Question

Problem 9

In a study of heart surgery, one issue was the effect of drugs called beta-blockers on the pulse

rate of patients during surgery. The available subjects were divided at random into two groups of

40 patients each. 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. The treatment

group had a mean pulse rate of 57 and standard deviation 7. For the control group, the mean

pulse rate was 60 and the standard deviation was 5.

(a) Construct and interpret a 93% confidence interval for the difference in mean pulse rates.

(b) Perform the 5 steps of Hypothesis Testing.

Answer 1

(a)

we are given:

Sample Mean 1(Xˉ1​) =	57
Sample Standard Deviation 1(s1​) =	7
Sample Size 1(N1​) =	40
Sample Mean 2 (Xˉ2​) =	60
Sample Standard Deviation 2 (s2​) =	5
Sample Size 2 (N2​)=	40

df=n1​+n2​−2=40+40−2=78.

critical value for α=0.07 and df=78 degrees of freedom is tc​=t_1−α/2;n−1​=1.837

s_p​​=​=

s_p=6.083

s_e==

s_e=1.36 --------------------(1)

CI​=​(−−tc​×s_{e ,}  −+tc​×s_e)

CI=(57−60−1.837×1.36 , 57−60+1.837×1.36)

CI=(−5.499,−0.501)​

(b)

Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho:μ1​ =μ2​

Ha: μ1​ ≠ μ2

Based on the information provided, the significance level is α=0.93, and the degrees of freedom are df = 78​.

tc​=0.088, for α=0.93 and df=78.

test statistic:

t= ~t_n1+n2-2

t=

t=

t=−2.206

Since it is observed that ∣t∣=2.206>tc​=0.088, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0.0304, and since p=0.0304<0.93, it is concluded that the null hypothesis is rejected.

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population mean μ1​ is different than μ2​, at the 0.93 significance level.

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