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

$H_0: \bar x = \bar y\\H_a: \bar x \neq \bar y$

(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 = $\frac{5}{2.3299}$ =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

4 0

4 years ago

Jean has a loan of $2,000 the rate of simple interest in 8% a year he has paid $640 interest how many years has he had the loan

Andreas93 [3]

Simple interest means the interest at the end of each year will always be 8% of 2000 (assuming he doesnt pay some of it off), and will not be compounded with the existing interest.

To get a percentage, multiply by the percentage/100, in this case i.e. multiply by 0.08.
2000*0.08 = $160.
Therefore he has $160 of interest each year to pay.
Now we just need to work out how many lots of 160 go into 640, i.e. divide through.
640/160 = 4 - therefore he has had the loan for 4 years.

8 0

3 years ago