Solved (Free): In an investigation of pregnancy induced hypertension, one group of women with this disorder was treated with low dose aspirin, and the second group was given a placebo

ByDr. Raju Chaudhari

Mar 30, 2021

In an investigation of pregnancy induced hypertension, one group of women with this disorder was treated with low dose aspirin, and the second group was given a placebo. A sample consisting of 23 women who received aspirin has mean arterial blood pressure 111 mm Hg and standard deviation 8 mm Hg; a sample of 24 women who were given the placebo has mean blood pressure 109 mm Hg and standard deviation 8 mm Hg.

a) At the 0.01 level of significance, test the null hypothesis that the two populations of women have the same mean arterial blood pressure.
b) Construct a 99% confidence interval for the true difference in population means. Does this interval contain the value 0?

Solution

Let $\mu_1$ be the population mean arterial blood pressure for women treated with aspirin, and $\mu_2$ be the mean for women treated with placebo.

a) The hypothesis testing problem is $H_0 : \mu_1 = \mu_2$ against $H_a:\mu_1\neq \mu_2$.

c) Given that the sample size $n_1 = 23$, $n_2 = 24$, sample mean $\overline{x}_1= 111$,
$\overline{x}_2= 109$, sample standard deviation $s_1 = 8$ and $s_2 = 8$.

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{(23-1)8^2 +(24-1)8^2}{23+24-2}}\\ & = 8. \end{aligned}

Speciffy the level of significance

The significance level is $\alpha = 0.01$.

Determine the critical value

As the alternative hypothesis is $\textit{two-tailed}$, the critical value of $t$ using $\alpha = 0.01$ and degrees of freedom $n_1+n_2-2=23+24-2=45$ $\text{are}$ $\text{-2.69 and 2.69}$.

The rejection region (i.e. critical region) is $\text{t < -2.69 or t > 2.69}$.

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{(111-109)-0}{8\sqrt{\big(\frac{1}{23}+\frac{1}{24}\big)}}\\ &= 0.8568 \end{aligned}

The rejection region (i.e. critical region) is $\text{t < -2.69 or t > 2.69}$.

The test statistic is $t =0.8568$ 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 ($t=0.8568$). That is p-value = $2*P(t\geq 0.8568 ) = 0.3961$.

The p-value is $0.3961$ which is $\textit{greater than}$ the significance level of $\alpha = 0.01$, we $\textit{fail to reject}$ the null hypothesis.

We do not have enough evidence to reject the null hypothesis. We conclude that there is no evidence that aspirin provides a change in mean blood
pressure compared to placebo.

b) The critical value $t_{\alpha/2,n_1+n_2-2} = t_{0.05,45} = 1.679$.

The margin of error for difference of means $\mu_1-\mu_2$ is

\begin{aligned} E & = t_{\alpha/2,n_1+n_2-2} \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\\ & = 1.679 \sqrt{\frac{8^2}{23}+\frac{8^2}{24}}\\ & = 3.919. \end{aligned}

$90$% confidence interval estimate for the difference $(\mu_1-\mu_2)$ is

\begin{aligned} (\overline{X}_1 -\overline{X}_2)- E & \leq (\mu_1-\mu_2) \leq (\overline{X}_1 -\overline{X}_2) + E\\ (111-109) - 3.919 & \leq (\mu_1-\mu_2) \leq (111-109) + 3.919\\ -1.919 & \leq (\mu_1-\mu_2) \leq 5.919. \end{aligned}

Thus, $90$% confidence interval estimate for the difference $(\mu_1-\mu_2)$ is $(-1.919,5.919)$. The interval contains the value 0.