The sodium ion level in a urine specimen was measured using an ion-selective electrode. The following values were obtained: 102, 97, 99, 98, 101, 106 mM. Calculate the mean and standard deviation of t

The mean is the sum of all values divided by the number of values.

$$\bar{x} = \frac{\sum x_i}{n}$$ $$\bar{x} = \frac{102 + 97 + 99 + 98 + 101 + 106}{6}$$ $$\bar{x} = \frac{603}{6}$$ $$\bar{x} = 100.5 \text{ mM}$$

The formula for the sample standard deviation is:

$$s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}$$

First, calculate $(x_i - \bar{x})^2$ for each data point:

• $(102 - 100.5)^2 = (1.5)^2 = 2.25$
• $(97 - 100.5)^2 = (-3.5)^2 = 12.25$
• $(99 - 100.5)^2 = (-1.5)^2 = 2.25$
• $(98 - 100.5)^2 = (-2.5)^2 = 6.25$
• $(101 - 100.5)^2 = (0.5)^2 = 0.25$
• $(106 - 100.5)^2 = (5.5)^2 = 30.25$ Now, sum these values:

$$\sum (x_i - \bar{x})^2 = 2.25 + 12.25 + 2.25 + 6.25 + 0.25 + 30.25 = 53.5$$

Substitute into the standard deviation formula:

$$s = \sqrt{\frac{53.5}{6-1}} = \sqrt{\frac{53.5}{5}} = \sqrt{10.7}$$ $$s \approx 3.271 \text{ mM}$$

The formula for the confidence limit (CL) is:

$$\text{CL} = \bar{x} \pm \frac{t_s \cdot s}{\sqrt{n}}$$

For 95% confidence, $t_s = 2.57$.

$$\text{CL}_{95\%} = 100.5 \pm \frac{2.57 \times 3.271}{\sqrt{6}}$$ $$\text{CL}_{95\%} = 100.5 \pm \frac{8.41547}{2.44949}$$ $$\text{CL}_{95\%} = 100.5 \pm 3.435$$ $$\text{CL}_{95\%} = (100.5 - 3.435, 100.5 + 3.435)$$ $$\text{CL}_{95\%} = (97.065, 103.935) \text{ mM}$$

For 99% confidence, $t_s = 4.03$.

$$\text{CL}_{99\%} = 100.5 \pm \frac{4.03 \times 3.271}{\sqrt{6}}$$ $$\text{CL}_{99\%} = 100.5 \pm \frac{13.17813}{2.44949}$$ $$\text{CL}_{99\%} = 100.5 \pm 5.380$$ $$\text{CL}_{99\%} = (100.5 - 5.380, 100.5 + 5.380)$$ $$\text{CL}_{99\%} = (95.120, 105.880) \text{ mM}$$

i) Mean and Standard Deviation:
Mean ($\bar{x}$): $\boxed{\text{100.5 mM}}$
Standard deviation ($s$): $\boxed{\text{3.271 mM}}$

ii) 95% and 99% Confidence Limits:
95% Confidence Limit: $\boxed{\text{(97.065, 103.935) mM}}$
99% Confidence Limit: $\boxed{\text{(95.120, 105.880) mM}}$