Regarding the hypothesis tested in this problem, it is found that:
a) The test statistic is: t = 0.49.
b) The p-value is: 0.6297..
c) The null hypothesis is not rejected, as the p-value is greater than the significance level.
d) It appears that the students are legitimately good at estimating one minute, as the sample does not give enough evidence to reject the null hypothesis.
<h3>What is the test statistic?</h3>
The test statistic of the t-distribution is given by the equation presented as follows:
In which the variables of the equation are given as follows:
is the sample mean.
is the value tested at the hypotheses.
- s is the standard deviation of the sample.
The value tested at the hypothesis is:
From the sample, using a calculator, the other parameters are given as follows:
Hence the test statistic is obtained as follows:
t = 0.49.
<h3>What are the p-value and the conclusion?</h3>
The p-value is obtained using a t-distribution calculator, with a two-tailed test, as we are testing if the mean is different a value, in this case 60, with:
- 15 - 1 = 14 df, as the number of degrees of freedom is one less than the sample size.
- t = 0.49, as obtained above.
Hence the p-value is of 0.6297.
Since the p-value is greater than the significance level of 0.01, the null hypothesis is not rejected, attesting to the ability of the students to estimate one minute.
More can be learned about the test of an hypothesis at brainly.com/question/13873630
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