A company claims that their new energy drink increases the average energy level of an individual by 1.5 points on a scale of 10. To test this claim, a random sample of 36 individuals are selected and given the energy drink. The average increase in energy level for these individuals was found to be 1.7 with a standard deviation of 0.5. Can we reject the company's claim at a 5% level of significance?

Step 1 ：Step 1: Set up the Hypotheses. The null hypothesis, \(H_0\), is that the mean increase in energy level is 1.5. The alternative hypothesis, \(H_1\), is that the mean increase in energy level is not 1.5. So, we have: \(H_0: \mu = 1.5\) and \(H_1: \mu \neq 1.5\).

Step 2 ：Step 2: Calculate the Test Statistic. The test statistic for a sample mean is given by \(Z = \frac{\bar{X} - \mu_0}{\sigma / \sqrt{n}}\), where \(\bar{X}\) is the sample mean, \(\mu_0\) is the mean under the null hypothesis, \(\sigma\) is the standard deviation of the population, and \(n\) is the sample size. Substituting the given values, we have \(Z = \frac{1.7 - 1.5}{0.5 / \sqrt{36}} = 2.4\).

Step 3 ：Step 3: Determine the Rejection Region. For a two-tailed test at a 5% level of significance, the critical values are \(Z = -1.96\) and \(Z = 1.96\). So, we reject the null hypothesis if \(Z < -1.96\) or \(Z > 1.96\).

Step 4 ：Step 4: Make the Decision. Since our calculated test statistic \(Z = 2.4\) falls in the rejection region, we reject the null hypothesis.