Answer to Question #83931 in Mechanics | Relativity for Victoria agwu

Question

Six measurements were made of the time for 20 oscillations of a mass hung from a spiral spring as follows: 20.1,20.2,20.3,20.1,20.0,20.2
Find the mean value of the period of oscillation and the standard error in this value

Answer 1

The mean value:

$$T=(20.1+20.2+20.3+20.1+20.0+20.2)/6=20.2$$

To calculate the standard error, first calculate the squared deviations of each measurement from the mean:

(20.1-20.2)^2=0.01

(20.2-20.2)^2=0

(20.3-20.2)^2=0.01

(20.1-20.2)^2=0.01

(20.0-20.2)^2=0.04

(20.2-20.2)^2=0

The standard deviation:

$$σ=sqrt[(0.01+0+0.01+0.01+0.04+0)/6]=0.4.$$

The standard error:

$$s=σ/sqrt(n)=0.4/sqrt(6)=0.16$$