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

` $$ \begin{aligned} t&=\frac{(\overline{x}_1 -\overline{x}_1)-(\mu_1-\mu_2)}{sp\sqrt{\big(\frac{1}{n_1}+\frac{1}{n_2}\big)}}\\ &= \frac{(111-109)-0}{8\sqrt{\big(\frac{1}{23}+\frac{1}{24}\big)}}\\ &= 0.8568 \end{aligned} $$ `

**Decision:Traditional approach**

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.

#### Further Reading

- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators