Chapter A1, Problem A1.8QAP

Interpretation Introduction

(a)

Interpretation:

The following data is obtained from the several food plant preparations for potassium ion-

Samples	${\text{precent K}}^{+}$
1	5.15,5.03,5.04,5.18,5.20
2	7.18,7.17,6.97
3	4.00,3.93,4.15,3.86
4	4.68,4.85,4.79,4.62
5	6.04,6.02,5.82,6.06,5.88

The value of mean and standard deviation for each sample is to be determined.

Concept introduction:

The following formula will be used-

Mean-

$\overline{x}=\frac{x_1+x_2+\mathrm{.........}+x_n}{N}$

Or

$\overline{x}=\frac{{\displaystyle \sum }_{i=1}^N{x}_i}{N}$

Deviation-

$s=\sqrt{\frac{{\displaystyle \sum }_{i=1}^N{x}_i^2-\frac{{\left({\displaystyle \sum }_{i=1}^N{x}_i\right)}^2}{N}}{N-1}}$

Interpretation Introduction

(b)

Interpretation:

The following data is obtained from the several food plant preparations for potassium ion-

Samples	${\text{precent K}}^{+}$
1	5.15,5.03,5.04,5.18,5.20
2	7.18,7.17,6.97
3	4.00,3.93,4.15,3.86
4	4.68,4.85,4.79,4.62
5	6.04,6.02,5.82,6.06,5.88

The value s pooled, $s_{pooled}$ is to be determined.

Concept introduction:

The following formula will be used-

$s_{pooled}=\sqrt{\frac{{\displaystyle \sum }_{i=1}^{N_1}{\left(x_i-{\overline{x}}_1\right)}^2+{\displaystyle \sum }_{i=1}^{N_1}{\left(x_i-{\overline{x}}_1\right)}^2+\mathrm{...}}{N_1+N_2+\mathrm{....}}}$

Interpretation Introduction

(c)

Interpretation:

The following data is obtained from the several food plant preparations for potassium ion-

Samples	${\text{precent K}}^{+}$
1	5.15,5.03,5.04,5.18,5.20
2	7.18,7.17,6.97
3	4.00,3.93,4.15,3.86
4	4.68,4.85,4.79,4.62
5	6.04,6.02,5.82,6.06,5.88

The reason for the $s_{pooled}$ to be a better estimate of $\sigma$ in comparison of standard deviation from any one sample needs to be explained.

Concept introduction:

The following formula will be used −

$s_{pooled}=\sqrt{\frac{{\displaystyle \sum }_{i=1}^{N_1}{\left(x_i-{\overline{x}}_1\right)}^2+{\displaystyle \sum }_{i=1}^{N_1}{\left(x_i-{\overline{x}}_1\right)}^2+\mathrm{...}}{N_1+N_2+\mathrm{....}}}$