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
h₀ μ = 60
h₁ μ ≠ 60
(type integers or decimals. do not round)
identify the test statistic
0.43
(round to two decimal places as needed)
identify the p - value
(round to three decimal places as needed)

Explanation:

Step1: Determine the degrees of freedom

The sample size \(n = 15\). The degrees of freedom \(df=n - 1=15 - 1 = 14\)

Step2: Find the P - value

We have a two - tailed t - test with test statistic \(t = 0.43\) and \(df=14\). Using a t - distribution table or a calculator (e.g., in R: 2*(1 - pt(0.43,14))), we calculate the P - value.

$$P - value=2\times(1 - P(T\leqslant0.43))$$

where \(T\) follows a \(t\) - distribution with \(df = 14\).

$$P - value\approx2\times(1 - 0.664)=2\times0.336 = 0.672\approx0.671$$

(rounded to three decimal places)

Answer:

$0.671$