Question 4.4

(a) What is the point estimate for the average height of active individuals? What about the median?
     Point estimate of mean = 171.1 and median = 170.3
(b) What is the point estimate for the standard deviation of the heights of active individuals? What about the IQR?
     SD height of active individuals is 9.4
     IQR = 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.

z1 <- (180 - 171.1)/9.4
z1

[1] 0.9468085

z2 <- (155 - 171.1)/9.4
z2

[1] -1.712766
     They are not unusual, the limit is with in 2 and -2
(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.
     Might be slight change in mean and SD because of 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 (Hint: recall that SD¯x = ! pn )? Compute this quantity using the data from the original sample under the condition that the data are a simple random sample.
     SD/sqrt(n) = 9.4/sqrt(507) = 0.42

Question 4.14

(a) We are 95% confident that the average spending of these 436 American adults is between $80.31 and $89.11.
     False, the 95% confidence interval is in range population,not on a sample size
(b) This confidence interval is not valid since the distribution of spending in the sample is right skewed.
     False, the sample size is 436 larger than 30, skewed right
(c) 95% of random samples have a sample mean between $80.31 and $89.11.
     False, Confidence interval is for population, not for other samples (d) We are 95% confident that the average spending of all American adults is between $80.31 and $89.11.
     True, it is 95% confidence interval that population mean is in the interval.
(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.
     True, 90% confidence interval is slender 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.
     False
(g) The margin of error is 4.4.

e1 <- (89.11-80.31)/2
e1

[1] 4.4

Question 4.24

(a) Are conditions for inference satisfied?
     Yes, Interference are satisfied because the sample data is from large city and independent
(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.
     H0: u=32
     H1: u<32

mean<-30.69
sd<-4.31
n<-36

se<-sd/sqrt(n)
z1<-(mean-32)/se

p <- pnorm(z1)
p

[1] 0.0341013
(c) Interpret the p-value in context of the hypothesis test and the data.
     0.1 is the significance level,the p-value should be greater than 0.05, else H0 would be rejected.
(d) Calculate a 90% confidence interval for the average age at which gifted children first count to 10 successfully.

z<-abs(qnorm(.1))
lwr <- mean - (z * se)
upr <- mean + (z * se)
lwr

[1] 29.76942

upr
[1] 31.61058
(e) Do your results from the hypothesis test and the confidence interval agree? Explain.
     Yes, Null Hypothesis u=32.
Question 4.26

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

mean = 118.2
n = 36
sd = 6.5
se = sd/sqrt(n)

z1 = (mean -100)/se
pnorm(z1)

[1] 1
     We can reject the null hypothesis, The average IQ of the mother is varies than the population
(b) Calculate a 90% confidence interval for the average IQ of mothers of gifted children.

lwr = mean - 1.645*se
upr = mean + 1.645*se
lwr

[1] 116.4179

upr

[1] 119.9821


(c) Do your results from the hypothesis test and the confidence interval agree? Explain.
     Yes, the average IQ of mothers of gifted children is between 116.4125 and 119.9875 which is above average of 100

Question 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.
     The sampling distribution of the mean is the mean of the population from which the scores were sampled, also the mean is close to the population mean when sample size increases,we can calibrate our sample mean, sd, se closer the the real, or population equivalents. The bell curve is used to describe a graphical depiction of a normal probability distribution
Question 4.40

CFLBs. A manufacturer of compact fluorescent light bulbs advertises that the distribution of the lifespans of these light bulbs is nearly normal with a mean of 9,000 hours and a standard deviation of 1,000 hours.
(a) What is the probability that a randomly chosen light bulb lasts more than 10,500 hours?

z_10500 <- (10500 - 9000) / 1000

p <- 1 - pnorm(z_10500)
p

[1] 0.0668072
(b) Describe the distribution of the mean lifespan of 15 light bulbs.

sd<-1000
mean<-9000
n<-15

life_15 <- rnorm(n, mean, sd)
mean(life_15)

[1] 8961.995

sd(life_15)

[1] 880.4527
(c) What is the probability that the mean lifespan of 15 randomly chosen light bulbs is more than 10,500 hours?

se <- sd / sqrt(n)

z_10500 <-(10500-mean)/se
p <- pnorm(z_10500, mean, sd)
p

[1] 1.189897e-19 (d) 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)

(e) Could you estimate the probabilities from parts (a) and (c) if the lifespans of light bulbs had a skewed distribution?
     No, Since it is assumed as normal districution also skewed a lot
Question 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.
     The p value will decrease as the sample size increases. The standard error decreases, the z-score increases