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November 12th, 2017, 08:59 AM  #1 
Newbie Joined: Nov 2017 From: Bulgaria Posts: 3 Thanks: 0  Probabilities
Hello guys, I recently started my course on statistics and probabilities but I am helpless with the problem below so any help or small tip is highly appreciated! In an experiment with a new tranquilizer, the pulse rates (per minute) of 12 patients were determined before they were given the tranquilizer and again 5 minutes later, and their pulse rates were found to be reduced on the average by 7.2 beats with a standard deviation of 1.8. At the level of significance 0.05, do we have significant evidence that the mean pulse reduction with this tranquilizer is less than 9.0 beats? Thanks in advance! 
November 13th, 2017, 03:15 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 
Though they don't use the word, the fact that you are given a average of 7.2 and a [I]standard deviation[/b] of 0.05 implies that this is a normal distribution. Since the sample mean and standard deviation are 7.2 and 0.05, and there were 12 measurements in the sample, the population mean and standard deviation are taken to be 7.2 and $\displaystyle 1.8\sqrt{12}= 6.235$, respectively. You want to find x such that the probability P(x< 0.9) is equal to 0.05. The "standard normal variable" is given by z= (x 7.2)/6.235. Look that up in a table or app of the standard normal distribution. There is probably a table in your text book or you can use or there is a simple app at Z table  Normal Distribution Calculator Compatible with iPhone and iPad.

November 14th, 2017, 07:49 AM  #3 
Math Team Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 878 Thanks: 60 Math Focus: सामान्य गणित  
November 14th, 2017, 10:12 AM  #4  
Senior Member Joined: Feb 2010 Posts: 688 Thanks: 130  Quote:
Let $\displaystyle \mu$ = the mean pulse rate per minute The null and alternate hypotheses are: $\displaystyle H_o: \mu = 9$ $\displaystyle H_a: \mu < 9$ $\displaystyle z = \dfrac{\overline{x} \mu}{\dfrac{\sigma}{\sqrt{n}}}$ Substituting in we get: $\displaystyle z = \dfrac{7.29}{\dfrac{1.8}{\sqrt{12}}} = 3.46$ This yields a $\displaystyle p$value of 0.0003. With a $\displaystyle p$value this small, the conclusion would be to reject the null hypothesis and conclude that there has been a reduction in mean pulse rate. However (and here I disagree with Country Boy), any set of numbers (normal or abnormal!) has a mean and standard deviation. The question is whether this sample of 12 numbers comes from a population that is normally distributed. Since we don't know that, I would have wanted a larger sample size (greater than about 30) so that the Central Limit Theorem would kick in. So in this case, since we don't know the population is normal and we have a small sample size, I would want to see the 12 actual measurements. At that point I would draw lots of pictures (including a normal quantile plot) to assess whether it is safe to assume that the sample comes from a normal population.  
November 15th, 2017, 09:40 AM  #5 
Newbie Joined: Nov 2017 From: Bulgaria Posts: 3 Thanks: 0 
I agree with you! Since Pvalue is less than 0.05 then we should reject the null hypothesis and assume that the mean pulse rate reduction is less than 9!


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