Question

O. Ehimwenma and M. Tagbo, researchers in Nigeria, were interested in how the characteristics of the spleen of residents in their tropical environment compare to those found elsewhere in the world. They published their findings in the article "Determination of Normal Dimensions of the Spleen by Ultrasound in an Endemic Tropical Environment" (Nigerian Medical Journal, Vol. 52, No. 3, pp. 198-203). The researchers randomly sampled $91$males and $109$females in Nigeria. The mean and standard deviation of the spleen lengths for the males were $11.1\mathrm{cm}$and $0.9\mathrm{cm}$respectively, and those for the females were $10.1\mathrm{cm}$and $0.7\mathrm{cm}$, respectively. At the$1\%$ significance level, do the data provide sufficient evidence to conclude that a difference exists in mean spleen lengths of male and female Nigerians?

Short answer

The presented data provide adequate evidence to infer that there is a difference in mean spleen lengths between male and female Nigerians at the significance level of $1\%$.

Step-by-step solution

Given Information

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

Population $1$: Male Nigerians ; ${\overline{x}}_1=11.1,s_1=0.9,n_1=91$

Population $2$: Female Nigerians; ${\overline{x}}_2=10.1,s_2=0.7,n_2=109$

Significance level is $1\%$.

Explanation

Population $1$: Male Nigerians : ${\overline{x}}_1=11.1,s_1=0.9,n_1=91$

Population $2$: Female Nigerians; ${\overline{x}}_2=10.1,s_2=0.7,n_2=109$

The major goal is to determine whether male and female Nigerians have different mean spleen lengths.

Define null and alternate hypotheses.

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

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

Hypotheses is two-tailed.

We determine the significance level as$1\%$.

Calculation

Pooled standard deviation,$s_p=\sqrt{\frac{\left(n_1-1\right)s_1^2+\left(n_2-1\right)s_2^2}{n_1+n_2-2}}$

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

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

$\Rightarrow s_p=0.797$

Test statistic,$t_0=\frac{{\overline{x}}_1-{\overline{x}}_2}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

$\Rightarrow t_0=\frac{11.1-10.1}{0.797\sqrt{\frac{1}{91}+\frac{1}{109}}}$

$\Rightarrow t_0=8.836$

We need to determine critical values

Here, localid="1651300988224" role="math" $$\begin{gathered} df=n_1+n_2-2 \\ =91+109-2 \\ =198 \end{gathered}$$

localid="1651301006222" role="math" $\Rightarrow \mathrm{df}=198$

Using table IV

localid="1651300997665" role="math" $$\begin{gathered} \text{Critical value,}\pm t_{\alpha /2}=\pm t_{0.01/2} \\ =\pm t_{0.005} \\ =\pm 2.601 \end{gathered}$$

Since $t_0=8.836$, the test statistic is in the two-tailed hypotheses test rejection zone. As a result, null hypotheses are ruled out.