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?

Average= 171.1 Median = 170.3

  1. What is the point estimate for the standard deviation of the heights of active individuals? What about the IQR?

SD=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.

Since the median is 171.1 and SD is 9.4 , 180cm falls withing a single SD from the mean. As such this person could not be considered unusally tall.

However 155cm almost 2SD away from the mean and can definitely be considered shorter than most

  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.

With a single sample it is very unlikely that the mean and sd would be the same. This is due to variability and the probability that the sample chosen will not be representative

  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.

Standard Error

#Standard error = Std dev/sqrt(n)
 (9.4)/sqrt(507)
## [1] 0.4174687

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.

False: Point estimate falls within the 95% Confidence Interval

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

False: Sample is large enough

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

False: 95% CI does not translate to 95% of the samples

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

True: CI denotes that the true population mean should fall between $80.31 and $89.11.

  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.

False : 90% CI is wider not narrower since this denotes that the user is comfortable with more uncertaininty.

  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.

False: The sample should be larger

  1. The margin of error is 4.4.

True:

#ME= CI range/2

(89.11-80.31)/2
## [1] 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?

Sample is large, random and as a result independent. The distribution is somewhat skewed but for the purposes of this exercise the conditions are satisfied.

  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.

Ho - mean= 32 Ha - mean < 32

x <- 32
n <- 36
min <- 21
mean <- 30.69
sig <- 4.31
max <- 39


StdErr <- sig/sqrt(n)
Z <- (mean - x)/(StdErr)

#Calculating p
pnorm(q=30.69, mean=32, sd = StdErr) * 2
## [1] 0.0682026
  1. Interpret the p-value in context of the hypothesis test and the data.

Since significance level was 10% and p value is 6.8% we reject our null. Gifted children do appear to count to 10 faster than the general population.

  1. Calculate a 90% confidence interval for the average age at which gifted children first count to 10 successfully.
#upper
mean-1.64*StdErr
## [1] 29.51193
#lower
mean+1.64*StdErr
## [1] 31.86807
  1. Do your results from the hypothesis test and the confidence interval agree? Explain.

The upper and lower limits range from 29.5-31.8 months, which agrees with our test above since both of these values are below 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. Performahypothesistesttoevaluateifthesedataprovideconvincingevidencethattheaverage 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.

Ho- mean IQ= 100 Ha- mean IQ is not equal to 100

x <- 100
n <- 36
min <- 101
mean <- 118.2
sig <- 6.5
max <- 131


StdErr <- sig/sqrt(n)
Z <- (mean - x)/(StdErr)

#Calculating p
(1- pnorm(118.2, mean =100, sd = StdErr)) * 2
## [1] 0

Since p=0 we have to reject the null

  1. Calculate a 90% confidence interval for the average IQ of mothers of gifted children.
#upper
mean-1.64*StdErr
## [1] 116.4233
#lower
mean+1.64*StdErr
## [1] 119.9767
  1. Do your results from the hypothesis test and the confidence interval agree? Explain.

Results agree since we rejected the null and the 90% CI range is 116.4 - 119.98

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.

According to the CLT if enough samples are taken from within a sample, then the means of these samples will be normally distributed.

This distribution of the means of multiple samples taken from the original sample is called the “sampling distribution”.

The larger the sample the more normal the distribution, the samller the sample the more skewed the distribution.

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?
# calculating the probability

1-pnorm(q=10500, mean=9000, sd=1000)
## [1] 0.0668072
  1. Describe the distribution of the mean lifespan of 15 light bulbs.
  1. nearly normal
  2. mean = 9000
  3. Standard Dev = 258.20.
  1. What is the probability that the mean lifespan of 15 randomly chosen light bulbs is more than 10,500 hours?
#probability
1 - pnorm(q=10500, mean=9000, sd=258.20)
## [1] 3.13392e-09
  1. Sketch the two distributions (population and sampling) on the same scale.
seq <- seq(5000,12000,100)
d1<- dnorm(seq, 9000,1000)
d2<- dnorm(seq, 9000, 258)



#Population
plot(seq, d1, type="l", main="Population")

#Sample
plot(seq, d2, type="l", main="Sample")

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

Without increasing the sample size a skewed distribution will not let us calculate probabilities accurately

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.

Since the sample size increased, the p value will decrease because of a reduction in uncertainity