed test), so we will reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the true average percentage of organic matter in such soil is different from 3%.