- Problem 4.4.
- Point estimate for average height is mean of sample distribution, or 171.1
- Point estimate for standard deviation is sample standard deviation of sample or 9.4, IQR will be Q3-Q1 or 14.
- I would not consider a 1m80cm person unusually tall, it is just a little bit over Q3 value of 177.8 or in another words ~25% of people are taller than this person. A 1m55cm person does appear to be unusaully short. He or she is ~2 SD’s away from the mean, 171.1-2*9.4 or 152.3 which is very close to 155. For normal distribution which this one looks to be, 2 SD’s away will be ~95%, or only ~2.5% of people are shorter than the person.
- It most likely will not be exactly the same but it should be reasonably close. It is very unlikely to pull the sample data that will have exactly the same mean/SD. There will be some slight variations from sample to sample.
- SD of mean of the sample is SD/sqrt(N) or 9.4/507^0.5 or 0.4174687
9.4/507**.5
## [1] 0.4174687
- Problem 4.14.
- Wrong. 95% interval applies to our population, not our sample.
- Wrong. Our sample is very big - 436 adults, so some right skewedness is not enough to disqualify the sample.
- Wrong. We can not say with certancy that this will be the case. This is our approximation.
- Correct.
- Correct.
- Wrong. It should be 9 times bigger. Squire root of 9 is 3.
- Correct
- Problem 4.24
- Yes. DIstribution is not strongly skewed.
- Our hypotesis is that 32 falls in 90% confidence interval. Mean+SE/sqrt(N)xZ or 30.69+4.31xZ/sqrt(36) or 31.87166. We will have to reject the hypotesis.
30.69+4.31*1.645/6
## [1] 31.87166
- P value of 0.05.If our value of 32 corrsponds to 1-p value then we reject the hypothesis, if it is less than we accept it.
- 30.69-4.31xZ/sqrt(36) or 29.50853 to 31.87166
30.69-4.31*1.645/6
## [1] 29.50834
- It agrees. The interval upper boundry is less than 32
- Problem 4.26
- Hypotesis is that 100 will falls in 90% confidence interval. Mean+SE/sqrt(N)xZ or 118.2-6.5xZ/sqrt(36) or 31.87166. We will have to reject the hypotesis.
118.2-6.5*1.645/6
## [1] 116.4179
- 31.87166 to 118.2+6.5xZ/sqrt(36) or 119.9821
118.2+6.5*1.645/6
## [1] 119.9821
- Problem 4.34
- If we take infinitive number of independent samples from population, the means of these samples will create the sampling distribution ofb the mean. The shape will become more centered, narrow. The mean will be the same. SD will become smaller as size increases.
- Problem 4.40
- z will be (10,500-9,000)/1,000 or 1.5. It means that for bulb to last more that 10,500 is ~7%
- Normal with mean of 10,500 and SD of 1,000/SQRT(15) or 258.1989
- (10,500-9,000)/258.1989 or 5.8095. It means that average to be more 10,500 is practically 0%
pnorm(1.5)
## [1] 0.9331928
1000/(15**.5)
## [1] 258.1989
1500/258.1989
## [1] 5.809475
pnorm(5.8095)
## [1] 1
mean<-10500
sd<-1000
x <- seq(7000, 14000, by = 10)
# Choose the mean as 2.5 and standard deviation as 0.5.
y <- dnorm(x, mean = 10500, sd = sd)
plot(x,y,col="red")
mean1<-10500
sd1<-258.1989
x1 <- seq(8000, 13000, by = 10)
# Choose the mean as 2.5 and standard deviation as 0.5.
y1 <- dnorm(x1, mean = 10500, sd = sd1)
lines(x1,y1,col="green")
* No, we could not. For (a) we need to know distribution, and for (c) we could not assume normality for sample size of 15 and skewed
- Problem 4.48
- P value will dicrease. If we increase N - sample size, then our distribution will become narrower with smaller Standard Deviation SD/sqrt(50) vs SD/sqrt(500), as a result, the same value will have smaller P value.