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Heights of adults. (7.7, p. 260) Researchers studying anthropometry collected body girth measurements and skeletal diameter measurements, as well as age, weight, height and gender, for 507 physically active individuals. The histogram below shows the sample distribution of heights in centimeters.

  1. What is the point estimate for the average height of active individuals? What about the median?
#point estimate
mean(bdims$hgt)
## [1] 171.1438
#median
median(bdims$hgt)
## [1] 170.3
  1. What is the point estimate for the standard deviation of the heights of active individuals? What about the IQR?
#standard deviation
sd(bdims$hgt)
## [1] 9.407205
#IQR
summary(bdims$hgt)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   147.2   163.8   170.3   171.1   177.8   198.1

Answer: standard deviation=9.4

IQR=Q3-Q1=177.8-163.8=14

  1. 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.
#180CM tall
#Z=(180-171.1)/9.4=0.9468085
pnorm(180,171.1,9.4)
## [1] 0.8281318
#155cm Tall
#Z=(155-171.1)/9.4=-1.712766
pnorm(155,171.1,9.4)
## [1] 0.0433778

Answer: If a person is 182cm, 82.81% of people shorter than the person, so I don’t think he is too tall. If a person is 155 cm, only 4.34% of people shorter than him, so I will consider he is too short.

  1. 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.

Answer: The men and the standard deviation should be different because the sample is randomly pick, so the data generate by the sample should be different.

  1. 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 \(SD_x = \frac{\sigma}{\sqrt{n}}\))? Compute this quantity using the data from the original sample under the condition that the data are a simple random sample.

Answer: Compute Standard Error for sample mean

#Standard Error
sd(bdims$hgt)/sqrt(507)
## [1] 0.4177887

Thanksgiving spending, Part I. The 2009 holiday retail season, which kicked off on November 27, 2009 (the day after Thanksgiving), had been marked by somewhat lower self-reported consumer spending than was seen during the comparable period in 2008. To get an estimate of consumer spending, 436 randomly sampled American adults were surveyed. Daily consumer spending for the six-day period after Thanksgiving, spanning the Black Friday weekend and Cyber Monday, averaged $84.71. A 95% confidence interval based on this sample is ($80.31, $89.11). Determine whether the following statements are true or false, and explain your reasoning.

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

Answer: False. The sample mean is in the 95% confident interval and the 95% confident interval bases on the population mean.

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

Answer: False. Confidence interval is still valid if the distribution is slightly right skewed.

  1. 95% of random samples have a sample mean between $80.31 and $89.11.

Answer: False. Sample mean might be different base on the size, but 95% of the sample mean might within the confident interval.

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

Answer: True. According the definition of the parameter value, 95% confident interval cover the average spending of all American adults.

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

Answer: True.The smaller the confidence interval the narrower the interval.

  1. 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.

Answer: False. We use \(SD_x = \frac{\sigma}{\sqrt{n}}\))to compute the error. If we need to decrease the margin of error to a third of what is now, we need 9 times of the n in the denominator.

  1. The margin of error is 4.4.

Answer: True. The margin of error is calculate by (89.11-80.31)/2=4.4

Gifted children, Part I. Researchers investigating characteristics of gifted children col- lected data from schools in a large city on a random sample of thirty-six children who were identified as gifted children soon after they reached the age of four. The following histogram shows the dis- tribution of the ages (in months) at which these children first counted to 10 successfully. Also provided are some sample statistics.

  1. Are conditions for inference satisfied?

Answer: The data is from schools in a large city on a random sample and it is independent. And the histogram is not heavily right skewed.

  1. 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.

Answer:

null hypothesis(\(H_0\)): children first count to 10 successfully when they are 32 months old alternative hypothesis(\(H_A\)): children first count to 10 successfully are not 32 months old

#Two tails test
sd <- 4.31/sqrt(36)
Z <- (30.69 - 32)/sd
p <- pnorm(Z)*2 
p
## [1] 0.0682026

Becuse the P value is small than 0.1, we reject \(H_0\).

  1. Interpret the p-value in context of the hypothesis test and the data.

Answer: Since we ha a p-value lower than the significance level, we can conclude that children first count to 10 successfully are less than 32 months old.

  1. Calculate a 90% confidence interval for the average age at which gifted children first count to 10 successfully.
higher<-30.69+1.645 *(4.31/sqrt(36))
lower<-30.69-1.645 *(4.31/sqrt(36))
c(lower,higher)
## [1] 29.50834 31.87166
  1. Do your results from the hypothesis test and the confidence interval agree? Explain.

Answer: Yes. It is agree. The upper age is 31.87 months which less than 32 months.

Gifted children, Part II. Exercise above describes a study on gifted children. In this study, along with variables on the children, the researchers also collected data on the mother’s and father’s IQ of the 36 randomly sampled gifted children. The histogram below shows the distribution of mother’s IQ. Also provided are some sample statistics.

  1. 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.

Answer: The hypothesis is:

\(H_0\): \(\mu = 100\)

\(H_1\): \(\mu \neq 100\).

n<-36 
min <-101
mean  <- 118.2
sd  <-6.5
max <-131

require(fastGraph)
## Loading required package: fastGraph
sd <- 4.31/sqrt(n)
Z <- (mean-100)/sd

#Two tails test
p <- (1-pnorm(Z))*2
p
## [1] 0

0 is smaller than 0.1, so we reject \(H_0\) and accept the Alternate hypothesis \(H_1\).

  1. Calculate a 90% confidence interval for the average IQ of mothers of gifted children.
higher<-118.2+1.645 *(6.5/sqrt(36))
lower<-118.2-1.645 *(6.5/sqrt(36))

c(lower,higher)
## [1] 116.4179 119.9821
  1. Do your results from the hypothesis test and the confidence interval agree? Explain.

Answer: Yes. It is agree. IQ 100 is not within the 116.41 and 119.98 range.

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.

Answer: The Sampling Distribution of the Mean is the mean of the population from where the items are sampled.As the sample size increases, the shape of the distribution becomes more normal, the center is closer to the true mean, and the spread of the samples are smaller and vice versa.

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.

  1. What is the probability that a randomly chosen light bulb lasts more than 10,500 hours?
z<-(10500-9000)/1000
1-pnorm(z)
## [1] 0.0668072
  1. Describe the distribution of the mean lifespan of 15 light bulbs.

Answer: The example is randomly pick and it is normal distribution.

  1. What is the probability that the mean lifespan of 15 randomly chosen light bulbs is more than 10,500 hours?
sd<-1000/sqrt(15)
z<-(10500-9000)/(1000/sd)
z
## [1] 387.2983

Answer: The z sore is 387.3 and it is impossible to randomly choose 15 light bulbs and they are all more than 10,500 hours.

  1. Sketch the two distributions (population and sampling) on the same scale.
n <- 15 
sample15 <- rnorm(n, mean = 9000, sd = 1000)
mean15 <- mean(sample15)
sd15 <- sd(sample15)


hist(sample15, probability = TRUE)
x <- 0:15000
y15 <- dnorm(x = x, mean = mean15, sd = sd15)
y <- dnorm(x = x, mean = 9000, sd = 1000)
lines(x = x, y = y15, col = "green")
abline(v=mean15,col="green")
lines(x = x, y = y, col = "blue")
abline(v=9000,col="blue")

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

Answer: No, we can do any estimation if the lifespans of light bulbs had a skewed distribution. All the estimation base on the normal distribution.

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.

Answer: P value will decrease. When the n size increase, sample sd decrease, and Z value increase because sample sd is the denominator. If Z value increase, then the p value decrease.