Question

Six measurements were made of the time in seconds 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.The standard error is

Answer 1

Answer:s=σ/sqrt(n)=0.4/sqrt(6)=0.16s=σ/sqrt(n)=0.4/sqrt(6)=0.16

Explanation:

The mean value:

T=(20.1+20.2+20.3+20.1+20.0+20.2)/6=20.2T=(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.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)=0.16s=σ/sqrt(n)=0.4/sqrt(6)=0.16The mean value:

T=(20.1+20.2+20.3+20.1+20.0+20.2)/6=20.2T=(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.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)=0.16s=σ/sqrt(n)=0.4/sqrt(6)=0.16

Answer 2

Answer:

Approx = 0.05 = 5×10⁻²

Explanation:

Mean = (20.1 + 20.2 + 20.3 + 20.1 + 20.0 + 20.2) ÷ 6

= 120.9 ÷ 6

= 20.15

Standard Deviation(SD) = √[ ∑(X − μ)² / N]

$\sqrt{x} 1/(6 - 1) (20.1 - 20.1)² + (20.1 - 20.3)² + (20.1 - 20.3)²$

i.e √1/(6 - 1) (20.1 - 20.1)² + (20.1 - 20.3)² + (20.1 - 20.3)²

= √1/5(0.0001 + 0.04 + 0.04)

= √1/5(0.0801)

= √0.01602

= 0.1266

SD = 0.1266

Standard Error = SD/√(n)

Where (n) = number of samples = 6

Standard Error = SD/√6

Standard Error = 0.1266/√6

Standard Error = 0.05168 ≅ 5.2×10⁻²