Question

A group of students 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?

71

79

36

65

43

26

57

66

69

46

61

72

95

88

65

1. what are the null and alternative hypotheses?

2. Determine the test statistic.

3. Determine the P- value

Answer 1

Q1) As we are testing here whether times are from a population with a mean equal to 60 seconds, therefore the null and the alternative hypothesis here are given as:

$H_0: \mu = 60$

097 11:11

Q2) The computations are made using the following table here:

X	(X - Mean(X))^2
	71	70.56
79	268.96
36	707.56
65	5.76
43	384.16
26	1339.56
57	31.36
66	11.56
69	40.96
46	275.56
61	2.56
72	88.36
95	1049.76
88	645.16
65	5.76
939	4927.6

The sample mean and sample standard deviation here are computed as:

$\bar X = \frac{\sum X_i}{n} = \frac{939}{15} = 62.6$

$s = \sqrt{\frac{\sum (X_i - \bar X)^2}{n - 1}} = \sqrt{\frac{4927.6}{15-1}} = 18.7609$

The test statistic here is computed as:

$t^* =\frac{\bar X - \mu_0}{\frac{s}{\sqrt{n}}} = \frac{62.6 - 60}{\frac{18.7609}{\sqrt{15}}} = 0.5367$

Therefore 0.5367 is the test statistic value here.

Q3) For n - 1 = 14 degrees of freedom, the p-value here is obtained from t distribution tables as:
p = 2P( t₁₄ > 0.5367) = 2*0.2999 = 0.5998

Therefore 0.5998 is the required p-value here.