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The standard deviation of the contents in the sample of 27 filled cups is 0.17 oz.

The significance level is 0.05.

The standard deviation of the specific machine design is 0.15 oz.

For applying the hypothesis test, first, set up a null and an alternative hypothesis.

The null hypothesis is the statement about the value of a population parameter, which is equal to the claimed value. It is denoted by\({H_0}\).

The alternate hypothesis is a statement that the parameter has a value opposite to the null hypothesis. It is denoted by\({H_1}\).

The claim states that the machine dispenses amounts with a standard deviation greater \(\left( \sigma \right)\) than the standard deviation of 0.15.

From the claim, the null and alternative hypotheses are as follows.

\(\begin{array}{l}{{\rm{H}}_0}:\sigma = 0.15\\{{\rm{H}}_1}:\sigma > 0.15\end{array}\)

Here, \(\sigma \)is the standard deviation of the dispensed amounts.

To conduct a hypothesis test of a claim about a population standard deviation\(\sigma \) or population variance\({\sigma ^2}\),the test statistic is as follows.

\(\begin{array}{c}{\chi ^2} = \frac{{\left( {{\rm{n}} - 1} \right) \times {s^2}}}{{{\sigma ^2}}}\\ = \frac{{\left( {27 - 1} \right) \times {{0.17}^2}}}{{{{0.15}^2}}}\\ = 33.396\end{array}\).

Thus, the value of the test statistic is 33.396.

The degree of freedom is as follows.

\(\begin{array}{c}df = n - 1\\ = 27 - 1\\ = 26\end{array}\)

The test is right-tailed.

The probability of the right-tailed areas is as follows.

\(\begin{array}{c}P\left( {{\chi ^2} > \chi _\alpha ^2} \right) = \alpha \\P\left( {{\chi ^2} > \chi _{0.05}^2} \right) = 0.05\end{array}\)

Referring to the chi-square table, the critical value of\({\chi ^2}\)is obtained as 38.885 from the area to the right of 0.05, and it corresponds to a degree of freedom of 26.

The decision rule for the test is stated as follows.

If the test statistic is greater than the critical value, reject the null hypothesis at the given level of significance.

As it is observed that \({\chi ^2}\left( {33.396} \right)\, < \,\chi _{0.05}^2\left( {38.885} \right)\), the null hypothesis is failed to be rejected.

Therefore, it is concluded that there is not enough evidence to support the claim that the standard deviation of the content that the machine dispenses is greater than 0.15 oz.