In analyzing a lot with random sample variation, you find a sampling standard deviation of \({\bf{65}}\% .\)Assuming negligible error in the analytical procedure, how many samples must be analyzed to give \(9{\bf{5}}\% \)confidence that the error in the mean is within\(64\% \)of the true value? Answer the same question for a confidence level of \(90\% \).

• 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.

In this task we have a sampling standard deviation of \(65\% \). We will calculate the number of samples that must be analyzed in order to give \(95\% \) and \(90\% \) confidence that the error in the mean is within \( \pm 4\% \) of the true value.

\(95\% \)Confidence

Here we will use the equation \(28 - 7\) where \({s_s} = 0.05\)and \(e = 0.04\) :

\(\begin{aligned}{}n &= \frac{{{t^2}s_s^2}}{{{e^2}}}\\n &= \frac{{{{1.960}^2} \times {{0.05}^2}}}{{{{0.04}^2}}}\\n &= 6\end{aligned}\)

For \(n = 6.\) there are \({5^\circ }\) of freedom so \(t = 2.571\) which would give us the following value of \(n\) :

\(\begin{aligned}{}n &= \frac{{{t^2}s_8^2}}{{{e^2}}}\\n &= \frac{{{{2.571}^2} \times {{0.05}^2}}}{{{{0.04}^2}}}\end{aligned}\)

\(n = 10.3\)

For \(n = 10.3\) there are \({9^\circ }\) of freedom so \(t = 2.262\) which would give us the following value of \(n\) :

\(\begin{aligned}{}n& = \frac{{{t^2}s_s^2}}{{{e^2}}}\\n &= \frac{{{{2.262}^2} \times {{0.05}^2}}}{{{{0.04}^2}}}\\n &= 8\end{aligned}\)

Find that for \(t = 2.365\) the \(n = 8.74\) so in this case we would use a total of \(8\) samples

\(90\% \)Confidence:

Here we will also use the equation 28-7 where \({s_s} = 0.05\)and \(e = 0.04\) :

\(\begin{aligned}{}n&= \frac{{{t^2}s_s^2}}{{{e^2}}}\\n &= \frac{{{{1.645}^2} \times {{0.05}^2}}}{{{{0.04}^2}}}\\n &= 6\end{aligned}\)

For \(n = 6\) there are \({5^\circ }\) of freedom so \(t = 2.015\) which would give us the following value of \(n\) :

\(\begin{aligned}{}n &= \frac{{{t^2}s_s^2}}{{{e^2}}}\\n &= \frac{{{{2.015}^2} \times {{0.05}^2}}}{{{{0.04}^2}}}\\n &= 6\end{aligned}\)

When,\(t = 1.833\)then \(n = 5.25\),so in this case we would use a total of \(6\)samples.

Therefore, the samples in \(95\% \) Confidence is \(n = 6,10.3,8\) and the samples in \(90\% \) Confidence is\(n = 6,6,5.25\).