Chapter 1, Problem 1.10QAP

Interpretation Introduction

(a)

Interpretation:

The least-squares estimates of the slope and intercept for the given data set should be calculated.

Concept introduction:

The least-squares estimates of a set of data behaving linearly are defined as follows:

$y=mx+b$

$\text{When }y\text{ and }x\text{ are variables, }m\text{ is the slope of the line and }b\text{ is the }y\text{ intercept}\text{.}$

$m$ can be calculated using:

$\begin{array}{l}m=\frac{S_{xy}}{S_{xx}}\\ \text{where,}\\ S_{xy}={\displaystyle \sum x_i}{y}_i-\frac{{\displaystyle \sum x_i{\displaystyle \sum y_i}}}{N}\text{ and }{S}_{xx}={\displaystyle \sum x_i^2-\frac{({\displaystyle \sum x_i{)}^2}}{N}}\end{array}$

$\begin{array}{l}\text{Similarly, }b\text{ can be calculated using:}\\ b=\overline{y}-m\overline{x}\\ \text{where,}\\ \overline{y}={\displaystyle \sum \frac{y_i}{N}}\text{ and }\overline{x}={\displaystyle \sum \frac{x_i}{N}}\\ (x_i\text{ and }{y}_i\text{ are individual pairs of data for x and y and }N\text{ is the number of data pairs)}\end{array}$

Answer to Problem 1.10QAP

$\begin{array}{l}m~0.0701\\ b~0.0083\end{array}$

Explanation of Solution

Given:

First, the data should be utilized to obtain $x_i^2\text{, }{y}_i^2\text{, and }{x}_i{y}_i$ values and their sums (appear in the raw after the dark line) as shown in the table below.

$\begin{array}{lllll}\text{Glucose Concentration, mM (}{x}_i)\hfill & \text{Absorbance }(y_i)\hfill & x_i^2\hfill & y_i^2\hfill & x_i{y}_i\hfill \\ 0.0\hfill & 0.002\hfill & 0\hfill & 0.000004\hfill & 0\hfill \\ 2.0\hfill & 0.150\hfill & 4\hfill & 0.0225\hfill & 0.3\hfill \\ 4.0\hfill & 0.294\hfill & 16\hfill & 0.086436\hfill & 1.176\hfill \\ 6.0\hfill & 0.434\hfill & 36\hfill & 0.188356\hfill & 2.604\hfill \\ 8.0\hfill & 0.570\hfill & 64\hfill & 0.3249\hfill & 4.56\hfill \\ 10.0\hfill & 0.704\hfill & 100\hfill & 0.495616\hfill & 7.04\hfill \\ 30.0\hfill & 2.154\hfill & 220\hfill & 1.117812\hfill & 15.68\hfill \end{array}$

How to calculate $m$:

The above calculated values in the table are used to calculate $S_{xy}\text{, and }{S}_{xx}$ as shown below:

$\begin{array}{l}{S}_{xy}={\displaystyle \sum x_i}{y}_i-\frac{{\displaystyle \sum x_i{\displaystyle \sum y_i}}}{N}\\ S_{xy}=15.68-\frac{30.0\times 2.154}{6}=4.91\\ S_{xx}={\displaystyle \sum x_i^2-\frac{({\displaystyle \sum x_i{)}^2}}{N}}\\ S_{xx}=220-\frac{{(30.0)}^2}{6}=70\\ \text{Now, let's calculate }m\text{ as follows:}\\ m=\frac{S_{xy}}{S_{xx}}=\frac{4.91}{70}=0.0701428571\sim 0.0701\end{array}$

How to calculate $b$:

First, use the data in the above table to obtain $\overline{y}\text{ and }\overline{x}$ as shown below:

$\begin{array}{l}\overline{y}={\displaystyle \sum \frac{y_i}{N}}\\ \overline{y}=\frac{2.154}{6}=0.359\\ \overline{x}={\displaystyle \sum \frac{x_i}{N}}=\frac{30.0}{6}=5\\ \text{Now, }b\text{ can be calculated as follows:}\\ b=\overline{y}-m\overline{x}=0.359-0.0701\times 5=0.0082857143\sim 0.0083\end{array}$

Interpretation Introduction

(b)

Interpretation:

Standard deviations of the slope and intercept and the standard error of the estimates for the given data set should be calculated using the LINEST function in EXCEL.

Concept introduction:

In order to use the LINEST function in EXCEL to obtain statistical values for a given data set, the following command should be typed on an EXCEL spread sheet, in which the given data is already been typed in two columns.

$\text{=(known\_}y\text{s,[known\_}x\text{s],[constant],[stats])}$

Answer to Problem 1.10QAP

$\begin{array}{l}\text{Standard deviation of the slope (}{S}_m\text{)}=0.0007\\ \text{Standard deviation of the intercept (}{S}_b\text{)}=0.004\\ \text{Standard error of the estimates (}{S}_r\text{) = 0}\text{.0056}\end{array}$

Explanation of Solution

First, the data set should be entered in to an EXCEL spread sheet (See the figure below).

Then on a different cell, type the following command (See figure below). In order to enter the cell numbers of the unknown $x\text{ and }y$ data, select the area where the data have entered.

$\text{=LINEST(A2:A7,B2:B7,,TRUE)}$

Once the formula command is written and the parenthesis is closed, then hit the ENTER button. Now, the value of $m$ will be generated as shown in the figure below.

Now, select few more cells including the cell in which the command is typed and then click on the cursor bar (See below).

Then hit $\text{CTRL+SHIFT+ENTER}$. Several numerical values will be generated as shown below.

Each of these numerical values is relevant to a statistical data (mentioned in BOLD, See below). Now pick the apocopate answers required.

Interpretation Introduction

(c)

Interpretation:

The $95\%$ confidence intervals for the slope and the intercept should be determined.

Concept introduction:

The general formula to calculate the $95\%$ confidence intervals for a statistic relevant to a regression analysis is as follows:

$\begin{array}{l}\text{95\% CI for a given statistic}=\\ \text{staistic value}\pm \text{ critical value for the confidence interval}\times \text{standard error of the statistic}\\ \text{For slope (}m\text{), the equation can be rewritten as,}\\ \text{95\% CI}=m\pm t\times S_m\\ \text{For intercept (}b\text{), the equation can be rewritten as,}\\ \text{95\% CI}=b\pm t\times S_b\end{array}$

Answer to Problem 1.10QAP

$\begin{array}{l}\text{95\% CI for }m=0.070\pm 0.002\\ \text{95\% CI for }b=0.08\pm 0.01\end{array}$

Explanation of Solution

$\begin{array}{l}\text{95\% CI}=m\pm t\times S_m\\ \text{95\% CI}=b\pm t\times S_b\end{array}$

The t value of the above equation depends on the confidence level (95 % in this question) and the degrees of freedom of the data set.

The degrees of freedom for a regression analysis = $N-2$. Therefore, in this question, the degrees of freedom = 4.

Once the degree of freedom is calculated the t value can be extracted from a standard table similar to the table shown below:

Based on the table, when the degree of freedom is 4 and the cofidence interval is 95 %, the t value is $\text{2}\text{.776 (}\sim \text{2}\text{.78)}$.

$\begin{array}{l}\text{For slope (}m\text{),}\\ \text{95\% CI}=m\pm t\times S_m=0.0701\pm 2.78\times 0.0007=0.0701\pm \text{0}\text{.001946}\\ \text{95\% CI for }m=0.0701\pm 0.002\\ \\ \text{For intercept (}b\text{),}\\ \text{95\% CI}=b\pm t\times S_b=0.0083\pm 2.78\times 0.004=0.0083\pm 0.01112\\ \text{95\% CI for }b=0.08\pm 0.01\end{array}$

Interpretation Introduction

(d)

Interpretation:

The glucose concentration and its standard deviation of a serum sample, which gave an absorbance of 0.350 should be calculated.

Concept introduction:

To determine an unknown glucose concentration $c$ from the least-squares line, the valueof the absorbancey, the slope ( $m$ ) and intercept ( $b$ ) can be used as follows,

$c_x=\frac{y_c-b}{m}$

The standard deviation of $c$ ( $S_c$ ) is calculated by,

$\begin{array}{l}{S}_c=\frac{S_y}{m}\sqrt{\frac{1}{M}+\frac{1}{N}+\frac{{({\overline{y}}_c-\overline{y})}^2}{m^2{S}_{xx}}}\\ \text{Where,}\\ M=\text{number of replicate results}\\ N=\text{number of standards}\\ {\overline{y}}_c=\text{mean response for the unknown}\\ S_y,\overline{y},\text{ and }{S}_{xx}\text{ were defined in early steps}\text{.}\end{array}$

Answer to Problem 1.10QAP

$\begin{array}{l}{c}_x=0.487\text{ mM}\\ S_c=0.09\text{ mM}\end{array}$

Explanation of Solution

Plug in the appropriate numerical values in the below equation to determine the $c_x$ value.

$c_x=\frac{y_c-b}{m}=\frac{0.350-0.0083}{0.0701}=4.87446505\sim 4.87$

How to calculate $S_c$:

$\begin{array}{l}\text{In the given question,}\\ M=1\text{, because only one data has obtained}\\ N=\text{6}\\ {\overline{y}}_c=y_c=0.350,\text{bacuase only one data has obtained}\\ S_y,\overline{y},\text{ and }{S}_{xx}\text{ were calculated in early steps}\text{.}\\ \text{Now, plug in the values in the below equation to calculate the standard deviation of the glucose concentration}\\ \text{ of the serum sample}\text{.}\\ \\ S_c=\frac{S_r}{m}\sqrt{\frac{1}{M}+\frac{1}{N}+\frac{{({\overline{y}}_c-\overline{y})}^2}{m^2{S}_{xx}}}=\frac{0.0056}{0.0701}\sqrt{\frac{1}{1}+\frac{1}{6}+\frac{{(0.350-0.359)}^2}{{0.0701}^{2}70}}=0.0862953169\sim 0.09\end{array}$