Question

How many 2.8-g samples must be analyzed to give 95% confidence that the mean is known to within ±4%?

Short answer

6 no. of 2.8-g samples must be analyzed to give 95% confidence and the mean is known to within ±4.

Step-by-step solution

Assumption

From table 28-4 it can be determined that the value of sampling constant (Ks=mR²) is approximately constant for large samples, but this phenomenon is not true for the smallest sample. Therefore, for fig 28-3 we assign K_s =36 g. This is the average from the 1.3g and 5.8g samples in Table 28-4. Therefore, in this case for 2.8 g samples we can assume K_s =36.

Let,

K_s= Sampling constant

m = mass of particle

R= relative standard deviation

s_s = Standard deviation

e = mean

n= number of particles or elements

Determine the number of samples

$$\begin{gathered} K_s=m{R}^2 \\ {R}^2=\frac{K_s}{m} \\ {R}^2=\frac{36}{2.8} \\ R=\pm 3.6\% \end{gathered}$$

For this case standard deviation should be ±3.6%

A 2.8g sample gives standard deviation (s_s=3.6%) and expecting a mean value of (e=4%). Both the uncertainties must be expressed in relative form.

$n=\frac{t^2\times {s_s}^2}{e^2}$

Assume t=1.960 (Table 4-4 for 95% confidence interval and infinite degree of freedom)

$n=\frac{{\left(1.960\right)}^2{\left(0.036\right)}^2}{{\left(0.04\right)}^2}=3.1\approx 3$

For n=3 there will be 2 degrees of freedom. So, for second trial from table 4-4 we can assume students t value 4.303.

$n=\frac{{\left(4.303\right)}^2{\left(0.036\right)}^2}{{\left(0.04\right)}^2}=14.99\approx 15$

For n=15 there will be 14 degrees of freedom. So, for third trial from table 4-4 we can assume students t value 2.150

$n=\frac{{\left(2.15\right)}^2{\left(0.036\right)}^2}{{\left(0.04\right)}^2}=3.744\approx 4$

For n=4 there will be 3 degrees of freedom. So, for fourth trial from table 4-4 we can assume students t value 3.182.

$n=\frac{{\left(3.182\right)}^2{\left(0.036\right)}^2}{{\left(0.04\right)}^2}=8.20\approx 8$

For n=8 there will be 7 degrees of freedom. So, for fifth trial from table 4-4 we can assume students t value 2.365.

$n=\frac{{\left(2.365\right)}^2{\left(0.036\right)}^2}{{\left(0.04\right)}^2}=4.5\approx 5$

For n=5 there will be 4 degrees of freedom. So, for sixth trial from table 4-4 we can assume students t value 2.776.

$n=\frac{{\left(2.776\right)}^2{\left(0.036\right)}^2}{{\left(0.04\right)}^2}=6.24\approx 6$

For n=6 there will be 5 degrees of freedom. So, for seventh trial from table 4-4 we can assume students t value 2.571.

$n=\frac{{\left(2.571\right)}^2{\left(0.036\right)}^2}{{\left(0.04\right)}^2}=5.5\approx 6$

The trial reaches a constant value near n=6. Therefore, 6 no. of 2.8-g samples must be analyzed to give 95% confidence and the mean is known to within ±4.