In an investigation of pregnancy-induced hypertension, one group of women with this disorder was treated with low-dose aspirin, and a second group was given a placebo. A sample consisting of 50 women who received aspirin has mean arterial blood pressure 120mmHg and standard deviation 10mmHg; a sample of 42 women who were given the placebo has mean blood pressure 115mmHg and standard deviation 12mmHg. Assume that the underlying population variances are equal. a. At the 0.05 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?

Given that in an investigation of pregnancy-induced hypertension, one group of women with this disorder was treated with low-dose aspirin, and a second group was given a placebo. A sample consisting of 50 women who received aspirin has mean arterial blood pressure 120mmHg and standard deviation 10mmHg; a sample of 42 women who were given the placebo has mean blood pressure 115mmHg and standard deviation 12mmHg.

Population variances are equal

[tex]H_0: \bar x = \bar y\\H_a: \bar x \neq \bar y[/tex]

(two tailed test at 5% significance level)

Here x denotes group I and Y group II

N Mean StDev SE Mean

Sample 1 50 120 10 1.4142

Sample 2 42 115 12 1.8516

Pooled std deviation = 10.9565

df=80

Mean difference= 5

Test statistic t = mean diff/std error = [tex]\frac{5}{2.3299}[/tex]=2.1803

p value = 0.0318

since p <0.05 we reject H0

b) We find p >0.01 hence at 1% significance level we accept H0

This implies that 99% confidence interval contains mean difference =0