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 (from "Engineering Properties of Soil," Soil Science, 1998: 93–102).1.10 5.09 0.97 1.59 4.60 0.32 0.55 1.450.14 4.47 1.20 3.50 5.02 4.67 5.22 2.693.98 3.17 3.03 2.21 0.69 4.47 3.31 1.170.76 1.17 1.57 2.62 1.66 2.05The values of the sample mean, sample standard deviation,and (estimated) standard error of the mean are2.481, 1.616, and .295, respectively. Does this data suggestthat the true average percentage of organic matterin such soil is something other than 3%? Carry out atest of the appropriate hypotheses at significance level.10. Would your conclusion be different if a 5 .05 hadbeen used? [Note: A normal probability plot of the datashows an acceptable pattern in light of the reasonablylarge sample size.]

Answer 1

We conclude that the true average percentage of organic matter in such soil is something other than 3% at 10% significance level.

We conclude that the true average percentage of organic matter in such soil is 3% at 5% significance level.

Explanation:

We are given a random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specimen;

1.10, 5.09, 0.97, 1.59, 4.60, 0.32, 0.55, 1.45, 0.14, 4.47, 1.20, 3.50, 5.02, 4.67, 5.22, 2.69, 3.98, 3.17, 3.03, 2.21, 0.69, 4.47, 3.31, 1.17, 0.76, 1.17, 1.57, 2.62, 1.66, 2.05.

Let
`\mu` = true average percentage of organic matter

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

Alternate Hypothesis,
`H_A` :
`\mu \\eq` 3% {means that the true average percentage of organic matter in such soil is something other 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. =
`(\bar X-\mu)/((s)/(√(n) ) )` ~
`t_n_-_1`

where,
`\bar X` = sample mean percentage of organic matter = 2.481%

s = sample standard deviation = 1.616%

n = sample of soil specimens = 30

So, the test statistics =
`(2.481-3)/((1.616)/(√(30) ) )` ~
`t_2_9`

= -1.76

The value of t-test statistics is -1.76.

(a) Now, at 10% level of significance the t table gives a critical value of -1.699 and 1.699 at 29 degrees of freedom for the two-tailed test.

Since the value of our test statistics doesn't lie within the range of critical values of t, so we have sufficient evidence to 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 something other than 3% at 10% significance level.

(b) Now, at 5% level of significance the t table gives a critical value of -2.045 and 2.045 at 29 degrees of freedom for the two-tailed test.

Since the value of our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that the true average percentage of organic matter in such soil is 3% at 5% significance level.