Respuesta :
Answer:
Step-by-step explanation:
Given that a group of students estimated the length of one minute without reference to a watch or clock, and the times (seconds) are listed below.
Data set is as ollows:
70 79 38 63 44 23 62 61 67 50 61 70 94 87 65
[tex]Mean = 62.27Variance =333.35std dev = 18.258std error = 4.714[/tex]
H0: mu = 60 sec
Ha: mu not equals 60 sec
Mean diff = 2.27
[tex]Mean = 62.27Variance =333.35std dev = 18.258std error = 4.714[/tex]
Test statistic = 2.27/SE =0.4815
p value =0.6376
Since p>0.05, we accept null hypothesis
i.e. there is statistical evidence to say that students are reasonably good at estimating one minute
Answer:
By the statistical solution below,It appears that students are reasonably good at estimating one minute.
Step-by-step explanation:
Given information
A group of student estimated the length of one minute
Sample data [tex](X)[/tex] = 70,79,38,63,44,23,62,61,67,50,61,70,94,87,65
Sample size [tex](n)[/tex] = 15
Now, Mean ([tex]X'[/tex]) = [tex](\sum X)/n[/tex]
[tex]X'=934/15\\X'=62.27[/tex]
Now calculate Variance [tex](v)[/tex] = [tex]v =\frac{\sum(X_i-X')^2}{n-1}[/tex]
on putting the value in above equation and solving
We get,
[tex]v=333.35[/tex]
Now we know that standard deviation [tex]\sigma=\sqrt{v}[/tex]
[tex]\sigma=\sqrt333.35\\\sigma=18.25\\[/tex]
Mean difference:
[tex]= 62.27-60.00\\=2.27\\[/tex]
Test statistics :
[tex]t=2.27/(\sqrt18.25)\\t= 0.4815\\[/tex]
For the test statistics the [tex]p[/tex] value is 0.6376
Since [tex]p>0.05[/tex] , We accept null hypothesis
That means , It appears that students are reasonably good at estimating one minute.
For more information visit
https://brainly.com/question/12808804?referrer=searchResults
