Homework Questions

Chapter 4 Foundations for Inference

Practice: 4.3, 4.13, 4.23, 4.25, 4.39, 4.47

Graded: 4.4, 4.14, 4.24, 4.26, 4.34, 4.40, 4.48

4.4

1. The point estimate is the mean which is 171.1 cm, the median is 170.3
2. the point estimate for the stddev is 9.4 cm, the IQR is 177.8 - 163.8 = 14cm c.I would say that the 180cm persion is not unusually tall as they fall within one standard deviation of the mean, however the person that is 155cm tall, I would consider this person this person to be unusually short because they are more than one standard deviation shorter than the mean
3. I would expect the mean and standard deviation of the sample of physically active humans heights to be similar, although I am very interested to see if this number may differ e.The standard error would be 9.4/507 = .019

4.14

1. FALSE, Having a 95% Confidence Interval would mean that we are 95% sure that the average consumer spending of all americans on Black friday weekend falls between 80.31 and 89.11, assuming that this is an unbiased sample
2. FALSE, The distribution is not skewed to the right
3. FALSE, the average mean for the population as a whole has a 95% chance of being in that confidence interval
4. TRUE
5. TRUE, a 90% confindence interval would be narrower because we would be less sure that the average would fall inside of it.
6. FALSE, we would need 9 times the sample size to cut the margin or error in half
7. TRUE, the margin of error is half the width of the interval

4.24

1. Yes the conditions for inference are satisfied the sample size is at least 30 and the population is normally distributed
2. We define a null and alternate hypothesis: Ho: average for gifted children to count to 10 is 32 months Ha: average for gifted children to count to 10 is lower than 32 months

samp_mean<-30.69
samp_sd<-4.31
n<-36
mu<-32

#Calculate the standard error
SE<-samp_sd/sqrt(n)
SE

## [1] 0.7183333

#Calculate the Z Score
z<-(mu-samp_mean)/SE
z

## [1] 1.823666

#Calculate the P-value
pvalue <- 1-pnorm(32,mean = 30.69, sd=SE)
pvalue

## [1] 0.0341013

2. cont.. Since the pvalue is below the .10 we reject the null hypothesis

3. The pvalue represents the probability that the average age a gifted child will learn to read is 32 months or more and the probability of this is only 3.4% which is within our constraints of significance

4. The 95% confidence interval is between 29.51 and 31.87

Z=qnorm(0.95)
samp_mean + Z*SE

## [1] 31.87155

samp_mean - Z*SE

## [1] 29.50845

5. Yes my results from my hypothesis test and my confidence interval do agree because there is only a 5% chance that the gifted child mean for reading is over 31.87 and only a 3.4% chance it is over 32 months

4.26

1. Hypothesis test to see if the IQ of gifted children’s mothers is higher than that of the average human Ho: IQ of moms = 100 Ha: IQ of moms > 100

mean = 118.2
sd = 6.5
SE= sd/sqrt(36)
pnorm(100,mean = 118.2, sd=SE)

## [1] 1.22022e-63

a.cont… the probability of the null hypothesis being true is almost completely zero, this beats the level of significance needed

Z=qnorm(0.95)
mean + Z*SE

## [1] 119.9819

mean - Z*SE

## [1] 116.4181

2. cont… the range is from 116.4 to around 12o
3. my results from the hypothesis test do match up, both prove that the average level of intelligence for mothers of gifted children is higher than that of the average human

4.34

The sampling distribution of the mean is the probability distribution of all possible samples of size n drawn from a population. This distribution is a normal distribution centered, that is with a mean equal to the mean of the population, as long as the population distribution is nor highly skewed rihgt or left. As the sample size increases, the shape becomes more normal, center remains the same (as long as the size is sufficient), and spread becomes narrower.

4.40

1. the probability of a randomly chosen bulb lasts more than 10,500 hours is 6.7%

1-pnorm(10500,9000,1000)

## [1] 0.0668072

b.the standard erro is 258 hours

samp_mean<-9000
n<-15
SE<-1000/sqrt(15)
SE

## [1] 258.1989

c.the probability is almost zero

1-pnorm(10500,9000,258)

## [1] 3.050719e-09

mean=9000; sd=1000

Popx <- seq(-4,4,length=100)*sd + mean
Poph <- dnorm(Popx,mean,sd)

mean=9000; sd=258
Samx <- seq(-4,4,length=100)*sd + mean
Samh <- dnorm(Samx,mean,sd)

library(data.table)
bulbdata = as.data.table(cbind(Popx,Poph,Samx,Samh))

library(ggplot2)
ggplot(data=bulbdata, aes(Popx,Poph)) + geom_line() +
geom_line(data=bulbdata, aes(Samx,Samh)) 

5. It is much harder to estimate probability when there is a skewed distribution

4.48

P-value is going to decrease since larger sample size will result in smaller Standard Error. the standard error decreases since it is inversely proportional the the square root of n, remember SE=sd/sqrt(n)