Lab4b - Data 606

load(url("http://www.openintro.org/stat/data/ames.RData"))

population <- ames$Gr.Liv.Area
samp <- sample(population, 60)

1. Describe the distribution of your sample. What would you say is the “typical” size within your sample? Also state precisely what you interpreted “typical” to mean.

hist(samp, breaks=25)

summary(samp)

##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##     800    1092    1447    1473    1687    2790

• samp = 1283, 1595, 1493, 2142, 1739, 1567, 2790, 2704, 2450, 1091, 864, 996, 1114, 1686, 1260, 1512, 866, 1472, 1468, 1575, 1050, 1077, 2046, 1008, 1187, 2482, 2161, 1281, 1601, 1385, 1501, 1092, 1606, 1824, 1964, 1053, 1426, 1344, 1482, 864, 800, 1961, 874, 1055, 1690, 1414, 1302, 1122, 1846, 864, 1416, 1350, 919, 1898, 1362, 1553, 1556, 1790, 948, 1534
• the shape of this distribution looks right skewed
• mean_population = 1499.6904437 vs. mean_samp = 1472.5833333
• It’s not a “typical” sample size
• A “typical” size is when the sample’s values are most close to the population values

2. Would you expect another student’s distribution to be identical to yours? Would you expect it to be similar? Why or why not?

• No. It’s not the same but it’s very close to it.

sample_mean <- mean(samp)

se <- sd(samp) / sqrt(60)
lower <- sample_mean - 1.96 * se
upper <- sample_mean + 1.96 * se
c(lower, upper)

## [1] 1355.720 1589.447

3. For the confidence interval to be valid, the sample mean must be normally distributed and have standard error \(s / \sqrt{n}\). What conditions must be met for this to be true?

• the observations should be independent
• the sample size should be at least 30
• the shape of the distribution of the sample should be normal

Confidence levels

4. What does “95% confidence” mean? If you’re not sure, see Section 4.2.2.

• confidence: \(Point Estimate ± Confidence Interval * SE\)
• confidence_interval = 1.96 in case of 95% confidence

mean(population)

## [1] 1499.69

5. Does your confidence interval capture the true average size of houses in Ames? If you are working on this lab in a classroom, does your neighbor’s interval capture this value?

if (mean(population) > lower & mean(population) < upper){
  printo<-"TRUE"
} else{
  printo<-"FALSE"
}

• TRUE. My classmates’ interval should capture a very close value to this value.

6. Each student in your class should have gotten a slightly different confidence interval. What proportion of those intervals would you expect to capture the true population mean? Why? If you are working in this lab in a classroom, collect data on the intervals created by other students in the class and calculate the proportion of intervals that capture the true population mean.

sd_samp_classmates<-c()
mean_samp_classmates<-c()
for(i in 1:35){
  samp_classmates<-sample(samp, 60)
  sd_samp_classmates[i]<-sd(samp_classmates)
  mean_samp_classmates[i]<-mean(samp_classmates)
}
upper_samp_classmates<-mean_samp_classmates+(1.96*sd_samp_classmates)
lower_samp_classmates<-mean_samp_classmates-(1.96*sd_samp_classmates)

• Taking a sample of 35 with the classmates sampling, 95% confidence is expected to be captured closes to the population mean: (567.3622729, 2377.8043938)

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:

• Obtain a random sample.
• Calculate and store the sample’s mean and standard deviation.
• Repeat steps (1) and (2) 50 times.
• Use these stored statistics to calculate many confidence intervals.

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.

samp_mean <- rep(NA, 50)
samp_sd <- rep(NA, 50)
n <- 60

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.

c(lower_vector[1], upper_vector[1])

## [1] 1328.970 1550.997

---

On your own

• 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.

  plot_ci(lower_vector, upper_vector, mean(population))

  

df <- data.frame(lower_vector, upper_vector)

low <- sum(df$upper_vector < mean(population))
up <- sum(df$lower_vector > mean(population))

without_mean_proportion <- round((low + up)/60 ,2)

• 95 close to main confidence.

• Pick a confidence level of your choosing, provided it is not 95%. What is the appropriate critical value?

• Pickcing 90% confidence level will have a confidence level of 1.65.

• Calculate 50 confidence intervals at the confidence level you chose in the previous question. You do not need to obtain new samples, simply calculate new intervals based on the sample means and standard deviations you have already collected. Using the 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?

lower_vector <- samp_mean - 1.65 * samp_sd / sqrt(n) 
upper_vector <- samp_mean + 1.65 * samp_sd / sqrt(n)
plot_ci(lower_vector, upper_vector, mean(population))

df <- data.frame(lower_vector, upper_vector)

left <- sum(df$upper_vector < mean(population))
right <- sum(df$lower_vector > mean(population))

without_mean_proportion <- round((left + right)/60 ,2)

• 90 close to main confidence.