Nigerian Spleen Length. Refer to Exercise 10.49 and determine a 99% confidence interval for the difference between mean spleen lengths of Nigerian males and females.

Male and female Nigerians' spleen lengths are given as examples.

Male Nigerians

$$\begin{gathered} ;{\overline{x}}_1=11.1, \\ {s}_1=0.9, \\ {n}_1=91 \end{gathered}$$

Female Nigerians;

The significance level is 1%.

The main goal is to find a confidence interval of $99\%$for the difference between two population mean ${\mu }_1$and${\mu }_2$

The null and alternate hypotheses should be stated.

Null hypotheses: $H_0:{\mu }_1={\mu }_2$

Alternate hypotheses: $H_a:{\mu }_1\ne {\mu }_2$

Two-tailed hypotheses exist.

Use Table IV to find $t_{\alpha /2}$with $df=n_1+n_2-2$for a confidence level of $1-a$

alpha=0.01 for $99\%$confidence level,

localid="1651405723632" $$\begin{gathered} \text{df}=n_1+n_2-2 \\ =(91+109-2) \\ =198. \end{gathered}$$

When $\mathrm{df}=198$, using table IV for critical values, w

Critical value,

Determine the confidence interval's endpoints.

Pooled standard deviation,

$s_p=\sqrt{\frac{(91-1)(0.9{)}^2+(109-1\left)\right(0.7{)}^2}{91+109-2}}$

$s_p=\sqrt{\frac{90(0.81)+108(0.49)}{198}}$

Confidence interval $=\left({\overline{x}}_1-{\overline{x}}_2\right)\pm t_{\alpha /2}·s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$

Confidence interval$=(11.1-10.1)\pm 2.601\times 0.797\sqrt{\frac{1}{91}+\frac{1}{109}}$

Confidence interval$=1\pm 0.29$

Confidence interval$=0.71to1.29$