# Solved:A study was conducted to determine whether an expectant mother’s cigarette smoking has any effect on the bone mineral content of her otherwise healthy child. A sample of 77 newborns

#### ByRaju Chaudhari

Sep 26, 2020

A study was conducted to determine whether an expectant mother's cigarette smoking has any effect on the bone mineral content of her otherwise healthy child. A sample of 77 newborns whose mothers smoked during pregnancy has a mean bone mineral content $\overline{x}_1=0.098$ g/cm and standard deviation $s_1=0.026$ g/cm; a sample of 161 infants whose mothers did not smoke has mean $\overline{x}_2= 0.095$ g/cm and standard deviation $s_2=0.025$ g/cm. Assume that the underlying population variances are equal.

a) Are the two samples of data paired or independent?
b) State the null and Alternative hypothesis of the two sided test?
c) Conduct the test at the 0.05 level of significance. What do you conclude?

#### Solution

a) The two samples are independent. Independent because the mothers who smoked were not specifically matched to mothers who did not smoke during pregnancy.

b) The null and Alternative hypothesis of the two sided test are $H_0 : \mu_1 = \mu_2$ against $H_a:\mu_1\neq \mu_2$.

c) Given that the sample size $n_1 = 77$, $n_2 = 161$, sample mean $\overline{x}_1= 0.098$,
$\overline{x}_2= 0.095$, sample standard deviation $s_1 = 0.026$ and $s_2 = 0.025$.

Define test statistic

The test statistic is

 \begin{aligned} t& =\frac{(\overline{x}_1 -\overline{x}_1)-(\mu_1 - \mu_2)}{sp\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}} \end{aligned}
where

 \begin{aligned} s_p & = \sqrt{\frac{(n_1-1)s_1^2 +(n_2-1)s_2^2}{n_1+n_2-2}}\\ & = \sqrt{\frac{(77-1)0.026^2 +(161-1)0.025^2}{77+161-2}}\\ & = 0.0253. \end{aligned}

Speciffy the level of significance

The significance level is $\alpha = 0.05$.

Determine the critical value

As the alternative hypothesis is $\textit{two-tailed}$, the critical value of $t$ using $\alpha = 0.05$ and degrees of freedom $n_1+n_2-2=77+161-2=236$ $\text{are}$ $\text{-1.97 and 1.97}$.

The rejection region (i.e. critical region) is $\text{t < -1.97 or t > 1.97}$.

Computation

The test statistic under the null hypothesis is
 \begin{aligned} t&=\frac{(\overline{x}_1 -\overline{x}_1)-(\mu_1-\mu_2)}{sp\sqrt{\big(\frac{1}{n_1}+\frac{1}{n_2}\big)}}\\ &= \frac{(0.098-0.095)-0}{0.0253\sqrt{\big(\frac{1}{77}+\frac{1}{161}\big)}}\\ &= 0.8558 \end{aligned}

The rejection region (i.e. critical region) is $\text{t < -1.97 or t > 1.97}$.
The test statistic is $t =0.8558$ which falls $\text{outside}$ the critical region, we $\textit{fail to reject}$ the null hypothesis.

OR

$p$-value approach

The test is $\textit{two-tailed}$ test, so p-value is the area to the $\textit{extreme}$ of the test statistic ($t=0.8558$). That is p-value = $2*P(t\geq 0.8558 ) = 0.393$.

The p-value is $0.393$ which is $\textit{greater than}$ the significance level of $\alpha = 0.05$, we $\textit{fail to reject}$ the null hypothesis.

We fail to reject the null hypothesis that the mean bone mineral content in newborns whose mothers smoked during pregnancy is the same as the mean bone mineral content in those whose mothers did not smoke. Thus we conclude that we do not have suffiecient evidence to conclude that there is a significant difference in those means.