Testing Hypotheses. In Exercises 13–24, assume that a simple random sample has been selected and test the given claim. Unless specified by your instructor, use either the P-value method or the critical value method for testing hypotheses. Identify the null and alternative hypotheses, test statistic, P-value (or range of P-values), or critical value(s), and state the final conclusion that addresses the original claim.

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

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.

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 thestudents who estimated the length of one minuteis 15.

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,

$$\begin{gathered} H_0:\mu =60 \\ {H}_1:\mu \ne 60 \end{gathered}$$

The test is two-tailed.

Formula for test statistic is given by,

$t=\frac{\overline{x}-\mu }{\frac{s}{\sqrt{n}}}$

Where , $\overline{x}$is the sample mean and s is the standard deviation of the sample and is the value of population mean which is used in null hypothesis.

Using the given data, the mean is computed as,

$$\begin{gathered} \overline{x}=\frac{\sum x_i}{n} \\ =\frac{69+81+39+...+65}{15} \\ =62.67 \end{gathered}$$

The sample standard deviation is computed as,

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

By substituting this values, test statistics is given by,

$$\begin{gathered} t=\frac{\overline{x}-\mu }{\frac{s}{\sqrt{n}}} \\ =\frac{62.66-60}{\frac{19.48}{\sqrt{15}}} \\ =0.5301 \end{gathered}$$

The degree of freedom is,

$$\begin{gathered} \text{df}=n-1 \\ =15-1 \\ =14 \end{gathered}$$

In the t-distribution table, find the value corresponding 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 ${\text{t}}_{\text{0.025}}$therefore use 2.145 as a critical value.

Thus, the critical value role="math" localid="1649067349226" ${\text{t}}_{\text{0.025}}$ is 2.145.

The rejection region is $\left(t:t>2.145\text{and}t<-2.145\right)$.

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.

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.