Heart Rates Sampling Distributions

Load the necessary packages

library(ggplot2)

According to some sources, heart rates are normally distributed with a mean of 76 bpm and a standard deviation of 8bpm.

mu_hr<-76
sigma_hr<-8

Generate 10,000 observations from this distribution:

heart_rate_population<-data.frame(rnorm(10000,76,8))
#plot the data
ggplot(heart_rate_population, aes(x = heart_rate_population)) + geom_dotplot(binwidth = .2)+
  scale_x_continuous(breaks=c(52,60,68,76,84,92,100))

Let’s imagine we take samples of 20 individuals from this population. What would we see?

mean_of_sample_means<-76 #the mean of sample means equals the population mean
mean_of_sample_means

[1] 76

sd_of_sample_means<- 8/sqrt(20)
sd_of_sample_means

[1] 1.788854

Generate 10,000 sample means from samples of size 20

sample_means<-data.frame(rnorm(10000,76,1.80))
ggplot(sample_means, aes(x = sample_means)) + geom_dotplot(binwidth = .05)+
  scale_x_continuous(breaks=c(70.6,72.4,74.2,76,77.8,79.6,81.4))

Now let’s take a look at some sample data from a real sample of 20

heart_rates=c(61,73,80,86,68,64,44,59,75,67,85,79,82,55,87,62,76,96,88,81)
heart_rates<-data.frame(heart_rates)
ggplot(heart_rates, aes(x = heart_rates)) + geom_dotplot(binwidth = 1)+
  scale_x_continuous(breaks=c(52,60,68,76,84,92,100))

heart_rates=c(61,73,80,86,68,64,44,59,75,67,85,79,82,55,87,62,76,96,88,81)
summary(heart_rates)

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  44.00   63.50   75.50   73.40   82.75   96.00 

sd(heart_rates)

[1] 13.12049

observed_sample_mean=73.4

Calculate the z-score for the sample mean

z=(73.4-76)/1.8
z

[1] -1.444444

Interpret this result: A sample mean of 73.4 is 1.44 standard deviations below the true mean.

There is a special function in R that allows us to calculate the percentile associated with any z-score. We will use dnorm(z) to find what proportion of values fall below our z-score of -1.44.

pnorm(-1.44)*100

[1] 7.49337

Interpretation: According to the normal model, a sample of 20 heart rates will be this far below the mean about 7.5% of the time.

Try a few more:

pnorm(-2)*100 

[1] 2.275013

pnorm(2)*100

[1] 97.72499

(pnorm(2)-pnorm(-2))*100  #should be about 95%

[1] 95.44997