\[ \begin{array}{ll} H_{0}: \mu= & \text { T } 5496 \text { not daim } \\ H_{1}: \mu< & \text { T } 5496 \text { daim } \end{array} \] This hypothesis test is a \( \square \) one-tailed test. Espablat \( \infty \) Part: \( 1 / 5 \) Part 2 of 5 (b) Find the critical value(s). Round the answer(s) to at least three decimal places. If there is more than one critical value, separate them with commas, Critical value(s): \( \square \) 2.467

We start with the hypotheses   H₀: μ = (T 5496 not daim)   H₁: μ < (T 5496 daim) so the test is a one‐tailed (left‐tailed) test because the alternative indicates a value less than the null. When working with a left‐tailed test, the rejection region is found in the lower tail of the test statistic’s distribution. (For example, if you are using the t–distribution, the critical value is the t–value corresponding to a cumulative probability equal to the significance level α.) If the answer “2.467” appears in your answer box it is important to check the direction. Because the test is left‐tailed, the critical value must be negative. That is, if your lookup or software provided t₍α,d.f.₎ = 2.467 then the critical value you use for a left–tailed test is –2.467. Thus, your answer is:  Critical value(s): –2.467 In words, if your test statistic is less than –2.467 you would reject the null hypothesis in favor of the alternative hypothesis.