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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, \(t = \frac{{\bar x - {\mu _0}}}{{s/\sqrt n }}\).

The given true average weight is \(200lb\) per price. Let the given boxplot be:

Because the majority of the boxplot lies to the right (and thus above) the \(200lb\) specification, the production criteria do not appear to have been met.

Let the given be:

\(\begin{array}{c}n = 30\\\mathop x\limits^\_ = 206.73\\s = 6.35\end{array}\)

Assume: \(\alpha = 0.05\).Claim that the average is equal to \(200lb\) per price. The given claim is either the null hypothesis or the alternative hypothesis.

The value of the test statistic:

\(\begin{array}{c}t = \frac{{\bar x - {\mu _0}}}{{s/\sqrt n }}\\ = \frac{{206.73 - 200}}{{6.35/\sqrt {30} }}\\ \approx 5.805\end{array}\)

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 \(\begin{array}{c}df = n - 1\\ = 30 - 1\\ = 29\end{array}\) for the student.

\(P < 2 \times 0.0005 = 0.001\)

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

\(P < 0.05 \Rightarrow {\rm{Reject }}{H_0}\)

The true average weight is different from \(200lb\) per price is supported by appropriate evidence.