Question

The desired percentage of silicon dioxide in a certain type of cement is \(5.0 \%\). A random sample of \(n=36\) specimens gave a sample average percentage of \(\bar{x}=5.21\). Let \(\mu\) be the true average percentage of silicon dioxide in this type of cement, and suppose that \(\sigma\) is known to be \(0.38\). Test \(H_{0}: \mu=5\) versus \(H_{a}\) : \(\mu \neq 5\) using a significance level of \(.01\).

Short answer

The null hypothesis, which states that the population mean of silicon dioxide in the cement is 5%, is rejected based on the Z-test conducted at a 0.01 level of significance. This implies that the true average percentage of silicon dioxide in cement is different from 5%.

Step-by-step solution

State the hypotheses

The first step in hypothesis testing is to set up the null hypothesis and the alternative hypothesis. Here, they are given by: \[H_{0}: \mu=5\] This is the null hypothesis and it suggests that the true average percentage of silicon dioxide in cement is 5%. \[H_{a}: \mu \neq 5\] This is the alternative hypothesis and it suggests that the true average percentage of silicon dioxide in cement is not 5%.

Compute the Test Statistic

With σ known, the test statistic for the Z-test is given by: \[Z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}}\]. Substituting the given values we get, Z = \(\frac{5.21 - 5}{0.38 / \sqrt{36}} = 3.34\). So, calculated Z score is 3.34.

Determine the Critical Value

The problem states that we should use a significance level of .01. Since this is a two-tailed test, we split the alpha into two so that each tail contains 0.005 of the sampling distribution. That places the critical values at -2.57 and 2.57.

Make the Decision

To make the decision for the test, compare the test statistics with the critical value. Here, the calculated z score of 3.34 is beyond the critical z value of 2.57. Therefore, we reject the null hypothesis.