Question

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4) In a study of heart surgeryone issue was the effect of drugs called beta blockers on the pulse rate of patients during surgery. The available subjects were randomly assigned into two groups. One group received a beta blocker: the other group received a placebo. The pulse rate of each patient at a critical point during the operation was recorded. Here are the data in the summary form: Group Beta blocker Placebo 30 70.3 Part 1 Based off the results, is there statistical evidence that beta blockers lower the pulse rate on average? Use a 5% significance level. Step 1: H __to=0 : _l 200 Step 2: Test Statistic -2.9 5 or 0.017 0.017? Step 3: P-Value = _0. 008 a Step 4: a = 0.05 Decision: _0.008 20.05 reject ko Step 5: Conclusion: choue evidence to Subhout that bela bockers . lower the pulse rate on averape Part 2 Construct the appropriate confidence interval to estimate the true difference in means. Does this interval coincide with the decision from step 4? Explain Hint: CL-1 -2(alpha) - Hotectos -0.937 Confidence interval: -9.263 < Decision/explain: confidence interval Contains only nepates

Answer 1

here, we will do independent 2 sample t test.( i have not assumed equal variance)

hypothesis:-

$H_{0}:\mu_{1}-\mu_{2}=0$

$H_{1}:\mu_{1}-\mu_{2}< 0$

where, $\mu_{1}$ is the average pulse rate of person who receive beta blocker and $\mu_{2}$ is the average pulse rate of persons who receive placebo.

given data are:-

$\bar{x_{1}}=65.2,s_{1}=7.8,n_{1}=30$

$\bar{x_{2}}=70.3,s_{2}=8.3,n_{2}=30$

2).the test statistic be:-

$t=\frac{\bar{x_{1}}-\bar{x_{2}}}{\sqrt{\frac{s_{1}^2}{n_{1}}&plus;\frac{s_{2}^2}{n_{2}}}}=\frac{65.2-70.3}{\sqrt{\frac{7.8^2}{30}&plus;\frac{8.3^2}{30}}}=-2.45$

the degrees of freedom be:-

$df=\left [ \frac{(\frac{s_{1}^2}{n_{1}}&plus;\frac{s_{2}^2}{n_{2}})^2}{\frac{s_{1}^4}{n_{1}^2(n_{1}-1)}&plus;\frac{s_{2}^4}{n_{2}^2(n_{2}-1)}}\right ]$

$=\left [ \frac{(\frac{7.8^2}{30}&plus;\frac{8.3^2}{30})^2}{\frac{7.8^4}{30^2(30-1)}&plus;\frac{8.3^4}{30^2(30-1)}}\right ]$

$=57$

3)p value = 0.0086

[ using p value calculator for t = -2.45,  df = 57,left tailed test]

4).decision:-

p value = 0.0086 < 0.05(alpha)

so,at $\alpha$ = 0.05 , we reject the null hypothesis.

5).conclusion:-

there is sufficient statistical evidence to claim that beta blockers lower the pulse rate on average at 0.05 level of significance.

b). t critical value for df=57, 95% confidence level be:-

$t_{crit}=2.002$ [ from t distribution table, for both tailed test.]

the 95% confidence interval for $(\mu_{1}-\mu_{2})$ be:-

$=(\bar{x_{1}}-\bar{x_{2}})\pm t_{crit}\sqrt{\frac{s_{1}^2}{n_{1}}&plus;\frac{s_{2}^2}{n_{2}}}$

$=(65.2-70.3)\pm 2.002*\sqrt{\frac{7.8^2}{30}&plus;\frac{8.3^2}{30}}$

= ( -9.263, -0.937 )

decision:-

as 0 is not contained in the confidence interval, we reject the null hypothesis.

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