Chapter 1, Problem 1.74QP

(a)

Interpretation Introduction

Interpretation: The mean, standard deviation for the given set of data and the given value comes under the $95\%$ confidence interval or not is to be determined.

Concept introduction: The mean of the given set of data is calculated by using the formula,

$\overline{x}=\frac{{\displaystyle \sum x_{\text{i}}}}{n}$

The standard deviation for the given sets of data is calculated by using the formula,

$s=\sqrt{\frac{{\displaystyle \sum {\left(x_{\text{i}}-\overline{x}\right)}^2}}{n-1}}$

To determine: The mean and standard deviation of the given set of data.

Answer to Problem 1.74QP

Solution

The mean of the given set of data is $105.4mg/dL$ and the standard deviation is $3.91mg/dL$.

Explanation of Solution

Explanation

Given

The data for the diabetes from the given series of blood samples gives the results: $106,99,109,108$ and $105\text{mg/dL}$.

The mean of the given set of data is calculated by using the formula,

$\overline{x}=\frac{{\displaystyle \sum x_{\text{i}}}}{n}$

Where,

• $\overline{x}$ is the mean of the given set of data.
• $x_{\text{i}}$ is the i^th value of the distribution.
• $n$ is the number of values in the data.

Substitute the values of $x_{\text{i}}$ and $n$ for the given data in the above equation.

$\begin{array}{c}\overline{x}=\frac{106+99+109+108+105}{5}\text{mg/dL}\\ =\underset{\_}{105.4mg/dL}\end{array}$

The standard deviation for the given sets of data is calculated by using the formula,

$s=\sqrt{\frac{{\displaystyle \sum {\left(x_{\text{i}}-\overline{x}\right)}^2}}{n-1}}$

Where,

• $\overline{x}$ is the mean of the given set of data.
• $x_{\text{i}}$ is the i^th value of the distribution.
• $n$ is the number of values in the data.
• $s$ is the standard deviation of the data.

Substitute the values of $x_{\text{i}}$, $\overline{x}$ and $n$ for manufacturer 1 in the above equation.

$\begin{array}{c}s=\sqrt{\frac{\begin{array}{l}{\left(106-105.4\right)}^2+{\left(99-105.4\right)}^2\\ +{\left(109-105.4\right)}^2+{\left(108-105.4\right)}^2+{\left(105-105.4\right)}^2\end{array}}{5-1}}\\ =\sqrt{\frac{0.36+40.96+12.96+6.76+0.16}{4}}\\ =\underset{\_}{3.91mg/dL}\end{array}$

(b)

Interpretation Introduction

To determine: If the given value $120\text{mg/dL}$ is in the $95\%$ confidence interval of the given data or not.

Answer to Problem 1.74QP

Solution

The value $120\text{mg/dL}$ is much higher than the confidence interval of the given data.

Explanation of Solution

Explanation

Given

The mean of the given data is $105.4\text{mg/dL}$.

The standard deviation of the given set of data is $3.91\text{mg/dL}$.

The value of $n$ is $5$.

The value of confidence level at $95\%$ is $2.776$.

The confidence interval of the given set of data is calculated by using the formula,

$\mu =\overline{x}\pm \frac{ts}{\sqrt{n}}$

Where,

• $\mu$ is the confidence interval.
• $\overline{x}$ is the mean of the given data.
• $t$ is the value of confidence level.
• $s$ is the standard deviation of the given data.
• $n$ is the number of values in the data.

Substitute the values of $\overline{x}$, $t$, $s$ and $n$ in the above equation.

$\begin{array}{c}\mu =105.4\pm \frac{2.776\times 3.91}{\sqrt{5}}\\ =\underset{\_}{105.4\pm 4.9}\end{array}$

Conclusion

The mean of the given set of data is $105.4mg/dL$ and the standard deviation is $3.91mg/dL$.

The value $120\text{mg/dL}$ is much higher than the confidence interval of the given data