Question

According to a certain chemical theory, the percentage of iron in a certain compound should be 12.1. In order to test this theory it was decided to analyze nine different random samples of the compound to see if the measurements would differ significantly from 12.1%. Suppose the nine different samples gave the following results:

11.7 12.2 10.9 11.4 11.3 12.0 11.1 10.7 11.6

Use either classical or p-value approach to verify the hypothesis at level of significance = 0.05. Assume that the population is normally distributed


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Answer 1

Using the t-distribution, it is found that since the p-value of the test is 0.0036 < 0.05, there is enough evidence to conclude that the percentage of iron in a certain compound is different of 12.1.

At the null hypothesis, it is tested if the mean is of 12.1%, hence:

$H_0: \mu = 12.1$

At the alternative hypothesis, it is tested if the mean is different of 12.1%, that is:

$H_1: \mu \neq 12.1$

We have the standard deviation for the sample, thus, the t-distribution is used. The test statistic is given by:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

The parameters are:

$\overline{x}$ is the sample mean.

$\mu$ is the value tested at the null hypothesis.

s is the standard deviation of the sample.

n is the sample size.

For this problem, the values of the parameters are: $\mu = 12, n = 9$.

• Using a calculator, it is found that: $s = 0.495, \overline{x} = 11.43$

The value of the test statistic is:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{11.43 - 12.1}{\frac{0.495}{\sqrt{9}}}$

$t = -4.06$

The p-value is found using a two-tailed test, as we are testing if the mean is less than a value, with t = -4.06 and 9 - 1 = 8 df.

• Using a calculator, this p-value is of 0.0036.

Since the p-value of the test is 0.0036 < 0.05, there is enough evidence to conclude that the percentage of iron in a certain compound is different of 12.1.

A similar problem is given at https://brainly.com/question/16767703