Question

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 0.10 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?
67 80 40 64 41 19 62 61
66 47 64 72 95 91 64
perform the test assuming that the requirements are met. identify the null and alternative hypotheses.
$h_0$ $mu$ = 60
$h_1$ $mu$ $
eq$ 60
(type integers or decimals. do not round.)
identify the test statistic.

• 0.66

(round to two decimal places as needed)

Explanation:

Step1: State the hypotheses

The claim is that the mean of the population is \(60\) seconds. The null hypothesis \(H_0\) is the statement of equality, so \(H_0:\mu = 60\). The alternative hypothesis \(H_1\) is the statement that contradicts the null hypothesis for a two - tailed test, so \(H_1:\mu
eq60\)

Step2: Calculate the test statistic

The formula for the \(t\) - test statistic for a one - sample \(t\) - test is \(t=\frac{\bar{x}-\mu}{s/\sqrt{n}}\)

First, calculate the sample mean \(\bar{x}\):

$$
LATEXBLOCK0
$$

Calculate the sample standard deviation \(s\):

$$
LATEXBLOCK1
$$

$$
LATEXBLOCK2
$$

$$ s=\sqrt{\frac{5418.21}{14}}\approx19.77 $$

Then \(t=\frac{\bar{x}-\mu}{s/\sqrt{n}}=\frac{63.13 - 60}{19.77/\sqrt{15}}\approx\frac{3.13}{5.11}\approx - 0.66\) (rounded to two decimal places)

Answer:

The null hypothesis \(H_0:\mu = 60\) and the alternative hypothesis \(H_1:\mu
eq60\). The test - statistic is \(t=- 0.66\)