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The null hypothesis states that the population mean is equal to the value mentioned in the claim. If the null hypothesis is the claim, then the alternative hypothesis states the opposite of the null hypothesis.

\(\begin{array}{l}{H_0}:\mu = 0\\{H_a}:\mu \ne 0\end{array}\)

The formula for the value of the test statistic is given by, \(t = \frac{{\bar x - {\mu _0}}}{{s/\sqrt n }}\).

Let the given be:

\(\begin{array}{c}n = 15\\\mathop x\limits^\_ = 0.0453333\\s = 0.0899100\\\alpha = 0.01,0.05,0.10\end{array}\)

Claim that the average is equal to zero. The given claim is either the null hypothesis or the alternative hypothesis.

The value of the test statistic:

\(\begin{array}{c}t = \frac{{\bar x - {\mu _0}}}{{s/\sqrt n }}\\ = \frac{{0.0453333 - 0}}{{0.0899100/\sqrt {15} }}\\ \approx 1.953\end{array}\)

The P-value is the chance of getting the test statistic's result, or a number that is more severe. The P-value is the number (or interval) in the column header of the T table in the appendix that contains the t-value in the row\(\begin{array}{c}df = n - 1\\ = 15 - 1\\ = 14\end{array}\)for the student.

\(0.05 = 2 \times 0.025 < P < 2 \times 0.05 = 0.10\)

As the P-value is smaller than the significance level, so the null hypothesis is rejected.

\(\begin{array}{l}P > 0.01 \Rightarrow {\rm{Fail to reject }}{H_0}\\P > 0.05 \Rightarrow {\rm{Fail to reject }}{H_0}\\P < 0.10 \Rightarrow {\rm{Reject }}{H_0}\end{array}\)

At the significance levels of\(0.01\)and\(0.05\), the amount of impurity must be adjusted.

At the significance level \(0.10\), the amount of impurity must not be altered.