Chapter Seven Homework
Nathan Cook, Corrinne Van Dorpe, Marc Settin, Elijah Smith, Simon Everett
Wednesday, Dec 06, 2017 - 05:01:33 PM
This is entirely our own work except as noted at the end of the document.
1 Import the data set Spruce into R.
*Note: data set information can be found at http://www1.appstate.edu/~arnholta/Data/
Spruce <- read.csv("http://www1.appstate.edu/~arnholta/Data/Spruce.csv")
head(Spruce) Tree Competition Fertilizer Height0 Height5 Diameter0 Diameter5
1 1 NC F 15.0 60.0 1.984375 7.4
2 2 NC F 9.0 45.2 1.190625 5.2
3 3 NC F 12.0 42.0 1.785937 5.7
4 4 NC F 13.7 49.5 1.587500 6.4
5 5 NC F 12.0 47.3 1.587500 6.2
6 6 NC F 12.0 56.4 1.587500 7.4
Ht.change Di.change
1 45.0 5.415625
2 36.2 4.009375
3 30.0 3.914062
4 35.8 4.812500
5 35.3 4.612500
6 44.4 5.812500summary(Spruce) Tree Competition Fertilizer Height0 Height5
Min. : 1.00 C :36 F :36 Min. : 9.00 Min. :24.00
1st Qu.:18.75 NC:36 NF:36 1st Qu.:13.28 1st Qu.:37.92
Median :36.50 Median :14.75 Median :45.30
Mean :36.50 Mean :14.57 Mean :45.51
3rd Qu.:54.25 3rd Qu.:16.00 3rd Qu.:53.25
Max. :72.00 Max. :18.70 Max. :68.00
Diameter0 Diameter5 Ht.change Di.change
Min. :1.191 Min. : 2.700 Min. : 8.30 Min. :1.019
1st Qu.:1.786 1st Qu.: 4.450 1st Qu.:23.20 1st Qu.:2.712
Median :1.984 Median : 5.750 Median :30.10 Median :3.915
Mean :1.935 Mean : 5.931 Mean :30.93 Mean :3.996
3rd Qu.:1.984 3rd Qu.: 7.100 3rd Qu.:38.17 3rd Qu.:5.116
Max. :2.381 Max. :11.300 Max. :51.50 Max. :8.919 1.1 Create exploratory plots to check the distribution of the variable Ht.change.
EDA.hist(Spruce$Ht.change)`stat_bin()` using `bins = 30`. Pick better value with `binwidth`.EDA.qq(Spruce$Ht.change)
1.2 Find a 95% \(t\) confidence interval for the mean height change over the 5-year period of the study and give a sentence interpreting your interval.
spruce_t <- t.test(Spruce$Ht.change)
head(spruce_t)$statistic
t
23.7549
$parameter
df
71
$p.value
[1] 1.636099e-35
$conf.int
[1] 28.33685 33.52982
attr(,"conf.level")
[1] 0.95
$estimate
mean of x
30.93333
$null.value
mean
0 spruce_t$conf.int[1][1] 28.33685spruce_t$conf.int[2][1] 33.52982SOLUTION: We are 95% confident that the mean height change for a spruce tree over the 5-year period is within (28.33685, 33.52982) .
1.3 Create exploratory plots to compare the distributions of the variable Ht.change for the seedlings in the fertilized and nonfertilized plots.
#With Fertilizer
EDA.hist(Spruce$Ht.change[Spruce$Fertilizer == "F"])`stat_bin()` using `bins = 30`. Pick better value with `binwidth`.EDA.qq(Spruce$Ht.change[Spruce$Fertilizer == "F"])
#Without Fertilizer
EDA.hist(Spruce$Ht.change[Spruce$Fertilizer != "F"])`stat_bin()` using `bins = 30`. Pick better value with `binwidth`.EDA.qq(Spruce$Ht.change[Spruce$Fertilizer != "F"])
- Find the 95% one-sided lower \(t\) confidence bound for the difference in mean heights (\(\mu_F - \mu_{NF}\)) over the 5-year period of the study and give a sentence interpreting your interval.
spruce_t2 <- t.test(Spruce$Ht.change~Spruce$Fertilizer,alternative="greater")
head(spruce_t2)$statistic
t
7.558587
$parameter
df
69.6971
$p.value
[1] 6.067505e-11
$conf.int
[1] 11.46664 Inf
attr(,"conf.level")
[1] 0.95
$estimate
mean in group F mean in group NF
38.28889 23.57778
$null.value
difference in means
0 spruce_t2$conf.int[1][1] 11.46664SOLUTION: We are 95% confident that the difference in heights between spruce trees that were fertilized and those that were not during a 5 year period is at least 11.46664.
2 Consider the data set Girls2004 with birth weights of baby girls born in Wyoming or Alaska.
Girls2004 <- read.csv("http://www1.appstate.edu/~arnholta/Data/Girls2004.csv")
Girls2004 ID State MothersAge Smoker Weight Gestation
1 1 WY 15-19 No 3085 40
2 2 WY 35-39 No 3515 39
3 3 WY 25-29 No 3775 40
4 4 WY 20-24 No 3265 39
5 5 WY 25-29 No 2970 40
6 6 WY 20-24 No 2850 38
7 7 WY 20-24 No 2737 38
8 8 WY 25-29 No 3515 37
9 9 WY 25-29 No 3742 39
10 10 WY 35-39 No 3570 40
11 11 WY 20-24 No 3834 41
12 12 WY 20-24 Yes 3090 39
13 13 WY 25-29 Yes 3350 40
14 14 WY 30-34 No 3292 37
15 15 WY 15-19 No 3317 40
16 16 WY 30-34 No 2485 37
17 17 WY 20-24 No 3215 39
18 18 WY 20-24 No 3230 40
19 19 WY 30-34 No 3345 39
20 20 WY 25-29 No 3050 41
21 21 WY 30-34 No 2212 37
22 22 WY 35-39 No 3605 39
23 23 WY 30-34 No 2722 39
24 24 WY 30-34 No 2880 39
25 25 WY 20-24 No 3610 39
26 26 WY 30-34 No 3355 39
27 27 WY 20-24 No 3995 41
28 28 WY 20-24 Yes 2948 39
29 29 WY 35-39 No 3345 41
30 30 WY 30-34 Yes 2892 39
31 31 WY 20-24 No 2466 37
32 32 WY 20-24 Yes 3290 39
33 33 WY 25-29 No 3310 39
34 34 WY 40-44 No 3175 37
35 35 WY 25-29 No 2715 38
36 36 WY 25-29 No 3540 38
37 37 WY 25-29 No 3402 38
38 38 WY 25-29 Yes 3923 39
39 39 WY 20-24 No 3204 37
40 40 WY 15-19 Yes 2495 37
41 41 AK 20-24 No 4337 41
42 42 AK 20-24 No 2948 40
43 43 AK 30-34 No 3269 39
44 44 AK 20-24 No 3608 38
45 45 AK 30-34 No 4016 39
46 46 AK 25-29 No 2919 40
47 47 AK 20-24 No 2608 37
48 48 AK 40-44 No 4309 39
49 49 AK 20-24 No 3288 39
50 50 AK 25-29 No 3742 38
51 51 AK 15-19 No 4394 41
52 52 AK 20-24 No 2182 37
53 53 AK 25-29 No 4592 40
54 54 AK 20-24 No 3090 39
55 55 AK 30-34 No 3770 40
56 56 AK 20-24 No 3977 39
57 57 AK 25-29 No 3153 40
58 58 AK 25-29 No 3458 41
59 59 AK 15-19 No 3912 38
60 60 AK 20-24 Yes 2863 40
61 61 AK 35-39 No 3190 39
62 62 AK 25-29 Yes 3515 38
63 63 AK 25-29 No 3288 39
64 64 AK 15-19 No 3114 40
65 65 AK 30-34 Yes 3543 41
66 66 AK 20-24 No 3825 39
67 67 AK 25-29 No 3458 39
68 68 AK 30-34 No 3698 41
69 69 AK 20-24 No 3572 39
70 70 AK 30-34 Yes 2352 40
71 71 AK 20-24 No 3175 40
72 72 AK 25-29 No 3742 41
73 73 AK 20-24 No 3997 39
74 74 AK 25-29 No 2576 38
75 75 AK 30-34 No 3572 40
76 76 AK 35-39 No 3968 39
77 77 AK 20-24 No 4564 42
78 78 AK 20-24 No 4210 40
79 79 AK 25-29 No 3260 38
80 80 AK 20-24 No 3600 40summary(Girls2004) ID State MothersAge Smoker Weight
Min. : 1.00 AK:40 15-19: 6 No :69 Min. :2182
1st Qu.:20.75 WY:40 20-24:29 Yes:11 1st Qu.:3076
Median :40.50 25-29:22 Median :3331
Mean :40.50 30-34:15 Mean :3362
3rd Qu.:60.25 35-39: 6 3rd Qu.:3709
Max. :80.00 40-44: 2 Max. :4592
Gestation
Min. :37.00
1st Qu.:38.00
Median :39.00
Mean :39.14
3rd Qu.:40.00
Max. :42.00 2.1 Create exploratory plots and compare the distribution of weight between the babies born in the two states.
#Wyoming
EDA.hist(Girls2004$Weight[Girls2004$State == "WY"])`stat_bin()` using `bins = 30`. Pick better value with `binwidth`.EDA.qq(Girls2004$Weight[Girls2004$State == "WY"])
#Alaska
EDA.hist(Girls2004$Weight[Girls2004$State == "AK"])`stat_bin()` using `bins = 30`. Pick better value with `binwidth`.EDA.qq(Girls2004$Weight[Girls2004$State == "AK"])
SOLUTION: On average, the distribution of weight between the babies born in the two states is greater in Alaksa than in Wyoming.
2.2 Find a 95% \(t\) confidence interval for the difference in mean weights for girls born in these two states. Give a sentence interpreting this interval.
girls_t <- t.test(Girls2004$Weight[Girls2004$State == "AK"],Girls2004$Weight[Girls2004$State == "WY"])
girls_t
Welch Two Sample t-test
data: Girls2004$Weight[Girls2004$State == "AK"] and Girls2004$Weight[Girls2004$State == "WY"]
t = 2.7316, df = 71.007, p-value = 0.007946
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
83.29395 533.60605
sample estimates:
mean of x mean of y
3516.35 3207.90 girls_t$conf.int[1][1] 83.29395girls_t$conf.int[2][1] 533.606SOLUTION: We are 95% confident that the difference in weights between girls born in Wyoming and girls born in Alaska is within (83.29395, 533.606).
2.3 Create exploratory plots and compare the distribution of weights between babies born to nonsmokers and babies born to smokers.
#Smokers
EDA.hist(Girls2004$Weight[Girls2004$Smoker == "Yes"])`stat_bin()` using `bins = 30`. Pick better value with `binwidth`.EDA.qq(Girls2004$Weight[Girls2004$Smoker == "Yes"])
#Nonsmokers
EDA.hist(Girls2004$Weight[Girls2004$Smoker == "No"])`stat_bin()` using `bins = 30`. Pick better value with `binwidth`.EDA.qq(Girls2004$Weight[Girls2004$Smoker == "No"])
SOLUTION: On avereage, the weights of babies born for smokers is less than the weights of babies born for nonsmokers.
2.4 Find a 95% \(t\) confidence interval for the difference in mean weights between babies born to nonsmokers and smokers. Give a sentence interpreting this interval.
girls_t2 <- t.test(Girls2004$Weight[Girls2004$Smoker == "Yes"],Girls2004$Weight[Girls2004$Smoker == "No"])
girls_t2
Welch Two Sample t-test
data: Girls2004$Weight[Girls2004$Smoker == "Yes"] and Girls2004$Weight[Girls2004$Smoker == "No"]
t = -1.8552, df = 14.35, p-value = 0.08423
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-617.9197 44.0330
sample estimates:
mean of x mean of y
3114.636 3401.580 girls_t2$conf.int[1][1] -617.9197girls_t2$conf.int[2][1] 44.033SOLUTION: We are 95% confident that the difference in weights between girls born to smokers and girls born to non-smokers is within (-617.9197, 44.033)
3 Import the FlightDelays data set into R. Although the data represent all flights for United Airlines and American Airlines in May and June 2009, assume for this exercise that these flights are a sample from all flights flown by the two airlines under similar conditions. We will compare the lengths of flight delays between the two airlines.
FlightDelays <- read.csv("http://www1.appstate.edu/~arnholta/Data/FlightDelays.csv")
summary(FlightDelays) ID Carrier FlightNo Destination DepartTime
Min. : 1 AA:2906 Min. : 71.0 BNA: 172 4-8am : 699
1st Qu.:1008 UA:1123 1st Qu.: 371.0 DEN: 264 4-8pm : 972
Median :2015 Median : 691.0 DFW: 918 8-Mid : 257
Mean :2015 Mean : 827.1 IAD: 55 8-Noon :1053
3rd Qu.:3022 3rd Qu.: 787.0 MIA: 610 Noon-4pm:1048
Max. :4029 Max. :2255.0 ORD:1785
STL: 225
Day Month FlightLength Delay Delayed30
Fri:637 June:2030 Min. : 68.0 Min. :-19.00 No :3432
Mon:630 May :1999 1st Qu.:155.0 1st Qu.: -6.00 Yes: 597
Sat:453 Median :163.0 Median : -3.00
Sun:551 Mean :185.3 Mean : 11.74
Thu:566 3rd Qu.:228.0 3rd Qu.: 5.00
Tue:628 Max. :295.0 Max. :693.00
Wed:564 head(FlightDelays) ID Carrier FlightNo Destination DepartTime Day Month FlightLength Delay
1 1 UA 403 DEN 4-8am Fri May 281 -1
2 2 UA 405 DEN 8-Noon Fri May 277 102
3 3 UA 409 DEN 4-8pm Fri May 279 4
4 4 UA 511 ORD 8-Noon Fri May 158 -2
5 5 UA 667 ORD 4-8am Fri May 143 -3
6 6 UA 669 ORD 4-8am Fri May 150 0
Delayed30
1 No
2 Yes
3 No
4 No
5 No
6 No3.1 Create exploratory plots of the lengths of delays for the two airlines.
#United Airlines
EDA.hist(FlightDelays$Delay[FlightDelays$Carrier == "UA"])`stat_bin()` using `bins = 30`. Pick better value with `binwidth`.EDA.qq(FlightDelays$Delay[FlightDelays$Carrier == "UA"])
#American Airlines
EDA.hist(FlightDelays$Delay[FlightDelays$Carrier == "AA"])`stat_bin()` using `bins = 30`. Pick better value with `binwidth`.EDA.qq(FlightDelays$Delay[FlightDelays$Carrier == "AA"])
3.2 Find a 95% \(t\) confidence interval for the difference in mean flight delays between the two airlines and interpret this interval.
flight_t <- t.test(FlightDelays$Delay[FlightDelays$Carrier == "UA"],FlightDelays$Delay[FlightDelays$Carrier == "AA"])
flight_t
Welch Two Sample t-test
data: FlightDelays$Delay[FlightDelays$Carrier == "UA"] and FlightDelays$Delay[FlightDelays$Carrier == "AA"]
t = 3.8255, df = 1843.8, p-value = 0.0001349
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
2.868194 8.903198
sample estimates:
mean of x mean of y
15.98308 10.09738 flight_t$conf.int[1][1] 2.868194flight_t$conf.int[2][1] 8.903198SOLUTION: We are 95% confident that the difference in delay times between United Airlines and American Airlines is within (2.868194, 8.903198).
4 Run a simulation to see if the \(t\) ratio \(T = (\bar{X} -\mu)/(S/\sqrt{n})\) has a \(t\) distribution or even an approximate \(t\) distribution when the samples are drawn from a nonnormal distribution. Be sure to superimpose the appropriate \(t\) density curve over the density of your simulated \(T\). Try two different nonnormal distributions \(\left( Unif(a = 0, b = 1), Exp(\lambda = 1) \right)\) and remember to see if sample size makes a difference (use \(n = 15\) and \(n=500\)).
4.1 Class Notes:
\(T = \frac{\bar{X} - \mu}{\frac{s}{\sqrt{n}}}\) ~ \(t_{n-1}\) Assuming t distribution \(t_{n-1}\) is a normal distribution
sims <- 10^4
xbar <- numeric(sims)
SD <- numeric(sims)
#Example given in class,
## Start
Mu <- 100
sigma <- 10
n<- 15
## End
for (i in 1: sims)
{
xn <- rnorm(n, Mu, sigma)
xbar[i] <- mean(xn)
SD[i] <- sd(xn)
}
T <- (xbar - Mu) - (SD / sqrt(n))
hist(T)
#alternative way with hist graphics of density historgram
hist(T, freq = FALSE)
# Need to add t distribution according to the 14 degrees of freedom
curve(dt(x,15))
curve(dt(x,15), -5, 5)
# Class notes
junk <- rt(10000, 15)
hist(junk)
hist(junk, freq = FALSE)
hist(junk, freq = FALSE, breaks = "Scott")
curve(dt(x, 15), add = TRUE)
SOULTION: Unbalanced sample sizes, pooled CI’s are not capturing true mean differenceat 95%. Balanced case, pooled CI does better.
5 One question is the 2002 General Social Survey asked participants whom they voted for in the 2000 election. Of the 980 women who voted, 459 voted for Bush. Of the 759 men who voted, 426 voted for Bush.
- Find a 95% confidence interval for the proportion of women who voted for Bush.
prop.test(459,980)$conf.int[1][1] 0.4368038prop.test(459,980)$conf.int[2][1] 0.5001819SOLUTION: We are 95% confident that the proportion of women who voted for Bush from the 2002 General Social Survey is within (0.4368038,0.5001819).
- Find a 95% confidence interval for the proportion of men who voted for Bush. Do the intervals for the men and women overlap? What, if anything, can you conclude about gender difference in voter preference?
prop.test(426,759)$conf.int[1][1] 0.5250797prop.test(426,759)$conf.int[2][1] 0.5968214SOLUTION: We are 95% confident that the proportion of men who voted for Bush from the 2002 General Social Survey is is within (0.5250797, 0.5968214). The intervals for the proportion of men who voted for Bush and the proportion of women who voted for Bush do not overlap but since we tested the proportion of means seperately, we cannot conclude anything about their apparent difference. To be able to conclude something, we would need to test a difference of proportions explicitly.
# Your code hereSOLUTION:
Prob6 - A retail store wishes to conduct a marketing survey of its customers to see if customers would favor longer store hours. How many people should be in their sample if the marketers want their margin of error to be at most 3% with 95% confidence, assuming
- they have no preconceived idea of how customers will respond, and
library(PASWR2)
Attaching package: 'PASWR2'The following objects are masked from 'package:PASWR':
bino.gen, checking.plots, EPIDURAL, EURD, FCD, interval.plot,
ksdist, normarea, nsize, ntester, oneway.plots, SBIQ, SDS4,
SIGN.test, tsum.test, twoway.plots, z.test, zsum.testnsize(b = .03, p = .5, type = "pi")
The required sample size (n) to estimate the population
proportion of successes with a 0.95 confidence interval
so that the margin of error is no more than 0.03 is 1068 . # Your code hereSOLUTION: A sample from 1068 people would lead to a 95% confidence and a 3% margin of error.
- a previous survey indicated that about 65% of customers favor longer store hours.
library(PASWR2)
nsize(b = .03,p = .65, type = "pi")
The required sample size (n) to estimate the population
proportion of successes with a 0.95 confidence interval
so that the margin of error is no more than 0.03 is 972 . SOLUTION: 972 people are required.
Prob7 - Suppose researchers wish to study the effectiveness of a new drug to alleviate hives due to math anxiety. Seven hundred math students are randomly assigned to take either this drug or a placebo. Suppose 34 of the 350 students who took the drug break out in hives compared to 56 of the 350 students who took the placebo.
- Compute a 95% confidence interval for the proportion of students taking the drug who break out in hives.
prop.test(34, 350)
1-sample proportions test with continuity correction
data: 34 out of 350, null probability 0.5
X-squared = 225.6, df = 1, p-value < 2.2e-16
alternative hypothesis: true p is not equal to 0.5
95 percent confidence interval:
0.06913692 0.13429260
sample estimates:
p
0.09714286 SOLUTION: With a confidence level of 95% the proportion will be between .0691 and .1342.
- Compute a 95% confidence interval for the proportion of students taking the placebo who break out in hives.
prop.test(56, 350)
1-sample proportions test with continuity correction
data: 56 out of 350, null probability 0.5
X-squared = 160.48, df = 1, p-value < 2.2e-16
alternative hypothesis: true p is not equal to 0.5
95 percent confidence interval:
0.1240384 0.2036158
sample estimates:
p
0.16 SOLUTION: With a confidence level of 95% the proportion will be between .124 and .204
- Do the intervals overlap? What, if anything, can you conclude about the effectiveness of the drug?
SOLUTION: This intervals overlap but in such a small portion that it will not be relevant when evaluating the drug.
- Compute 95% confidence interval for the difference in proportions of students who break out in hives by using or not using this drug and give a sentence interpreting this interval.
i = c(350, 350)
t = c(34, 56)
prop.test(t, i)
2-sample test for equality of proportions with continuity
correction
data: t out of i
X-squared = 5.623, df = 1, p-value = 0.01773
alternative hypothesis: two.sided
95 percent confidence interval:
-0.11508783 -0.01062645
sample estimates:
prop 1 prop 2
0.09714286 0.16000000 SOLUTION: With a confidence level of 95% the proportion will be between -.115 and -.011
Prob8 - An article in the March 2003 New England Journal of Medicine describes a study to see if aspirin is effective in reducing the incidence of colorectal adenomas, a precursor to most colorectal cancers (Sandler et al. (2003)). Of 517 patients in the study, 259 were randomly assigned to receive aspirin and the remaining 258 received a placebo. One or more adenomas were found in 44 of the aspirin group and 70 in the placebo group. Find a 95% one-sided upper bound for the difference in proportions \((p_A - p_P)\) and interpret your interval.
# Your code here
g = c(44, 70)
t = c(259, 258)
prop.test(x = g, n = t, correct = FALSE, conf.level = .95, alternative = "greater")$conf[1] -0.1609853 1.0000000
attr(,"conf.level")
[1] 0.95SOLUTION: With a confidence level of 95% the proportion will be between -.161 and 1
Prob9 - The data set Bangladesh has measurements on water quality from 271 wells in Bangladsesh. There are two missing values in the chlorine variable. Use the following R code to remove these two observations.
> chlorine <- with(Bangladesh, Chlorine[!is.na(Chlorine)])
Bangladesh <- read_csv("http://www1.appstate.edu/~arnholta/Data/Bangladesh.csv")Parsed with column specification:
cols(
Arsenic = col_double(),
Chlorine = col_double(),
Cobalt = col_double()
)chlorine = with(Bangladesh, Chlorine[!is.na(Chlorine)])- Compute the numeric summaries of the chlorine levels and create a plot and comment on the distribution.
quantile(chlorine) 0% 25% 50% 75% 100%
1.0 5.0 14.2 55.5 1550.0 hist(chlorine)
SOLUTION: The distrution is skewed to the right
- Find a 95% \(t\) confidence interval for the mean \(\mu\) of chlorine levels in Bangladesh wells.
t.test(chlorine)
One Sample t-test
data: chlorine
t = 6.0979, df = 268, p-value = 3.736e-09
alternative hypothesis: true mean is not equal to 0
95 percent confidence interval:
52.87263 103.29539
sample estimates:
mean of x
78.08401 SOLUTION: With a confidence level of 95% the proportion will be between 52.87 and 103.29
- Find a 95% bootstrap percentile and bootstrap \(t\) confidence intervals for the mean chlorine level and compare results. Which confidence interval will you report?
SOLUTION: The bootstrap confidence interval is between 77.75 and 78.26.
- Johnson’s \(t\) confidence interval adjusts for skewness by shifting endpoints right or left for positive or negative skewness, respectively. The interval is \(\bar{X} + \hat{\kappa_3}/(6\sqrt{n})(1 + 2q^2) \pm q(S/\sqrt{n})\), where \(\hat{\kappa_3}\) is a sample estimate of the population skewness \(E(X - \mu)/\sigma^3\) and \(q\) denotes the \(1 - \alpha/2\) quantile for a \(t\) distribution with \(n-1\) degrees of freedom. Calculate Johnson’s \(t\) interval for the arsenic data (in
Bangladesh) and compare with the formula \(t\) and bootstrap \(t\) intervals.
SOLUTION: We’re 95% confident that the mean is between 124.8 and 125.51.
Prob10 - The data set MnGroundwater has measurements on water quality of 895 randomly selected wells in Minnesota.
MnGroundwater <- read_csv("http://www1.appstate.edu/~arnholta/Data/MnGroundwater.csv")Parsed with column specification:
cols(
County = col_character(),
Aquifer.Group = col_character(),
Water.Level = col_integer(),
Alkalinity = col_integer(),
Aluminum = col_double(),
Arsenic = col_double(),
Chloride = col_integer(),
Lead = col_double(),
pH = col_double(),
Basin.Name = col_character()
)head(MnGroundwater)# A tibble: 6 x 10
County Aquifer.Group Water.Level Alkalinity Aluminum Arsenic
<chr> <chr> <int> <int> <dbl> <dbl>
1 Aitkin surficial Quaternary 55 137000 0.059 1.810
2 Aitkin buried Quaternary 30 214000 2.380 0.059
3 Aitkin buried Quaternary 20 120000 0.410 1.440
4 Aitkin buried Quaternary 3 283000 158.190 6.340
5 Aitkin buried Quaternary 0 236000 0.059 10.170
6 Aitkin buried Quaternary 30 229000 0.059 6.900
# ... with 4 more variables: Chloride <int>, Lead <dbl>, pH <dbl>,
# Basin.Name <chr>- Create a histogram, a density, and a normal quantile plot of the alkalinity and comment on the distribution.
hist(MnGroundwater$Alkalinity)
qqnorm(MnGroundwater$Alkalinity)
mean(MnGroundwater$Alkalinity)[1] 290682.7sd(MnGroundwater$Alkalinity)[1] 108334.3
SOLUTION:
- Find the 95% \(t\) confidence interval for the mean \(\mu\) of alkalinity levels in Minnesota wells.
t.test(MnGroundwater$Alkalinity)
One Sample t-test
data: MnGroundwater$Alkalinity
t = 80.272, df = 894, p-value < 2.2e-16
alternative hypothesis: true mean is not equal to 0
95 percent confidence interval:
283575.6 297789.8
sample estimates:
mean of x
290682.7 SOLUTION: We’re 95% confident that the mean alkalinity is between 283575.6 and 297789.8
- Find the 95% bootstrap percentile and bootstrap \(t\) confidence intervals for the mean alkalinity level and compare the results. Which confidence interval will you report?
SOLUTION: The bootstrap t confidence interval is more accurate and is the best method.
Prob11 Consider the babies born in Texas in 2004 (TXBirths2004). We will compare the weights of babies born to nonsmokers and smokers.
Texas <- read_csv("http://www1.appstate.edu/~arnholta/Data/TXBirths2004.csv")Parsed with column specification:
cols(
ID = col_integer(),
MothersAge = col_character(),
Smoker = col_character(),
Gender = col_character(),
Weight = col_integer(),
Gestation = col_integer(),
Number = col_integer(),
Multiple = col_character()
)head(Texas)# A tibble: 6 x 8
ID MothersAge Smoker Gender Weight Gestation Number Multiple
<int> <chr> <chr> <chr> <int> <int> <int> <chr>
1 1 20-24 No Male 3033 39 1 No
2 2 20-24 No Male 3232 40 1 No
3 3 25-29 No Female 3317 37 1 No
4 4 25-29 No Female 2560 36 1 No
5 5 15-19 No Female 2126 37 1 No
6 6 30-34 No Female 2948 38 1 No- How many nonsmokers and smokers are there in this data set?
smokers <- filter(Texas, Texas$Smoker == "Yes")
nonsmokers <- filter(Texas, Texas$Smoker == "No")
count(smokers)# A tibble: 1 x 1
n
<int>
1 90count(nonsmokers)# A tibble: 1 x 1
n
<int>
1 1497SOLUTION: There are 1497 nonsmokers & 90 smokers.
- Create exploratory plots of the weights for the two groups and comment on the distributions.
ggplot(data = smokers, aes(sample = Weight)) + stat_qq() + theme_bw() +
ggtitle("Smokers")
ggplot(data = nonsmokers, aes(sample = Weight)) + stat_qq() + theme_bw() + ggtitle("Non-Smokers")
smokersExp <- c(mean(Texas$Weight[Texas$Smoker == "Yes"]), sd(Texas$Weight[Texas$Smoker == "Yes"]))
smokersExp[1] 3205.9889 504.2439nonSmokersExp <- c(mean(Texas$Weight[Texas$Smoker == "No"]), sd(Texas$Weight[Texas$Smoker == "No"]))
nonSmokersExp[1] 3287.4937 554.4829
SOLUTION: Smokers is normal shaped with a mean of 3205.9889 and a standard deviation of 504.2439. Nonsmokers is left skewed with a mean of 3287.4937 and a standard deviation of 554.4829.
- Compute the 95% confidence interval for the difference in means using the formula \(t\), bootstrap percentile, and bootstrap \(t\) methods and compare your results. Which interval would you report?
SE <- sqrt(var(smokers$Weight)/length(smokers$Weight) +
var(nonsmokers$Weight)/length(nonsmokers$Weight))
thetaHat <- mean(smokers$Weight) - mean(nonsmokers$Weight)
x <- length(smokers$Weight)
y <- length(smokers$Weight)
N <- 10^4
TStar <- numeric(N)
DM <- numeric(N)
for(i in 1:N)
{
xBS <- sample(smokers$Weight, x, replace=TRUE)
yBS <- sample(nonsmokers$Weight, y, replace=TRUE)
TStar[i] <- (mean(xBS) - mean(yBS) - thetaHat) /
sqrt(var(xBS)/x + var(yBS)/y)
DM[i] <- mean(xBS) - mean(yBS)
}
Boot <- thetaHat - quantile(TStar, c(.975, .025)) * SE
Boot 97.5% 2.5%
-187.27412 30.70641 Boot2 <- quantile(DM, c(0.025, 0.975))
Boot2 2.5% 97.5%
-233.87889 74.95667 t.test(smokers$Weight,nonsmokers$Weight)
Welch Two Sample t-test
data: smokers$Weight and nonsmokers$Weight
t = -1.4806, df = 102.38, p-value = 0.1418
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-190.69149 27.68196
sample estimates:
mean of x mean of y
3205.989 3287.494 SOLUTION: Report the Bootstrap interval.
- Modify your result from the previous question to obtain a one-sided 95% \(t\) confidence interval (hypothesizing that babies born to nonsmokers weigh more than babies born to smokers).
Boot <- thetaHat - quantile(TStar, .975) * SE
Boot 97.5%
-187.2741 test <- t.test(smokers$Weight, nonsmokers$Weight, alternative="greater")
test
Welch Two Sample t-test
data: smokers$Weight and nonsmokers$Weight
t = -1.4806, df = 102.38, p-value = 0.9291
alternative hypothesis: true difference in means is greater than 0
95 percent confidence interval:
-172.8809 Inf
sample estimates:
mean of x mean of y
3205.989 3287.494 SOLUTION: We are 95% sure that the difference in the mean is greater than -172.88094.