A 90 percent confidence level is equivalent to an alpha level of 0.10. Because extreme values in one rather than two directions will lead to rejection of the null hypothesis, this is a one‐tailed test, and you do not divide the alpha level by 2. The number of degrees of freedom for the problem is 6 – 1 = 5. The value in the t‐table for t .10,5 is 1.476. Because the computed t‐value of 1.71 is larger than the critical value in the table, the null hypothesis can be rejected, and the professor has evidence that the class mean on the math test would be at least 70.