Confidence Intervals

The data

download.file("http://www.openintro.org/stat/data/ames.RData", destfile = "ames.RData")
load("ames.RData")

Random sample size n = 60

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

Exercise 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, main = "Histogram of House Sizes in Ames, Iowa", xlab = "House size")

median(samp)

## [1] 1456

mean(samp)

## [1] 1519.467

The distribution is slightly skewed right. Since there isn’t much skew, both median or mean would be adequate measures to find then “typical” house size. In fact, both measures are fairly similar: mean = 1409.517 and median = 1407. For this, I will use the mean, therefore, the “typical” house is 1409.517 (unsure of units)

Exercise 2

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

I wouldn’t expect it to be identical to my sample, but I would expect it to be fairly similar. This is because the sample size is 60 which is considered large enough.

Confidence intervals

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] 1405.916 1633.017

Exercise 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?

If these are the same rules for a t-distribution, then the sample must be random, the observations must be independent, and the n has to be large and if n is small, the population must be normally distributed.

Confidence levels

Exercise 4

What does “95% confidence” mean?

95% confidence means that we are 95% sure the population mean falls within the range outlined.

mean(population)

## [1] 1499.69

Exercise 5

Does your confidence interval capture the true average size of houses in Ames?

Yes, the average - 1499.69 - is within the confidence interval. 1467.756 < 1499.69 < 1738.711

Exercise 6

What proportion of the intervals would you expect to capture the true population mean? Why?

I would expect 95 out of 100 to capture the true value (or 95%). Since my class size is around 30, I would expect only 1 or 2 confidence intervals to NOT capture the true mean.

Start by creating empty vectors:

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

Now for the “loop”:

for(i in 1:50) {
  samp <- sample(population, n)
  samp_mean[i] <- mean(samp)
  samp_sd[i] <- sd(samp)
}

Finally, construct the confidence intervals:

lower_vector <- samp_mean - 1.96 * samp_sd / sqrt(n)
upper_vector <- samp_mean + 1.96 * samp_sd / sqrt(n)

c(lower_vector, upper_vector)

##   [1] 1424.514 1351.310 1396.667 1367.645 1403.375 1373.508 1292.319 1346.705
##   [9] 1450.274 1379.100 1299.914 1434.638 1549.134 1466.914 1366.673 1368.764
##  [17] 1375.325 1325.794 1358.073 1393.510 1444.535 1431.244 1429.687 1345.474
##  [25] 1405.105 1415.175 1375.313 1295.793 1301.863 1380.670 1347.793 1444.591
##  [33] 1294.120 1355.745 1432.832 1335.027 1283.136 1328.442 1341.111 1489.160
##  [41] 1437.870 1476.534 1421.674 1409.941 1320.915 1292.525 1363.032 1296.496
##  [49] 1283.813 1449.983 1703.853 1601.223 1655.000 1686.155 1592.525 1663.225
##  [57] 1501.048 1551.995 1692.859 1626.833 1519.753 1674.562 1833.732 1714.153
##  [65] 1576.794 1632.770 1641.542 1524.639 1599.893 1659.223 1709.898 1695.423
##  [73] 1785.313 1555.693 1654.295 1656.158 1641.354 1522.741 1536.737 1626.464
##  [81] 1604.141 1707.109 1543.314 1597.955 1717.768 1549.507 1526.831 1561.891
##  [89] 1541.856 1774.407 1696.930 1741.066 1863.493 1646.159 1555.318 1506.342
##  [97] 1615.968 1545.304 1514.920 1699.917

On your own

plot_ci(lower_vector, upper_vector, mean(population))

That was a really cool function. It looks like 4 out of our 50 confidence intervals did NOT contain the actual population mean

Now let’s try 90% confidence

qnorm(0.975, 0, 1)

## [1] 1.959964

qnorm(0.95, 0, 1)

## [1] 1.644854

So this time we will use the critical value 1.645

lower_vector2 <- samp_mean - 1.645 * samp_sd / sqrt(n)
upper_vector2 <- samp_mean + 1.645 * samp_sd / sqrt(n)

c(lower_vector2, upper_vector2)

##   [1] 1446.961 1371.392 1417.426 1393.240 1418.575 1396.789 1309.092 1363.202
##   [9] 1469.768 1399.007 1317.579 1453.918 1572.004 1486.781 1383.558 1389.978
##  [17] 1396.717 1341.773 1377.505 1414.862 1465.859 1452.473 1458.264 1362.367
##  [25] 1425.129 1434.540 1396.691 1314.030 1320.737 1400.421 1368.392 1465.687
##  [33] 1314.144 1375.208 1455.728 1352.262 1302.719 1347.201 1357.242 1512.082
##  [41] 1458.687 1497.791 1457.177 1428.923 1339.751 1309.706 1383.358 1316.490
##  [49] 1302.384 1470.067 1681.406 1581.141 1634.241 1660.560 1577.325 1639.944
##  [57] 1484.275 1535.498 1673.366 1606.926 1502.087 1655.282 1810.863 1694.286
##  [65] 1559.909 1611.555 1620.149 1508.661 1580.461 1637.871 1688.574 1674.194
##  [73] 1756.736 1538.800 1634.271 1636.794 1619.975 1504.504 1517.863 1606.712
##  [81] 1583.541 1686.013 1523.289 1578.492 1694.872 1532.272 1507.248 1543.132
##  [89] 1525.725 1751.485 1676.113 1719.809 1827.989 1627.177 1536.482 1489.160
##  [97] 1595.642 1525.310 1496.349 1679.833

plot_ci(lower_vector2, upper_vector2, mean(population))

This time 6 out of 50 were “misses”. It is hard to tell, but our confidence intervals should be narrower at the expence of our confidence getting lower. This is reflected by the increase in the number of intervals which do not contain the actual population mean.

Confidence Intervals

Linnea Lundh

3/13/2021

Confidence Intervals

The data

Exercise 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.

Exercise 2

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

I wouldn’t expect it to be identical to my sample, but I would expect it to be fairly similar. This is because the sample size is 60 which is considered large enough.

Confidence intervals

Exercise 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?

If these are the same rules for a t-distribution, then the sample must be random, the observations must be independent, and the n has to be large and if n is small, the population must be normally distributed.

Confidence levels

Exercise 4

What does “95% confidence” mean?

95% confidence means that we are 95% sure the population mean falls within the range outlined.

Exercise 5

Does your confidence interval capture the true average size of houses in Ames?

Yes, the average - 1499.69 - is within the confidence interval. 1467.756 < 1499.69 < 1738.711

Exercise 6

What proportion of the intervals would you expect to capture the true population mean? Why?

I would expect 95 out of 100 to capture the true value (or 95%). Since my class size is around 30, I would expect only 1 or 2 confidence intervals to NOT capture the true mean.

On your own

That was a really cool function. It looks like 4 out of our 50 confidence intervals did NOT contain the actual population mean

Now let’s try 90% confidence

So this time we will use the critical value 1.645

This time 6 out of 50 were “misses”. It is hard to tell, but our confidence intervals should be narrower at the expence of our confidence getting lower. This is reflected by the increase in the number of intervals which do not contain the actual population mean.