In a study of heart surgery, one tested the effect of drugs called beta-blockers

Question

In a study of heart surgery, one tested the effect of drugs called beta-blockers on the pulse of patients during surgery. Thirty available subjects received a beta blocker and the pulse rate of each patient was recorded during a critical point during the procedure. The treatment group had a mean pulse rate of 65.2 beats per second and a sample standard deviation of s=7.8. Patients without the beta-blocker have a known average pulse rate of 69.2 during the same critical point of surgery.

We want to test the hypothesis that the mean pulse rate for the treatment group of patients with beta-blockers reduced the known average pulse rate.
Which of the following is the correct results of the test for significance at the

We accept the alternative hypothesis and conclude that the mean pulse rate for the treatment group of patients with beta-blockers reduced the known average at the 5% level, because the p-value is less than 5%.

We accept the alternative hypothesis and conclude that the mean pulse rate for the treatment group of patients with beta-blockers reduced the known average at the 5% level, because the p-value is more than 5%.

We reject the null hypothesis and conclude that the mean pulse rate for the treatment group of patients with beta-blockers reduced the known average at the5% level, because the p-value is less than 5%.

We reject the null hypothesis and conclude that the mean pulse rate for the treatment group of patients with beta-blockers reduced the known average at the 5% level, because the p-value is more than 5%.

Explanation / Answer

here null hypothesis: mean=69.2

alternate hypothesis: mean<69.2

std error =std deviation/(n)1/2 =7.8/(30)1/2 =1.4241

therefore test stat =(X-mean)/std error =(65.2-69.2)/1.4241=-2.8088

for above p value =0.0044

therefore

We reject the null hypothesis and conclude that the mean pulse rate for the treatment group of patients with beta-blockers reduced the known average at the5% level, because the p-value is less than 5%.

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