The CdSe content (g/L) of six different samples of nanocrystals was measured by two methods. Do the two methods differ significantly at the 95 % confidence level?

⦁ The standard deviation (SD), like the average deviation, is a measure of spread around the mean.

⦁ Standard deviation is defined as the average deviation of data from the mean. To put it another way, the standard deviation is about the same as the average deviation.

We'll utilise Case 3 of the t test, often known as the paired t test, because the challenge is comparing two approaches with several samples each. Below is a tabular representation of the supplied values, plus an additional column to indicate individual differences $\left({\mathrm{d}}_{\mathrm{i}}\right)$

Before we start working on the problem, let's have a look at some of the ${\mathrm{t}}_{\mathrm{calce}}$Let's start with the average difference $\left(\overline{\mathrm{d}}\right)$and the difference standard deviation $\left({\mathrm{s}}_{\mathrm{d}}\right)$

$$\begin{gathered} \overline{\mathrm{d}}=\frac{{\mathrm{d}}_{\mathrm{i}}}{\mathrm{n}} \\ =\frac{0.05+0.11+\left(-0.17\right)+0.02+0.09+0.20}{6} \\ =0.05 \end{gathered}$$

$$\begin{gathered} s_d=\sqrt{\frac{{\left(0.05-0.05\right)}^2+{\left(0.11-0.05\right)}^2+{\left(-0.17-0.05\right)}^2{\left(0.02-0.05\right)}^2+{\left(0.20-0.05\right)}^2}{6-1}} \\ =0.1240967365 \end{gathered}$$

We can already compute the cost now ${\mathrm{t}}_{\mathrm{calc}}$ and contrast it with ${\mathrm{t}}_{\mathrm{table}}$ :

$$\begin{gathered} {\mathrm{t}}_{\mathrm{table}}=\frac{\left|\overline{\mathrm{d}}\right|}{{\mathrm{s}}_{\mathrm{d}}}\sqrt{\mathrm{n}} \\ =\frac{\left|0.05\right|}{0.1240967365}\sqrt{6} \\ =0.9869 \end{gathered}$$

${\mathrm{t}}_{\mathrm{table}}=2.571$ (based 4 degrees of freedom on Table 4$-\mathrm{n}-1-6-1-5;95\%$ level of assurance)