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 height
mean(bdims$hgt)
## [1] 171.1438
#median height
median(bdims$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$hgt)
## [1] 9.407205
#IQR
IQR(bdims$hgt)
## [1] 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.
#calculate mean +or - 2 SD

171.1+2*9.4
## [1] 189.9
171.1-2*9.4
## [1] 152.3
#this produces [152.3, 189.9] 

#180 cm unusually tall?

# from the rough interval above, 180 not unusual

#assume sample mean = pop, sample sd=pop
pnorm(180,mean=171.1,sd=9.4)
## [1] 0.8281318
#No, 82th percentile

#155 cm unusually short?
# from the rough interval above, 155, not unusual
# [152.3, 189.9] 

#assume sample mean = pop, sample sd=pop
pnorm(155,mean=171.1,sd=9.4)
## [1] 0.0433778
# Yes, p=.04.

At 5% significance level, the height of 155cm falls in the 4th percentile, which although can happen by chance, can be considered unusual.

  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.

DISCUSSION:

A simple random sample of physically active individuals should produce similar results. Of course, sample size is an issue, with a larger sample size given a better point estimate. There will be sampling variation from sample to sample, therefore we do not expect exactly the same results.

  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.

DISCUSSION: Use standard error = sigma/ sqr(n)

sd(bdims$hgt)/(length(bdims$hgt)^.5)
## [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.

DISCUSSION:

False. Confidence interval statements are about the population parameter. F. The statement should read: We are 95% confident that the average spending of American adults is between $80.31 and $89.11.

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

DISCUSSION: If we bootstrap and sample with replacement, calculate the bootstrap statistic, (mean), Repeat many times to create a bootstrap distribution. Then, we can calculate the bounds of the confidence interval using the infer package.

This is the central limit theorem, that a not normal distribution becomes normal when looking at the sampling distribution.

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

DISCUSSION: False. Confidence Intervals mean we are 95% confident that the interval captures the population proportion. When generating samples and then a confidence interval, we would expect 95% of the confidence intervals to contain the population mean.

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

DISCUSSION:

TRUE - This is the proper way to make a statement regarding confidence intervals.

  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.

DISCUSSION - True. If we want an interval with lower confidence, for instance 90%, use a narrower interval because we are not as confident compared to 95% to capture the value.

  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.

DISCUSSION:

False. Margin of error= Z* SE where \(\frac{\sigma}{sqrt{n}}\). So for this to be 1/3, use \(1\sqrt{9}\)

  1. The margin of error is 4.4.

DISCUSSION:

The margin of error is \(\frac{\sigma}{\sqrt{n}}\). Also, this is equivalent to 1/2 of the confidence interval.

(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?

DISCUSSION:

Is the sample size sufficiently large? Yes. Are the observations independent and selected at random. Yes.

  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.
sd=4.31
n=36
mean=30.69

SE=sd/sqrt(n)
SE
## [1] 0.7183333
#Test null mu=32
Z=(mean-32)/SE
p=pnorm(Z)
p
## [1] 0.0341013

Reject the null mu=32 at alpha=.1.

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

DISCUSSION: p<.1 so reject the null.

  1. Calculate a 90% confidence interval for the average age at which gifted children first count to 10 successfully.
Z=1.645

LB=mean-SE
LB
## [1] 29.97167
UB=mean+SE
UB
## [1] 31.40833
#CI=[29.97, 31.41]
  1. Do your results from the hypothesis test and the confidence interval agree? Explain. DISCUSSION:

The results from the hypothesis test and confidence interval agree. We reject the null hypothesis that the mean =32 in the hypothesis test. The value of 32 falls outside of the confidence interval [29.97, 31.41].

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.

DISCUSSION:

H0:mu=100 HA:mu not equal to 100

n=36
mean=118.2
sd=6.5

#standard error
SE=sd/sqrt(n)

SE
## [1] 1.083333
#z score
Z=(mean-100)/SE
Z
## [1] 16.8
#probability
1-pnorm(Z)
## [1] 0

DISCUSSION:
Reject null.

  1. Calculate a 90% confidence interval for the average IQ of mothers of gifted children.
#LB
mean-1.645*(SE)
## [1] 116.4179
#UB
mean+1.645*(SE)
## [1] 119.9821
#Confidence interval
c(mean-1.645*(SE),mean+1.645*(SE))
## [1] 116.4179 119.9821
  1. Do your results from the hypothesis test and the confidence interval agree? Explain.

DISCUSSION: The hypothesis test result was to reject the null of mu=100.

The confidence interval does not contain 100. Therefore the two methodologies are in agreement.

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.

DISCUSSION:

The sampling distribution of the mean is the probability distribution of means for ALL possible random samples OF A GIVEN SIZE from some population.

As the sample size increases, the shape becomes bell-shaped, the mean approaches the population mean, the peak becomes higher and the spread decreases.

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?
1 - pnorm(10500, mean=9000, sd=1000)
## [1] 0.0668072
  1. Describe the distribution of the mean lifespan of 15 light bulbs.
# If 15 light bulbs are randomly selected....

mean=9000
mean
## [1] 9000
#SE
SE=1000/sqrt(15)
SE
## [1] 258.1989

DISCUSSION:

The distribution is N(9000,258.1989).

  1. What is the probability that the mean lifespan of 15 randomly chosen light bulbs is more than 10,500 hours?
#probability

1 - pnorm(10500, mean, 1000/sqrt(15))
## [1] 3.133452e-09
  1. Sketch the two distributions (population and sampling) on the same scale.
lifespan <- 5500:12500
plot(lifespan, dnorm(lifespan, 9000, 1000), type = "l", col = "red",ylim = c(0,0.002))
lines(lifespan, dnorm(lifespan, 9000, 1000/sqrt(15)), col = "blue" )
legend(10500, 0.0015, legend=c("Population", "Sample15"),
      col=c("red", "blue"), lty=1:2, cex=0.6)

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

DISCUSSION:

If the light bulbs exhibited a skewed distribution, the probabilities could not be estimated using normal assumptions.

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.

DISCUSSION:

The standard error SE=sd/sqrt(n) will decrease as n increases.

Z=x-mu/SE will increase as n increases.

The area in the tails will decrease so p will decrease.