Question

Got a Minute? Students of the author estimated the length of one minute without reference to a watch or clock, and the times (seconds) are listed below. 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?

69 81 39 65 42 21 60 63 66 48 64 70 96 91 65

Short answer

There is not sufficient evidence to accept the claim that these times are from a population with a mean equal to 60 seconds; and it appears that the students are good at estimating one minute.

Step-by-step solution

Given information

Sample times which the students took to estimate the length of one minute are given (in seconds) 69 81 39 65 42 21 60 63 66 48 64 70 96 91 65.

Check the requirements

The t-distribution would be used here assuming that the population is normally distributed and the samples are randomly selected.

Thus, the sample size (n) of the students who estimated the length of one minute is 15.

Describe the hypothesis

Null hypothesis\(\left( {{H_0}} \right)\)is a statement of claim that thetimes are from a population with a mean equal to 60 seconds.

Alternate hypothesis\(\left( {{H_1}} \right)\)is a statement of claim that thetimes are from a population with a mean is not equal to 60 seconds.

Assume\(\mu \)be the true mean times for the population.

Mathematically, it can be expressed as,

\(\begin{array}{l}{H_0}:\mu = 60{\rm{ }}\\{H_1}:\mu \ne 60\end{array}\)

The test is two-tailed.

Calculate the test statistic

Formula for test statistic is given by,

\(t = \frac{{\bar x - \mu }}{{\frac{s}{{\sqrt n }}}}\)

Where ,\(\bar x\)is the sample mean and s is the standard deviation of the sample and\(\mu \)is the value of population mean which is used in null hypothesis.

Using the given data, the mean is computed as,

\(\begin{array}{c}\bar x = \frac{{\sum {{x_i}} }}{n}\\ = \frac{{69 + 81 + 39 + ... + 65}}{{15}}\\ = 62.67\end{array}\)

The sample standard deviation is computed as,

\(\begin{array}{c}s = \sqrt {\frac{{\sum {{{\left( {{x_i} - \bar x} \right)}^2}} }}{{n - 1}}} \\ = \sqrt {\frac{{{{\left( {69 - 62.67} \right)}^2} + {{\left( {81 - 62.67} \right)}^2} + ... + {{\left( {65 - 62.67} \right)}^2}}}{{15 - 1}}} \\ = 19.48\end{array}\)

By substituting this values, test statistics is given by,

\(\begin{array}{c}t = \frac{{\bar x - \mu }}{{\frac{s}{{\sqrt n }}}}\\ = \frac{{62.66 - 60}}{{\frac{{19.48}}{{\sqrt {15} }}}}\\ = 0.5301\end{array}\)

Calculate the critical value.

The degree of freedom is,

\(\begin{array}{c}{\rm{df}} = n - 1\\ = 15 - 1\\{\rm{ }} = 14\end{array}\)

In the t-distribution table, find the valuecorresponding to the row value of degree of freedom 14 and column value of area in one tail 0.025 is 2.145 which is critical value\({{\rm{t}}_{{\rm{0}}{\rm{.025}}}}\)therefore use 2.145 as a critical value.

Thus, the critical value\({{\rm{t}}_{0.025}}\)is 2.145.

The rejection regionis\(\left( {t:t > 2.145\;{\rm{and}}\;t < - 2.145} \right)\) .

Compare test statistic and critical value.

Test statistic is 0.5301 and the critical values are\( \pm 2.145\).

According to this, we can conclude that the test statistics 0.5301 will not fall in the rejection region.

Therefore, we fail to reject the null hypothesis.

Conclusion

Therefore, there is not sufficient evidence to support the claim that there is significant difference between mean of times from true population and the hypothesis mean.

Thus, students are reasonably good at estimating one minute as the mean time is not significantly different from 60 seconds.