To compare the treatment effect of two medications two indep

(To compare the treatment effect of two medications, two independent samples of patient performance data were collected. The sample sizes are 13 and 17 respectively. Assuming that the two populations shared the same variance, the researchers decided to conduct a two independent samples test for the means. They found that the difference between the two sample means was 3.6 and the pooled sample standard deviation of the two samples was 4.8. What would be the obtained t-value under the null hypothesis that assumes no difference between the two population means?

Fourteen undergraduate students were randomly selected from a university and their systolic blood pressure was tested. The sample mean and standard deviation were 125.8 mmHg and 10.8 mmHg respectively. What is the obtained t-score for this sample under the null hypothesis Ho:

Solution

t test = mean diffeence/ (s*sqrt(1/n1+1/n2))

=3.6/(4.8*sqrt(1/13+1/17))

=2.03562

Answer: 2.04

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t = (xbar-mu)/(s/vn)

=(125.8-120)/(10.8/sqrt(14))

=2.009409

Answer: 2.01

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p1=70/220 = 0.3181818

p2=105/200 =0.525

So the lower bound is

(p1-p2) - Z*sqrt(p1*(1-p1)/n1+p2*(1-p2)/n2)

=(0.3181818-0.525) -1.96*sqrt(0.3181818*(1-0.3181818)/220 +0.525*(1-0.525)/200)

=-0.2994368

So the upper bound is

(p1-p2) + Z*sqrt(p1*(1-p1)/n1+p2*(1-p2)/n2)

=(0.3181818-0.525) +1.96*sqrt(0.3181818*(1-0.3181818)/220 +0.525*(1-0.525)/200)

=-0.1141996

Answer: [-0.299, -0.114]