Chapters 9 and 10 homework

library(openintro)

9.2.5

339 infants, 69 had reactions

1. 95% confidence interval ˜p = 71/343 = 0.2069970845 SEp = sqrt (0.2069970845(1-0.2069970845)/(343)) = sqrt 0.000478569363 = 0.02187622826

0.2069970845+(1.96*0.02187622826)

## [1] 0.2498745

0.2069970845-(1.96*0.02187622826)

## [1] 0.1641197

2. We are 95% confidence that the probability of having ad adverse reaction to the whooping cough vaccine is between 0.1641197 and 0.2498745.

3. We can be mostly confident the probability of an adverse reaction is below 0.25

4. This is a one-sided confidence level

9.4.4

932 births in 20 consecutive weeks, 216 on weekends, 716 weekdays birth frequency weekend: 932 * 2/7 = 266.2857142857 birth frequency weekday: 932* 5/7 = 665.7142857143

birthmatrix<-matrix(c(216, 266, 716, 666),ncol=2,byrow=TRUE)
rownames(birthmatrix)<-c("weekend","weekday")
colnames(birthmatrix)<-c("observed","expected")
birthmatrix <- as.table(birthmatrix)
birthmatrix

##         observed expected
## weekend      216      266
## weekday      716      666

null.probs = c(266/932, 666/932)
birthdist = c(216,716)
chisq.test(birthdist, p=null.probs)

## 
##  Chi-squared test for given probabilities
## 
## data:  birthdist
## X-squared = 13.152, df = 1, p-value = 0.0002872

1. X-squared = 13.152

2. Null hypothesis: there is no difference in the timing of births Alternate hypothesis: there is a difference in the timing of births between weekdays and weekends.

3. The p-value is less than the level of significance so we can reject the null hypothesis.

9.4.5

190 * 9/16 = 106.875 190 * 3/16 = 35.625 190 * 3/16 = 35.625 190 * 1/16 = 11.875

birdmatrix <-matrix(c(111,107,27,36,34,36,8,12),ncol=2,byrow=TRUE)
rownames(birdmatrix)<-c("whitefeathersmallcomb","whitefeatherlargecomb", "darkfeathersmallcomb", "darkfeatherlargecomb")
colnames(birdmatrix)<-c("observed","expected")
birdmatrix <- as.table(birdmatrix)
birdmatrix

##                       observed expected
## whitefeathersmallcomb      111      107
## whitefeatherlargecomb       27       36
## darkfeathersmallcomb        34       36
## darkfeatherlargecomb         8       12

null.probs = c(106.875/190, 35.625/190,35.625/190,11.875/190)
birddist = c(111,37,34,8)
chisq.test(birddist, p=null.probs)

## 
##  Chi-squared test for given probabilities
## 
## data:  birddist
## X-squared = 1.5509, df = 3, p-value = 0.6706

X-squared = 1.5509 p-value = 0.6706

We cannot reject the null hypothesis that there is no deviation from the Mendelian expected ratio.

10.2.5

wilt <- c(11,17) 
nowilt <- c(15,4)
test <- data.frame(wilt,nowilt)
chisq.test(test)

## 
##  Pearson's Chi-squared test with Yates' continuity correction
## 
## data:  test
## X-squared = 5.6885, df = 1, p-value = 0.01708

1. Mite infestation has no effect on the chance of getting wilt disease.

2. Mite infestation effects rates of wilt disease.

3. Wilt & mites = 11(11/26) No wilt & mites = 15(15/26) Wilt & no mites = 17(17/21) No wilt & no mites = 4(4/21)

4. X-squared = 5.6885, df = 1, p-value = 0.01708

5. The p-value is greater than the degree of significance so we cannot reject the null hypothesis.

10.2.7

seizures <- c(14,11) 
noseizures <- c(6,6)
test1 <- data.frame(seizures,noseizures)
chisq.test(test1)

## 
##  Pearson's Chi-squared test with Yates' continuity correction
## 
## data:  test1
## X-squared = 2.6469e-31, df = 1, p-value = 1

1. Null hypothesis: there is no difference in number of seizures between patients that take phenoytoin and valproate. Alternative hypothesis: there is a difference between number of seizures in patients taking phenoytoin and valproate.

2. X-squared = 2.6469e-31, df = 1, p-value = 1

3. The p-value is larger than the degree of significance so we cannot reject the null hypothesis.

2. These drugs are equally effective at preventing seizures.

10.3.3

died <- c(83,106)
alive <- c(264,242)
test2 <- data.frame(died, alive)
test2

##   died alive
## 1   83   264
## 2  106   242

chisq.test(test2)

## 
##  Pearson's Chi-squared test with Yates' continuity correction
## 
## data:  test2
## X-squared = 3.431, df = 1, p-value = 0.06399

1. Pr {D|S} = 83/347 = 0.2391930836 Pr {D|WW} = 106/347 = 0.3054755043

2. The p-value is larger than the level of significance (0.05) so we cannot reject the null hypothesis.

10.9.1

1. (25/(25+23)) / (492/(492+614)) = 0.5208333333/0.4448462929 = 1.1708163957
2. 25 * 614 / 23 * 492 = 15,350/11,316 = 1.356486391

1. (12/(12+8)) / (93/(93+84)) = 0.6/0.5254237288 = 1.1419354839
2. 12 * 84 / 8 * 93 = 1,008/744 = 1.3548387097

10.9.3

3995/(3995 + 221) / 42946/(42946+5007) = 0.9475806452/0.8955852606 = 1.0580574367

10.9.4a

1. 3995 * 5007 / 42946 * 221 = 20,002,965/9,491,066 = 2.1075572544

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