Problem 3

a

0.00135

mean<-120

sd<-10

z<-(150-mean)/sd

1-pnorm(z)
## [1] 0.001349898

b

0.20366

N_sqrt<-16**0.5

z1<-(126-120)*N_sqrt/sd

z2<-(122-120)*N_sqrt/sd

pnorm(z1)-pnorm(z2)
## [1] 0.2036579

Problem 4

a

0.061662

n_k<-choose(7,3)

n_k
## [1] 35
n_k*0.15**3*0.85**4
## [1] 0.06166199

b

0.7581

n<-100

p<-0.15

std<-(n*p*(1-p))**0.5

mean<-p*n

z<-(17.5-mean)/std

pnorm(z)
## [1] 0.7580801

Problem 5

a

{4.55325,5.04675}

mean<-4.8

std<-1.5

n<-100

sample_std<-1.5/(n**0.5)

ci_l<-mean-sample_std*1.645

ci_u<-mean+sample_std*1.645

ci_l
## [1] 4.55325
ci_u
## [1] 5.04675

b

{0.46202,0.51798}

mean<-0.49

n<-1226

std<-(mean*(1-mean))**0.5

sample_std<-std/(n**0.5)

ci_l<-mean-sample_std*1.96

ci_u<-mean+sample_std*1.96

ci_l
## [1] 0.462017
ci_u
## [1] 0.517983

Problem 6

a

30|0 20|0 10|00 5|00 4|00 3|00 2|0 -5|0 -6|0 -7|0 -9|9 -12|0

b

Mean is 35.06. The range is 420.

c

Minimum is -120. Maximum is 300. Median is 35. 1st Q is -52.5. 3rd Q is 62.5.

vec<-c(-120,-99,-70,-60,-50,20,30,30,40,40,50,50,100,100,200,300)

quantile(vec)
##     0%    25%    50%    75%   100% 
## -120.0  -52.5   35.0   62.5  300.0

d

Distribution is right skewed.

boxplot(vec)

e

300 is a suspected outlier.

IQR<-62.5+52.5

35+1.5*IQR
## [1] 207.5
35-1.5*IQR
## [1] -137.5

f

The number of observations is less than 30, so we have to use t distribution with 15 DF. Data is skewed and we have a potential outlier, so we might have a problem with accuracy.

H0: mean is 0 H1: mean is >0

Z score of -1.295 is more than t value at 0.05 at -1.753, so we cannot reject H0 at 0.05 significance level. We do mot have enough evidence to accept the hypothesis that mean is more than 0 at 0.05 significance level.

std<-sd(vec)

hist(vec)

n<-16

sample_std<-std/(n**0.5)

sample_std
## [1] 27.07481
mean<-mean(vec)

z<-(-1)*mean/sample_std

z
## [1] -1.295023
qt(0.05, df=15)
## [1] -1.75305

Problem 7

Degrees of freedom is 4 (the lowest number of observations minus one)

H0: The right and left lane cars have the same speed

H1: The left lane cars are faster

At 0.01, we cannot reject H0. So, we do not have enough evidence to claim that the right lane is slower than the left.

mean1<-65

mean2<-69

meand<-mean2-mean1

std1<-4.12

std2<-3.22

stdd<-(std1**2/5+std2**2/6)**0.5

stdd
## [1] 2.263393
z<-(0-meand)/stdd

z
## [1] -1.767258
qt(0.01, df=4)
## [1] -3.746947

Problem 8

a

H0: The proportion of adults who has heart attack is the same for active and inactive population

H1: The proportion is not the same

We got p value of 0.26, so we do not have enough evidence to reject H0, which means that we do not have enough evidence to claim that proportion is not the same

b

{-0.07644,0.12644}

HAA<-10

HAIA<-25

AA<-100

IAA<-200

Pooled<-(HAA+HAIA)/(AA+IAA)

Pooled
## [1] 0.1166667
std<-(Pooled*(1-Pooled)/IAA+Pooled*(1-Pooled)/AA)**0.5

std
## [1] 0.03931709
meand<-HAIA/IAA-HAA/AA

meand
## [1] 0.025
z<-(0-meand)/std

z
## [1] -0.6358559
pnorm(z)
## [1] 0.2624352
ci_l<-meand-std*2.58

ci_u<-meand+std*2.58

ci_l
## [1] -0.07643808
ci_u
## [1] 0.1264381

Exercise 9

Degree of freedom is 2.

H0: There is no association

H1: There is an association

We reject H0 for 0.05 confidence and 2 DF (chi squire for 0.05 and 2 DF is 5.991). So we have enough evidence to say there is an association.

chi_sqr<-(11-(19*60/111))**2/(19*60/111)+(28-(60*52)/111)**2/(60*52)/111+(21-60*40/111)**2/(60*40)/111+(8-19*51/111)**2/(19*51)/111+(24-52*51)**2/(51*52)/111+(19-51*40/111)**2/(51*40)/111

chi_sqr
## [1] 23.51327