Big babies: The National Health Statistics Reports described a study in which a sample of 310 one-year-old baby boys were weighed. Their mean weight was 24.4 pounds with standard deviation 5.3 pounds. A pediatrician claims that the mean weight of one-year-old boys is less than 25 pounds. Do the data provide convincing evidence that the pediatrician's claim is true? Use the $\alpha=0.10$ level of significance and the critical value method with the Critical Values for the Student's $t$ Distribution Table. Part: 0 / 5 Part 1 of 5 (a) State the appropriate null and alternate hypotheses. \[ \begin{array}{l} H_{0}: \mu=25 \\ H_{1}: \mu<25 \end{array} \] This hypothesis test is a left-tailed $\quad$ test. Part: $1 / 5$ Part 2 of 5 Find the critical value(s). Round the answer(s) to three decimal places. If there is more than one critical value, separate them with commas. Critical value(s):

Step 1 ：(a) State the appropriate null and alternate hypotheses. The null hypothesis \(H_{0}: \mu=25\) states that the mean weight of one-year-old boys is 25 pounds. The alternate hypothesis \(H_{1}: \mu<25\) states that the mean weight of one-year-old boys is less than 25 pounds. This hypothesis test is a left-tailed test.

Step 2 ：Find the critical value(s). The critical value is the point (or points) on the scale of the test statistic beyond which we reject the null hypothesis. It is typically determined by the level of significance (alpha). In this case, alpha is 0.10. Since this is a left-tailed test, we need to find the t-value such that the area to the left of it under the t-distribution curve is 0.10. The degrees of freedom for this test is n-1, where n is the sample size. Here, n=310, so the degrees of freedom is 309.

Step 3 ：The critical value for this left-tailed test with alpha=0.10 and degrees of freedom=309 is approximately -1.284. This means that if our test statistic is less than -1.284, we will reject the null hypothesis.

Step 4 ：Final Answer: The critical value is \(\boxed{-1.284}\).