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

Q4.4

(a) What is the point estimate for the average height of active individuals? What about the median? (See the next page for parts (b)-(e).)

Ans: Point estimate of the population mean = 171.1.

Median = 175; the median is the observation right in the middle. If there are an even number of observations, there will be two values in the middle, and the median is taken as their average.

(b) What is the point estimate for the standard deviation of the heights of active individuals? What about the IQR?

Ans: 9.4 IQR: Q3-Q1=177.8-163.8=14

(c) Is a person who is 1m 80cm (180 cm) tall considered unusually tall? And is a person who is 1m 55cm (155cm) considered unusually short? Explain your reasoning.

Ans: Yes, it is above Q3. Yes, it is below Q1.

(d) The researchers take another random sample of physically active individuals. Would you expect the mean and the standard deviation of this new sample to be the ones given above? Explain your reasoning.

Ans: Yes. It should be the same mean and same standard deviation if the random sample size is big enought and new sample are random sample.

(e) The sample means obtained are point estimates for the mean height of all active individuals,if the sample of individuals is equivalent to a simple random sample. What measure do we use to quantify the variability of such an estimate Compute this quantity using the data from the original sample under the condition that the data are a simple random sample.

Ans: point estimation measure.

Q4.14

(a) We are 95% confident that the average spending of these 436 American adults is between $80.31 and $89.11.

Aas:False, it shoule between 82.89 abd 86.53.

l<-84.71-(80.31-84.71)^2/(sqrt(436))*1.96
r<-84.71+(80.31-84.71)^2/(sqrt(436))*1.96
c(l,r)
## [1] 82.89274 86.52726

(b) This confidence interval is not valid since the distribution of spending in the sample is right skewed.

Ans: True,since the mode is on the left side.

(c) 95% of random samples have a sample mean between $80.31 and $89.11.

Ans:True, since sample mean = 84.71 which is between 80.31 and 89.11.

(d) We are 95% confident that the average spending of all American adults is between $80.31 and $89.11.

Ans: False, 95% confident is between 82.89 and 86.53.

(e) A 90% confidence interval would be narrower than the 95% confidence interval since we don’t need to be as sure about our estimate.

Ans:Yes, since the SE of 90% is lesser than 95%.

(f) In order to decrease the margin of error of a 95% confidence interval to a third of what it is now, we would need to use a sample 3 times larger.

Ans: True, since z is reduced, and new margin of error is less than the old one.

#margin of error= z*SE
old<-(80.31-84.71)^2/(sqrt(436))*1.96
new<-(80.31-84.71)^2/(sqrt(436*3))*1.15
c(old,new)
## [1] 1.817265 0.615601

(g) The margin of error is 4.4.

Ans: False, it should be 1.82.

4.24

(a) Are conditions for inference satisfied?

Ans:Yes.

(b) Suppose you read online that children first count to 10 successfully when they are 32 months old, on average. Perform a hypothesis test to evaluate if these data provide convincing evidence that the average age at which gifted children fist count to 10 successfully is less than the general average of 32 months. Use a significance level of 0.10.

Ans: H0: u=32; H1: u<32

z<-(32-30.69)/(4.31/sqrt(36))
pnorm(-abs(z))
## [1] 0.0341013

(c) Interpret the p-value in context of the hypothesis test and the data.

Ans: 0.1 is the significance level. If average of the gifted children is 32 months is TRUE, the p-value should be greter than 0.05, otherwise, H0 will be rejected.

(d) Calculate a 90% confidence interval for the average age at which gifted children first count to 10 successfully.

#90% confident population inerval
z<-qnorm(.95)
left<-30-z*4.31/sqrt(36)
right<-30+z*4.31/sqrt(36)
c(left,right)
## [1] 28.81845 31.18155

(e) Do your results from the hypothesis test and the confidence interval agree? Explain.

#No,I don't agree for both. 
#For the hypothesis test, since p=0.0341<0.05, so we reject H0: u=32.
#For 90% confidence sample interval 32 is greater than 31.18155.

Q4.26

(a) Perform a hypothesis test to evaluate if these data provide convincing evidence that the average IQ of mothers of gifted children is different than the average IQ for the population at large, which is 100. Use a significance level of 0.10.

Ans: H0: u=100; HA: U != 100

z<-(100-118.2)/(6.5/sqrt(36))
pnorm(-abs(z))
## [1] 1.22022e-63

Ans: since p is close to 0, we reject H0: u=100.

(b) Calculate a 90% confidence interval for the average IQ of mothers of gifted children.

#90% confident population interval:
z<-qnorm(.95)
left<-118.2-z*6.5/sqrt(36)
right<-118.2+z*6.5/sqrt(36)
c(left,right)
## [1] 116.4181 119.9819

(c) Do your results from the hypothesis test and the confidence interval agree? Explain.

#No,I don't agree for both. 
#For the hypothesis test, since p=0<0.05, so we reject H0: u=100.
#For 90% confidence interval, 100 < 116.4181  which is out of boudary.

Q4.34

4.34 CLT. Define the term “sampling distribution” of the mean, and describe how the shape, center, and spread of the sampling distribution of the mean change as sample size increases.

#sample distribution of the mean is close to the population mean when sample size increases. 
#The belt shape curve also becomes more symetric and the peak of the central also higher.

Q.40

  1. What is the probability that a randomly chosen light bulb lasts more than 10,500 hours?
pnorm(10500,mean = 9000,sd=1000,lower.tail = F,log.p = F)
## [1] 0.0668072
  1. Describe the distribution of the mean lifespan of 15 light bulbs.
#since size=15 <30, the distribution would be extreamly skew.
  1. What is the probability that the mean lifespan of 15 randomly chosen light bulbs is more than 10,500 hours?
z<-(1050-9000)/(6.5/sqrt(15))
pnorm(-abs(z))
## [1] 0
  1. Sketch the two distributions (population and sampling) on the same scale.
par(mfrow = c(2, 1))

hist(rnorm(n=1000, mean=9000, sd=1000),breaks = 20)
hist(rnorm(n=50, mean=9000, sd=1000),breaks = 20)

  1. Could you estimate the probabilities from parts (a) and (c) if the lifespans of light bulbs had a skewed distribution?

Ans: sample size increse, z will increase, then p-value will decrese.In (a), the probability picking a light bulbs which longer than 10500 hrs is 0.0668. In (c) picking 15 light bulbs in which has only 1 light bulbs would be longer than 10500hrs and the rest are less than 10500, the mean of 15 light bulbs will have 0 probability greater than 10500 hrs.

Q.4.48 Same observation, different sample size. Suppose you conduct a hypothesis test based on a sample where the sample size is n = 50, and arrive at a p-value of 0.08. You then refer back to your notes and discover that you made a careless mistake, the sample size should have been n = 500. Will your p-value increase, decrease, or stay the same? Explain.

Ans: Since sample size increse, z will increase, then p-value will decrese.For example:

z<-qnorm(.92)
z
## [1] 1.405072
a<-z/sqrt(50)
z1<-a*sqrt(500)
z1
## [1] 4.443226
pnorm(-z1)
## [1] 4.430991e-06