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?
  2. What is the point estimate for the standard deviation of the heights of active individuals? What about the IQR?
  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.
  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.
  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 \(SD_x = \frac{\sigma}{\sqrt{n}}\))? Compute this quantity using the data from the original sample under the condition that the data are a simple random sample.

Answers:

summary(bdims$hgt)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   147.2   163.8   170.3   171.1   177.8   198.1
hgt_sd <-sd(bdims$hgt)
hgt_sd
## [1] 9.407205
  1. Mean is 171.1, Median 170.3.
  2. St.dev = 9.407205, Q3-Q1 = 177.8-163.8 = 14
  3. Unusual is more than two standard deviations from the mean. 171.1 + 18.8 = 189.9. Therefore 180cm is not unusually tall. 171.1 - 18.8 = 152.3. Therefore 155cm is not unusually short.
  4. The sample size is high enough that generally another random sampling would have a similar mean and st.dev but it would not be exactly the same as the ones given above.
st_err <- hgt_sd / sqrt(507)
st_err
## [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.
  2. This confidence interval is not valid since the distribution of spending in the sample is right skewed.
  3. 95% of random samples have a sample mean between $80.31 and $89.11.
  4. We are 95% confident that the average spending of all American adults is between $80.31 and $89.11.
  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.
  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.
  7. The margin of error is 4.4.

Answers:
a. False, this isn’t necessarily true as the confidence interval covers the entire population rather than just the sample surveyed.
b. False, the confidence interval can be valid when the distribution is right skewed.
c. False, the 95% confidence interval refers to the population mean, we cannot make a confidence claim about the sample mean.
d. True, as the given confidence interval refers to the population as a whole.
e. True, a lower confidence interval is narrower than a higher one.
f. False, because sqrt(N) (N being the sample size) is the denominator in the formula for standard error, N would have to be 9 times bigger to decrease the the margin of error to a third of its current value.
g. True

margin_error <- (89.11 - 80.31) / 2
margin_error
## [1] 4.4

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?
  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.
  3. Interpret the p-value in context of the hypothesis test and the data.
  4. Calculate a 90% confidence interval for the average age at which gifted children first count to 10 successfully.
  5. Do your results from the hypothesis test and the confidence interval agree? Explain.

Answers:
a. Yes because the shape approximates the normal distribution without great outliers, sample size is greater than 30, and the sample population is a random, independent selection from a large overall population (large city).
b. The null hypothsis is H0: mean = 32 while the alternative is Ha: mean < 32.

Z <- (30.69 - 32) / (4.31 /sqrt(36))
Z
## [1] -1.823666
p_value <- pnorm(Z)
p_value
## [1] 0.0341013
  1. The p-value of 0.034 is less than the significance level of 0.10, so there appears to be evidence to reject the null hypothesis in favor of the alternative.
30.69 - (qnorm(0.95) * (4.31 / sqrt(36)))
## [1] 29.50845
30.69 + (qnorm(0.95) * (4.31 / sqrt(36)))
## [1] 31.87155

We can say with 90% confidence that the average age at which gifted children first count to 10 successfully is between 29.51 and 31.87 months of age.
e. Yes, the results agree because the upper limit of the confidence interval is less than the null hypothesis which we rejected in favor of the alternative hypothesis.

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.
  2. Calculate a 90% confidence interval for the average IQ of mothers of gifted children.
  3. Do your results from the hypothesis test and the confidence interval agree? Explain.

Answers:
a. The null hypothesis is H0: mean IQ = 100. The alternative hypothesis Ha: mean IQ =/= 100.

Z <- (118.2 - 100) / (6.5 / sqrt(36))
Z
## [1] 16.8
p_value <- 2 * (pnorm(Z,lower.tail = FALSE))
p_value
## [1] 2.44044e-63

The p-value is less than the significance level and therefore the null hypothesis can be rejected.
b.

118.2 - (qnorm(0.95) * (6.5 / sqrt(36)))
## [1] 116.4181
118.2 + (qnorm(0.95) * (6.5 / sqrt(36)))
## [1] 119.9819

The 90% confidence interval for the average IQ of mothers of gifted chidlren is 116.418 to 119.982.

  1. Yes because we rejected the null hypothesis that the mother’s average IQ is 100, and the 90% confidence interval’s lower limit is above that mean.

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:

A sampling distribution of the mean is a distribution from a population that was randomly and independently sampled. The shape, center, and spread of the sampling distribution of the mean should more closely resemble that of the overall population as the sample size increases.

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?
  2. Describe the distribution of the mean lifespan of 15 light bulbs.
  3. What is the probability that the mean lifespan of 15 randomly chosen light bulbs is more than 10,500 hours?
  4. Sketch the two distributions (population and sampling) on the same scale.
  5. Could you estimate the probabilities from parts (a) and (c) if the lifespans of light bulbs had a skewed distribution?

Answers:
a.

1-pnorm(10500,9000,1000)
## [1] 0.0668072

A 6.68% chance.
b. The population distribution is described as nearly normal, so the distribution of 15 lightbulbs should also be nearly normal unless some extreme outliers are part of the 15 lightbulbs. A sample size of at least 30 is preferred.
c.

1 - pnorm(10500,9000,(1000/sqrt(15)))
## [1] 3.133452e-09

The probability is close to 0.

pop <- rnorm(10000, 9000, 1000)
hist(pop, xlim = c(5000, 12000))

sample <- rnorm(15, 9000, 1000)
hist(sample, xlim = c(5000, 12000))

  1. A normal distribution is required to estimate probabilities with the given parameters.

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.

Answers:
Increasing sample size will decrease p-value. Standard error will decrease when sample size increases (as sample size is a denominator in the standard error formula). As standard error decreases, Z-score will increase – which has an inverse effect on probability. Therefore increasing sample size decreases p-value.