4.4 (Graded) Heights of adults. 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? The point estimate for average height is equivalent to the mean of 171.1cm. The median is 170.3cm.

  2. What is the point estimate for the standard deviation of the heights of active individuals? What about the IQR? The SD is 9.4cm. The IQR is 163.8 - 177.8cm.

  3. 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. In both cases yes, as those observations lie outside of the IQR.

  4. 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. No, I would expect them to deviate slightly. Each random sample will produce different point estimates.

  5. 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.

ht.SD <- 9.4
ht.n <- 507
ht.SE <- ht.SD / sqrt(ht.n)
paste0("The standard error is ", round(ht.SE,3))
## [1] "The standard error is 0.417"
4.14 (Graded) 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: We are 95% confident that the average spending of the full population (of which the 436 American adults is a sample) falls in the range of $80.31 and $89.11.

  2. This confidence interval is not valid since the distribution of spending in the sample is right skewed. Probably FALSE: The observations are independent (random sample), the sample is large (n = 436), and the distribution does not appear to be strongly skewed.

  3. 95% of random samples have a sample mean between $80.31 and $89.11. FALSE: The lower and upper bounds will vary based on the sample.

  4. We are 95% confident that the average spending of all American adults is between $80.31 and $89.11. TRUE: We are 95% confident the population mean will fall in that range (as per 4.14a).

  5. 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: A 90% confidence interval would be narrower.

  6. 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: Margin of error is computed as the point estimate plus-or-minus z multiplied by the standard error, and standard error is calculated as standard deviation divide by the square root of sample. To decrease the margin of error by a factor of three, you’d need 9 times the sample: sqrt(9) = 3.

  7. The margin of error is 4.4.

89.11 - 80.31
## [1] 8.8
89.11 - 4.4
## [1] 84.71

TRUE: The margin of error is 4.40; this is added and subtracted to the point estimate of the mean (84.71) to produce the confidence interval.

4.24 (Graded) Gifted children, Part I. Researchers investigating characteristics of gifted children collected 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 distribution 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? Yes: the sample is random, the sample size exceeds 30, and while there is some left-ward skew the sample does appear to be unimodal so we’ll assume normal model is alright.

  2. 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. H0: The average age gifted children first count to 10 is the same or more than that of regular children HA: The average age gifted children first count to 10 is less than that of regular children. alpha = .10

  3. Interpret the p-value in context of the hypothesis test and the data. We reject H0 in favor of HA as the p-value of .034 is less than alpha = .10.

count.n <- 36
count.min <- 21
count.mu <- 30.69
count.sd <- 4.31
count.max <- 39
count.xbar <- 32

count.se <- count.sd / sqrt(count.n)
count.z <- (count.mu - count.xbar) / count.se
round(pnorm(count.z),3)
## [1] 0.034
  1. Calculate a 90% confidence interval for the average age at which gifted children first count to 10 successfully.
count.margin <- qnorm(.90) * count.se
count.lower <- count.mu - count.margin
count.upper <- count.mu + count.margin
round(c(count.lower, count.upper), 2)
## [1] 29.77 31.61
  1. Do your results from the hypothesis test and the confidence interval agree? Explain. Yes, they concur. We have 90% confidence that the average age gifted children first count to 10 is between 29.77 and 31.61. This is less than 32, the average of regular children. The p-value is less than alpha, which means we reject the null hypothesis that the average age of gifted children is the same or greater than that of regular children.
4.26 (Graded) Gifted children, Part II. Exercise 4.24 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. H0: The average IQ of mothers of gifted children is the same or less than that of mothers of regular children. HA: The average IQ of mothers of gifted children is greater than that of mothers of regular children.
iq.n <- 36
iq.min <- 101
iq.mu <- 118.2
iq.sd <- 6.5
iq.max <- 131
iq.xbar <- 100

iq.se <- iq.sd / sqrt(iq.n)
iq.z <- (iq.mu - iq.xbar) / iq.se
round(1 - pnorm(iq.z), 3)
## [1] 0
  1. Calculate a 90% confidence interval for the average IQ of mothers of gifted children.
iq.se <- iq.sd / sqrt(iq.n)
iq.margin <- qnorm(.90) * iq.se
iq.lower <- iq.mu - iq.margin
iq.upper <- iq.mu + iq.margin
round(c(iq.lower, iq.upper), 2)
## [1] 116.81 119.59
  1. Do your results from the hypothesis test and the confidence interval agree? Explain. Yes, they concur. We have 90% confidence that the average IQ of mothers of gifted children falls between 116.81 and 119.59. This is considerably higher than 100, the average for the population at large. The p-value is practically 0, which means we reject the null hypothesis that the average IQ of mothers of gifted children is the same or greater than that of regular children.
4.34 (Graded) 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 distribution of the means of every sampling made of a population. As sample size increases, the shape more closely approximates the shape and spread of a normal distribution centered on the mean. Said another way, the normal model becomes more reasonable, and the condition on skew can be relaxed.

4.40 (Graded) 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? There’s a 6.7% chance that a chosen light bulb 10,500 hours (or is 3.35%).
bulb.mu <- 9000
bulb.sd <- 1000
bulb.xbar <- 10500

bulb.se <- bulb.sd # we'll approximate SE using SD
bulb.z <- (bulb.xbar - bulb.mu) / bulb.se
round(1 - pnorm(bulb.z),3)
## [1] 0.067
round(pnorm(bulb.xbar, bulb.mu, bulb.sd, lower.tail = F), 3) # experimenting just with the pnorm function also seems to work
## [1] 0.067
  1. Describe the distribution of the mean lifespan of 15 light bulbs. The distribution should be normal with a mean around 9000 and a standard error of 258.2.
bulb.n <- 15

bulb.se15 <- bulb.sd / sqrt(bulb.n)
round(bulb.se15, 3)
## [1] 258.199
  1. What is the probability that the mean lifespan of 15 randomly chosen light bulbs is more than 10,500 hours? Practically negligible.
pnorm(bulb.xbar, bulb.mu, bulb.se15, lower.tail = F)
## [1] 3.133452e-09
  1. Sketch the two distributions (population and sampling) on the same scale.
bounds <- seq(5500, 12500, .01)
plot(bounds, dnorm(bounds, bulb.mu, bulb.sd), type = "l", ylim = c(0, .0018), ylab = "", xlab = "hours", main = "Bulb Lifespans", col = "blue")
lines(bounds, dnorm(bounds, bulb.mu, bulb.se15), col = "red") # Success!

  1. Could you estimate the probabilities from parts (a) and (c) if the lifespans of light bulbs had a skewed distribution? If the skewed distribution did not approximate a normal distribution, then no, we could not estimate the probabilities.
4.48 (Graded) 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.

A larger sample size shrinks the standard error, yielding a smaller P-value.