2023-03-14

Point Estimation

  • Any statistic can be a point estimate.
  • A statistic is an estimator of some parameter in a population.
  • ex:
    • Variable s is a point estimate for the population parameter “standard deviation”.
    • Variable Xbar is a point estimate for the population parameter “mean”.
    • Variable s^2 is a point estimate for the population parameter “population variance”.

Example 1 (Mean)

  • Below is a sample of 12 mean times for a 100 yard dash. Find the best estimate of the population mean.

  • 15.22, 14.34, 18.12, 12.61, 15.61, 14.22, 19.41, 12.22, 17.12, 14.22, 12.91, 18.12

  • Here, we will be looking to find Xbar because the problem statement talks about finding the population mean.

Example 1 (Math text in Latex)

  • The equation we will use for this problem is \[Xbar = (X1 + X2 + Xn...) / n\]

\[15.22 + 14.34 + ... + 18.12 = 184.12\] \[n = 12\] \[Xbar = 184.12 / 12 = 15.34\]

ggplot for Example 1

R Code for creating Example 1’s ggplot

# create a vector of the data set
meanTimes <- c(15.22, 14.34, 18.12, 12.61, 15.61, 14.22, 19.41, 12.22,
               17.12, 14.22, 12.91, 18.12)

# calculate Xbar
Xbar <- mean(meanTimes)

# create a vector for the mean positions
means <- c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12)

# create a dataframe for Xbar
XbarDF <- data.frame(t(meanTimes), means)

# create a ggplot of the data frame
ggplot(XbarDF, aes(x = means, y = meanTimes, fill="red")) + geom_bar(
  stat="identity", fill="red") + ggtitle("Barplot of Mean Times") + 
  labs(caption = "This bar plot shows the 12 means given in the problem
       statement and their times. The Xbar of this data set is 15.34.")

Example 2 (standard deviation)

  • 5 crystals are grown from a solution and the length of each crystal is measured in millimeters. Here is the data:

    • 9, 2, 5, 4, 12

  • Calculate the sample standard deviation of the length of the crystals.

  • Here we will find the standard deviation by following certain steps.

Example 2 (Math text in Latex)

  • Step 1: Find the mean of the data, (Xbar). \[Xbar = (9 + 2 + 5 + 4 + 12) / 5 = 6.4\]
  • Step 2: Subtract the mean from each data point and square it. \[(9 - 6.4)^2 = 6.76\] \[(2 - 6.4)^2 = 19.36\] \[(5 - 6.4)^2 = 1.96\] \[(4 - 6.4)^2 = 5.76\] \[(12 - 6.4)^2 = 31.36\]

Example 2 (Math text in Latex)

  • Calculate the mean of the squared differences, (sample variance). \[6.76 + 19.36 + 1.96 + 5.76 + 31.36 = 65.2\] \[n - 1 = 4\] \[variance = 65.2 / 4 = 16.3\]
  • Step 4: Square root the sample variance to get the standard deviation. \[stddev = \sqrt{16.3} = 4.04\]

ggplot for Example 2

plotly plot for Example 2

References