Question

In an experiment analogous to that in Figure 28-3, the sampling constant is found to be \({K_{\rm{s}}} = 20\;{\rm{g}}.\)

(a) What mass of sample is required for a \( \pm 2\% \)sampling standard deviation?

(b) How many samples of the size in part (a) are required to produce \(90\% \)confidence that the mean is known to within\(1.5\% \)?

Short answer

(a) The mass of the sample for the given standard deviation is\(5\;{\rm{g}}\).

(b) In this case we should use \(7\) samples,\(n = 4.8,8.1,6.4,7.2,6.7\)

Step-by-step solution

Definition of Standard deviation.

• The standard deviation is a measure of how far something deviates from the mean (for example, spread, dispersion, or spread). A "typical" variation from the mean is represented by the standard deviation.
• Because it returns to the data set's original units of measurement, it's a common measure of variability.
• The standard deviation, defined as the square root of the variance, is a statistic that represents the dispersion of a dataset relative to its mean.

Determine the mass of the sample

a)

Calculate the mass of sample with the following equation

\(\begin{array}{l}m{R^2} = {K_S}\\m = \frac{{{K_S}}}{{{R^2}}}\\m = \frac{{20\;{\rm{g}}}}{{{2^2}}}\\m = 5\;{\rm{g}}\end{array}\)

Hence, the mass of the sample for the given standard deviation is \(5\;{\rm{g}}\).

Calculate the number of samples.

b)

Use the equation 28-7 with \({s_s} = 0.02\) and \(e = 0.015\), the initial value of \(t\) is found on the Pg.72 Table \(4 - 4\) for confidence \(90\% \) :

\(\begin{array}{l}n = \frac{{{t^2}s_s^2}}{{{e^2}}}\\n = \frac{{{{1.645}^2} \times {{0.02}^2}}}{{{{0.015}^2}}}\\n = 4.8\end{array}\)

Next, for \(n = 5\) there are \({4^^\circ }\) of freedom so \(t = 2.132\) which gives the following value of \(n\) :

\(n = \frac{{{t^2}s_s^2}}{{{e^2}}}\)

\(\begin{array}{l}n = \frac{{{{2.132}^2} \times {{0.02}^2}}}{{{{0.015}^2}}}\\n = 8.1\end{array}\)

Then if we use in the same equation the \({7^^\circ }\) of freedom \(t = 1.895\) this would give us the following value of \(n\) :

\(\begin{array}{l}n = \frac{{{t^2}s_s^2}}{{{e^2}}}\\n = \frac{{{{1.895}^2} \times {{0.02}^2}}}{{{{0.015}^2}}}\\n = 6.4\end{array}\)

Then with \(t = 2.015 \to n = 7.2\) and \(t = 1.943 \to n = 6.7\)So, we would use a total of \(7\) samples.

Therefore, in this case we should use \(7\) samples,\(n = 4.8,8.1,6.4,7.2,6.7\)