In: Math
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 0.01 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?
72 79 42 66 41 23 63 65 67 48 65 72 96 88 68
Assuming all conditions for conducting a hypothesis test are met, what are the null and alternative hypotheses?
Solution:
x | x2 |
72 | 5184 |
79 | 6241 |
42 | 1764 |
66 | 4356 |
41 | 1681 |
23 | 529 |
63 | 3969 |
65 | 4225 |
67 | 4489 |
48 | 2304 |
65 | 4225 |
72 | 5184 |
96 | 9216 |
88 | 7744 |
68 | 4624 |
∑x=955 | ∑x2=65735 |
Mean ˉx=∑xn
=72+79+42+66+41+23+63+65+67+48+65+72+96+88+68/15
=955/15
=63.6667
Sample Standard deviation S=√∑x2-(∑x)2nn-1
=√65735-(955)215/14
=√65735-60801.6667/14
=√4933.3333/14
=√352.381
=18.7718
This is the two tailed test .
The null and alternative hypothesis is ,
H0 : μ = 60
Ha : μ ≠ 60
Test statistic = t
= (x̅- μ ) / S / √ n
= (63.67-67) / 18.77 / √ 15
= 0.757
Test statistic = t = 0.757
P-value =0.4614
α = 0.01
P-value > α
0.4614 > 0.01
Fail to reject the null hypothesis .
There is not sufficient evidence to conclude that the original claim that the mean of the population of the estimate is 60 seconds equal to it that of a group the students who are reasonably good at estimating one minute.