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?
# mean
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$hgt)
## [1] 9.407205
# IQR
IQR(bdims$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.
mean_hgt = mean(bdims$hgt)
sd_hgt = sd(bdims$hgt)
# 180cm within one standard deviation, expected
print((180 - mean_hgt)/sd_hgt)
## [1] 0.9414287
# 155cm not within one standard deviation, less expected (but not particularly rare either)
print((155 - mean_hgt)/sd_hgt)
## [1] -1.716109
  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.
# If you think that being physically active is correlated with height, then you would
# expect the mean and standard deviation to be different. I don't expect this 
# to be the case however so I think the mean and standard deviation will be similar.
  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
sd_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.

    False. You are actually 100% confident of the 436 American adults because you sampled them and
those are the actual spending data. You are 95% confident that the true average spending of all
American adults is within the stated confidence interval

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

False. We have a large sample (436) so we are not as concerend with skewness. 
However, if we wanted a better technique, we would instead
do bootstrap sampling to construct confidence intervals.
  1. 95% of random samples have a sample mean between $80.31 and $89.11.
False. Our confidence interval statement is only about the true population mean.
It has nothing to do with sample means of other samples
  1. We are 95% confident that the average spending of all American adults is between $80.31 and $89.11.
True, when we are only concerned with the six-day period after Thanksgiving, as that is
the data we have. I would not say that average spending range is valid for all-year round
as that is a different population from what we collected.
  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.
True. If we are less confident (90% compared to 95%), we are fewer standard deviations away from the mean
which translates to a narrower coverage of the distribution
  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
In a confidence interval, z⋆ × SE is called the margin of error.
Of this formula, sampling number (n) appears in SE as sigma/root(n).
If we want to reduce SE by a third, we need n to be 9 times larger
  1. The margin of error is 4.4.
89.11 - 84.71
## [1] 4.4
84.71 - 80.31
## [1] 4.4
# True.

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?
Samples are independent and n > 30. Conditions for inference 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 first count to 10 successfully is less than the general average of 32 months. Use a significance level of 0.10.
# we are specifcally asked to test LESS
# H0 = no difference between children mean 32 and sample mean (gifted) 30.69
# HA < gifted
std_err = 4.31/sqrt(36)
z = (30.69 - 32)/std_err
pnorm(z)
## [1] 0.0341013
# p is .0341013, which is lower than the significance level of .1
# reject null hypothesis
  1. Interpret the p-value in context of the hypothesis test and the data.
Significance level was 10% and p-val was 3.4%.
We reject the null hypothesis that there is no difference between mean months to count
to 10 for gifted and all children.
  1. Calculate a 90% confidence interval for the average age at which gifted children first count to 10 successfully.
# .95 since 5% on each tail)
z = qnorm(.95)
mean = 30.69
mean + (z * std_err) # upper
## [1] 31.87155
mean - (z * std_err) # lower
## [1] 29.50845
  1. Do your results from the hypothesis test and the confidence interval agree? Explain.
Yes they agree. We rejected the null (that 32 was the mean for gifted children) and our 90%
confidence interval for gifted children does not include 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.
n = 36
mean = 118.2
sd = 6.5
# asked for different, so 2-sided test
# H0 = Average IQ of mothers of gifted children is 100
# HA = Average IQ of mothers of gifted children is not 100
std_err = sd/sqrt(n)
z = (mean - 100)/std_err
p_val = (1-pnorm(z)) * 2
p_val
## [1] 0
# p_val is 0%, less than 10% significance, so reject the null hypothesis
  1. Calculate a 90% confidence interval for the average IQ of mothers of gifted children.
# .95 since 5% on each tail)
z = qnorm(.95)
mean = 118.2
mean + (z * std_err) # upper
## [1] 119.9819
mean - (z * std_err) # lower
## [1] 116.4181
  1. Do your results from the hypothesis test and the confidence interval agree? Explain.
Yes they agree. We rejected the idea that the average IQ of mothers of gifted
children is 100. Our 90% confidence interval for the true average IQ of mothers
of gifted children is 116.4181 - 119.9819

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.

Sampling distribution of the mean refers to: if we sample for a mean from the population
many times, we get a distribution of calculated sample means. This is the
sampling distribution of the mean. It is always normal/bell-shaped, and the mean
should be near the true population mean. As n increases, it is still normal/bell-shaped,
the center barely moves (and gets closer to the true population mean), and
the distribution becomes taller and narrower (less spread/variance) around the center/mean.

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?
mean = 9000
std_dev = 1000
z = (10500 - mean)/std_dev
1 - pnorm(z)
## [1] 0.0668072
# 6.68% chance
  1. Describe the distribution of the mean lifespan of 15 light bulbs.
# It should be nearly normal, with mean around 9000
# std_err = std_dev/sqrt(15)
std_dev/sqrt(15)
## [1] 258.1989
# It should have standard error of 258.1989
  1. What is the probability that the mean lifespan of 15 randomly chosen light bulbs is more than 10,500 hours?
1 - pnorm(q = 10500, mean = 9000, sd = std_dev/sqrt(15))
## [1] 3.133452e-09
# nearly 0 probability
  1. Sketch the two distributions (population and sampling) on the same scale.
# go 5 std dev for each distribution
x1 = seq(mean - (5 * std_dev), mean + (5 * std_dev), length=101)
pop = dnorm(x1, mean, std_dev)
n = 15
x2 = seq(mean - (5 * std_dev/sqrt(n)), mean + (5 * std_dev/sqrt(n)), length=101)
samp = dnorm(x2, mean, std_dev/sqrt(n))

plot(x2, samp, xlim = c(mean - (5 * std_dev), mean + (5 * std_dev)), xlab='hours', ylab = '')
lines(x1, pop)

  1. Could you estimate the probabilities from parts (a) and (c) if the lifespans of light bulbs had a skewed distribution?
The sample size of 15 is too small for skewed. We would have to increase
the sample size to get more reliable estimates.

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.

Your p-value should decrease. Similar to the confidence interval, when you increase
n the "width" of failing to reject H0 decreases. This intuitively makes sense because
you are collecting more evidence with a higher n. Therefore if you collected more datapoints
showing you to reject H0, then that is a lower probability of happening and thus a lower
p-value