Chapter 18, Problem 18.21E

(a)

To determine

To explain:

The difference in mean SRSs score for the populations of the given sample.

Answer to Problem 18.21E

  $t=6.2993$

Explanation of Solution

Given:

Group	$n$	$\overline{x}$	$s$
Delayed clamping	$21$	$65.30952$	$7.704$
Early clamping	$20$	$81.95$	$11.55$

Concept used:

Formula

Standard error

  $\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$

Degree of freedom

  $=\frac{{\left[\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right]}^2}{\frac{{\left(\frac{s_1^2}{n_1}\right)}^2}{n_1-1}+\frac{{\left(\frac{s_1^2}{n_1}\right)}^2}{n_2-1}}$

Significance level

  $\alpha =1-\text{confidence}$

Calculation:

  $\begin{array}{l}{n}_1=21,\overline{x_1}=65.30952,s_1=7.704668\\ n_2=20,\overline{x_2}=81.95,s_2=11.55183\end{array}$

Standard error

  $\begin{array}{l}=\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\\ =\sqrt{\frac{{\left(21.6\right)}^2}{130}+\frac{{\left(27.6\right)}^2}{115}}\\ =3.196\end{array}$

Degree of freedom

  $\begin{array}{l}=\frac{{\left[\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right]}^2}{\frac{{\left(\frac{s_1^2}{n_1}\right)}^2}{n_1-1}+\frac{{\left(\frac{s_1^2}{n_1}\right)}^2}{n_2-1}}\\ =39\end{array}$

At $95\%$ confidence interval for the proportion $p$

Critical $z$ value

  $\begin{array}{l}=z_{\alpha /2}\\ =2.0227\end{array}$

Margin of error

  $\begin{array}{l}t=z_{\alpha /2}SE\\ t=1.96\left(3.971\right)\\ t=6.2993\end{array}$

(b)

To determine

To explain:

The difference in mean for the populations of the given sample.

Answer to Problem 18.21E

  $\left(-22.8,-10.5\right)$

Explanation of Solution

Given:

Group	$n$	$\overline{x}$	$s$
Delayed clamping	$21$	$65.30952$	$7.704$
Early clamping	$20$	$81.95$	$11.55$

Concept used:

Formula

Standard error

  $\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$

Degree of freedom

  $=\frac{{\left[\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right]}^2}{\frac{{\left(\frac{s_1^2}{n_1}\right)}^2}{n_1-1}+\frac{{\left(\frac{s_1^2}{n_1}\right)}^2}{n_2-1}}$

Significance level

  $\alpha =1-\text{confidence}$

Calculation:

  $\begin{array}{l}{n}_1=21,\overline{x_1}=65.30952,s_1=7.704668\\ n_2=20,\overline{x_2}=81.95,s_2=11.55183\end{array}$

Standard error

  $\begin{array}{l}=\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\\ =\sqrt{\frac{{\left(21.6\right)}^2}{130}+\frac{{\left(27.6\right)}^2}{115}}\\ =3.196\end{array}$

Degree of freedom

  $\begin{array}{l}=\frac{{\left[\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right]}^2}{\frac{{\left(\frac{s_1^2}{n_1}\right)}^2}{n_1-1}+\frac{{\left(\frac{s_1^2}{n_1}\right)}^2}{n_2-1}}\\ =39\end{array}$

At $95\%$ confidence interval for the proportion $p$

Critical $z$ value

  $\begin{array}{l}=z_{\alpha /2}\\ =2.0227\end{array}$

Margin of error

  $\begin{array}{l}E=z_{\alpha /2}SE\\ E=1.96\left(3.971\right)\\ E=6.2993\end{array}$

The $95\%$ confidence interval for difference

  $\begin{array}{l}=\overline{x_1}-\overline{x_2}\pm z_{\alpha /2}SE\\ =\left(65.30952-81.95\right)\pm 6.1744\\ =-16.6405\pm 6.1744\\ =\left(-22.8,-10.5\right)\end{array}$