Students estimated the length of one minute without reference to a watch or clock, and the times (seconds) are listed below. Assume that a simple random sample has been selected. Use a 0.05 significance level to test the claim that these times are from a population with a mean equal to 60 seconds. Does it appear that students are reasonably good at estimating one minute?
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Perform the test assuming that the requirements are met. Identify the null and alternative hypotheses.
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(Type integers or decimals. Do not round.)

Step 1 ：State the null hypothesis and the alternative hypothesis. The null hypothesis is that the population mean is equal to 60 seconds. The alternative hypothesis is that the population mean is not equal to 60 seconds. So, $H_0:\mu =60$ and $H_1:\mu \ne 60$.

Step 2 ：Calculate the sample mean and the sample standard deviation. The sample mean is 62.8 and the sample standard deviation is approximately 19.69.

Step 3 ：Calculate the test statistic. The test statistic is calculated as follows: $t=\frac{\text{{sample mean}}-\text{{population mean}}}{\text{{sample standard deviation}}/\sqrt{n}}$, where n is the sample size. The test statistic is approximately 0.55.

Step 4 ：Calculate the critical value at the 0.05 significance level. The critical value is approximately 2.14.

Step 5 ：Compare the test statistic to the critical value. The test statistic is less than the critical value, so we do not reject the null hypothesis. This means that we do not have enough evidence to say that the population mean is not equal to 60 seconds.

Step 6 ：Final Answer: The null hypothesis is $H_0:\mu =60$ and the alternative hypothesis is $H_1:\mu \ne 60$. The test statistic is 0.55 and the critical value is 2.14. Since the test statistic is less than the critical value, we do not reject the null hypothesis. Therefore, it appears that students are reasonably good at estimating one minute. $\boxed{{\displaystyle H_0:\mu =60,H_1:\mu \ne 60,t=0.55,\text{critical value}=2.14}}$.