If you have access to data on an entire population, say the size of every house in Ames, Iowa, it’s straight forward to answer questions like, “How big is the typical house in Ames?” and “How much variation is there in sizes of houses?”. If you have access to only a sample of the population, as is often the case, the task becomes more complicated. What is your best guess for the typical size if you only know the sizes of several dozen houses? This sort of situation requires that you use your sample to make inference on what your population looks like.
In the previous lab, ``Sampling Distributions’’, we looked at the population data of houses from Ames, Iowa. Let’s start by loading that data set.
In this lab we’ll start with a simple random sample of size 60 from the population. Specifically, this is a simple random sample of size 60. Note that the data set has information on many housing variables, but for the first portion of the lab we’ll focus on the size of the house, represented by the variable Gr.Liv.Area.
Answer
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 334 1092 1452 1524 1774 4676
The sample is a unimodal, right skewed normal distribution. The range is almost 600. Typical size is mean. Above histogram sample size is 60. Mean is 1513. I think, Typical mean is the value represents an average living area size.
Answer
Another student’s distribution may not be identical to mine. Because the sample what they would selected could be different subset of the population.
One of the most common ways to describe the typical or central value of a distribution is to use the mean. In this case we can calculate the mean of the sample using,
Return for a moment to the question that first motivated this lab: based on this sample, what can we infer about the population? Based only on this single sample, the best estimate of the average living area of houses sold in Ames would be the sample mean, usually denoted as \(\bar{x}\) (here we’re calling it sample_mean). That serves as a good point estimate but it would be useful to also communicate how uncertain we are of that estimate. This can be captured by using a confidence interval.
We can calculate a 95% confidence interval for a sample mean by adding and subtracting 1.96 standard errors to the point estimate (See Section 4.2.3 if you are unfamiliar with this formula).
se <- sd(samp) / sqrt(60)
lower <- sample_mean - 1.96 * se
upper <- sample_mean + 1.96 * se
c(lower, upper)## [1] 1359.237 1689.696
This is an important inference that we’ve just made: even though we don’t know what the full population looks like, we’re 95% confident that the true average size of houses in Ames lies between the values lower and upper. There are a few conditions that must be met for this interval to be valid.
Answer
Conditions: The samples must be random. The sample should be independent. Sample should be less than 10% of the population. The sample size greater than 30.
Answer
The 95% confidence means we are 95% confident that the population mean is between the lower and upper boundary or we can say, 95% of those boundaries will contain the true population mean.
In this case we have the luxury of knowing the true population mean since we have data on the entire population. This value can be calculated using the following command:
## [1] 1499.69
Answer
Confidence interval mostly captures the true poulation mean.
Answer
95% of the intervals captured the true population mean. In normal distribution 95% of observations are within the 1.96 times of standard deviation.
Using R, we’re going to recreate many samples to learn more about how sample means and confidence intervals vary from one sample to another. Loops come in handy here (If you are unfamiliar with loops, review the Sampling Distribution Lab).
Here is the rough outline:
But before we do all of this, we need to first create empty vectors where we can save the means and standard deviations that will be calculated from each sample. And while we’re at it, let’s also store the desired sample size as n.
Now we’re ready for the loop where we calculate the means and standard deviations of 50 random samples.
for(i in 1:50){
samp <- sample(population, n) # obtain a sample of size n = 60 from the population
samp_mean[i] <- mean(samp) # save sample mean in ith element of samp_mean
samp_sd[i] <- sd(samp) # save sample sd in ith element of samp_sd
}Lastly, we construct the confidence intervals.
lower_vector <- samp_mean - 1.96 * samp_sd / sqrt(n)
upper_vector <- samp_mean + 1.96 * samp_sd / sqrt(n)Lower bounds of these 50 confidence intervals are stored in lower_vector, and the upper bounds are in upper_vector. Let’s view the first interval.
## [1] 1409.653 1670.380
Using the following function (which was downloaded with the data set), plot all intervals. What proportion of your confidence intervals include the true population mean? Is this proportion exactly equal to the confidence level? If not, explain why.
proportion <- 1 - (5 / 50)
paste0("Proportion of your confidence intervals include the true population mean is ",proportion)## [1] "Proportion of your confidence intervals include the true population mean is 0.9"
No, the propertion is not exactly equal to confidentce interval. Because as shown above out of 50 confidence interval 4 confidence interval does not contain true value. So proportion is 92%. The proportion is not exactly equal to 95% confidence.
May be because of less samples.
If confidence is 99%, the appropriate critical value for 99% confidence is 2.58.
plot_ci function, plot all intervals and calculate the proportion of intervals that include the true population mean. How does this percentage compare to the confidence level selected for the intervals?# choose confidence = 90%, critical value = 1.65
set.seed(123)
n = 60
for (i in 1:50){
lower_vector[i] <- samp_mean[i] - 1.65 * samp_sd[i] / sqrt(n)
upper_vector[i] <- samp_mean[i] + 1.65 * samp_sd[i] / sqrt(n)
}
plot_ci(lower_vector, upper_vector, mean(population))## [1] "Proportion is 0.82"
I can see all the propertion inteval include the true population mean is 0.82. The proportion level is 0.82 and the confidence level is 0.90.