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The null hypothesis (H鈧�) is that the true mean weight of the cans is equal to the claimed weight on the label. The alternative hypothesis (H鈧�) is that the true mean weight is less than the claimed weight. Mathematically, we have: H鈧�: 渭 = 16 ounces H鈧�: 渭 < 16 ounces

Given the sample mean (\(\bar{x}\)), sample standard deviation (s), and sample size (n), we can calculate the test statistic (t) using the following formula: t = \(($$ \bar{x} - \mu $$)/($$ \frac{s}{\sqrt{n}} $$)\) Where: \(\bar{x}\) = 15.7 (sample mean) 渭 = 16 (claimed mean weight) s = 0.5 (sample standard deviation) n = 9 (sample size) t = (15.7 - 16) / (0.5 / \(\sqrt{9}\)) = -1.8

Since this is a one-tailed t-test, we will determine the critical value based on the desired significance level (伪). We can use a common 伪 value of 0.05 for this example. With a sample size of 9, we have 8 degrees of freedom (df = n - 1). Using a t-distribution table or calculator, we find the critical value (t伪) for a one-tailed test with 伪 = 0.05 and df = 8: t伪 = -1.860 The rejection region is the set of all values of the test statistic that would lead us to reject the null hypothesis. For a one-tailed test with the alternative hypothesis stating that the true mean is less than the claimed value, we reject the null hypothesis if the test statistic is less than the critical value: Rejection region: t < -1.860

We will now compare the test statistic (t) to the critical value (t伪) in order to make a decision about the null hypothesis: t = -1.8 t伪 = -1.860 Since our test statistic (-1.8) is greater than the critical value (-1.860), we fail to reject the null hypothesis (H鈧�). In conclusion, there is not sufficient evidence to support the claim that the mean weight of the cans is less than the claimed weight on the label.