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.
(a) The point estimate for the average height of active individuals is the sample mean, 171.1 cm. The point estimate for the median is the sample median, 170.3.
(b) The point estimate for the standard deviation of the heights of active individuals is the standard deviation of the sample, 9.4 cm. The point estimate for the IQR is \(177.8 - 163.8 = 14\).
(c) Because this distribution of heights is roughly normal, we can find the approximate fraction of people shorter than those described here using $.
r z_tall <- pnorm(180, mean = 171.1, sd = 9.4) z_short <- pnorm(155, mean = 171.1, sd = 9.4) z_tall
## [1] 0.8281318
r z_short
## [1] 0.0433778 A person with a height of 180 cm is taller than about 83% of people. This is not unusually tall, since this person will frequently encounter others taller than they are.
A person with a height of 155 cm is only taller than about 4.3% of people. This is quite short, but even a person this short will occasionally find others shorter than they are. I would not say they are “unusually” short.
(d) For a new sample of physically active individuals, I would not expect a mean and standard deviation equal to those from the sample described above. Because each sample is randomly selected from a large population, normal sampling variability implies that means and standard deviations will vary slightly from sample to sample. However, I would expect the new sample mean and standard deviations to be close to those from the initial sample.
(e) To quantify the variability of sample means, we use the standard error. The standard error of the sampling distribution is \(SE(\bar{x}) = s/\sqrt{n}\).
r se <- 9.4 / sqrt(507) se
## [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.
(a) We are 95% confident that the average spending of these 436 American adults is between $80.31 and $89.11.
(b) This confidence interval is not valid since the distribution of spending in the sample is right skewed.
(c) 95% of random samples have a sample mean between $80.31 and $89.11.
(d) We are 95% confident that the average spending of all American adults is between $80.31 and $89.11.
(e) A 90% confidence interval would be narrower than the 95% confidence interval since we don’t need to be as sure about our estimate.
(f) 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.
(g) The margin of error is 4.4.

  1. False. We are completely confident that the average spending of these 436 American adults is between $80.31 and $89.11. Since we measured the spending of each of them, we have an exact average, $84.71, and not merely an estimate.

  2. False. Because \(n = 436\), which is much larger than 30, we know that the sampling distribution for mean spending is nearly normal. The skewness seen here is from a single sample. The Central Limit Theorem tells us that even skewed samples (and samples from skewed population distributions) lead to symmetric, normal sampling distributions.

  3. False. If the population mean is exactly $84.71, then it’s true that 95% of random samples with \(n = 436\) lie between $80.31 and $89.11. However, based on this single sample, we don’t know the population mean, and so we can’t make this claim. What we can say is that 95% of confidence intervals constructed in this way will contain the population mean.

  4. True. 95% of the confidence intervals constructed from a sample of \(n = 436\) will contain the true mean spending of American adults.

  5. True. Confidence intervals face a tradeoff between width and confidence level. In order to make stronger claims (narrower confidence intervals), we must forego some of the certainty of our claims.

  6. False. Margin of error = \(t^*\times(s / \sqrt{n})\). In order to reduce this expression to a third of its value, \(n\) must increase by a factor of 9.

  7. True. The margin of error is half the width of the confidence interval.

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 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?
  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 fist 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.

  1. Conditions for inference are satisfied if the sample data are independent of each other, and if \(n > 30\). Because the children sampled were randomly selected, we can infer that these measurements are independent. And \(n = 36\), so the second condition is satisfied.

  2. \(H_{0}: \mu = 32\)
    \(H_{1}: \mu < 32\)
    \(\alpha = 0.10\)
    \(\text{df} = 35\)
    \(t = (30.69 - 32) / (4.31/\sqrt{36}) = -1.824\)

The area to the left of \(t^*\) under the T distribution with \(\text{df} = 35\) is less than \(0.05\). So we reject the null hypothesis with \(\alpha < 0.05\).

  1. The p value is less than the significance level, so we reject the null hypothesis, and conclude that the mean age at which gifted children learn to count to 10 is less than 32 months.

  2. 90% confidence interval for \(\mu\):
    \(\bar{x}\pm t^* \times SE(\bar{x})\)
    \(30.69 \pm 1.69 \times 0.7183\)
    \((29.48, 31.90)\)

  3. The results from the hypothesis test agree with the confidence interval. The confidence interval states that we can be 90% confident that the true mean age at which gifted children learn to count to 10 is between 29.48 and 31.9. Because the confidence interval does not include the null value, 32, we reject the null 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.
  1. \(H_{0}: \mu = 100\)
    \(H_{1}: \mu \neq 100\)
    \(\alpha = 0.10\)
    \(\text{df} = 35\)
    \(t = (118.2 - 100) / (6.5/\sqrt{36}) = 16.8\)

The area under the \(t\) distribution to the right of \(t=16.8\) and to the left of \(t = -16.8\) with \(\text{df} =35\) is much less than 0.01. So we reject the null hypothesis and conclude that the average IQ of the mothers of gifted children is greater than average.

  1. 90% confidence interval for \(\mu\):
    \(\bar{x}\pm t^* \times SE(\bar{x})\)
    \(118.2 \pm 1.69 \times 1.083\)
    \((116.37, 120.03)\)

  2. The results from the hypothesis test agree with the confidence interval. The confidence interval states that we can be 90% confidence that the true mean IQ of mothers of gifted children is between about 116 and 120. Because the confidence interval does not include the null value, 100, we reject the null hypothesis.

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.

A sampling distribution of the mean is the distribution of sample means for equal-sized samples drawn randomly from the same population with replacement. A sampling distribution is important because it helps us understand how the properties of our particular sample might compare to the many other samples of equal size drawn from the same population. As long as sample size is greater than 30, and the elements in the sample are independent, then the shape of a sampling distribution of the mean converges toward a normal curve. This claim is the Central Limit Theorem. The center of a sampling distribution of the mean is an unbiased estimator of the population mean. As sample size increases, the variability of the sample mean decreases, but its average value does not change. As sample size increases, the spread of the sampling distribution decreases. The standard error of the sampling distribution is proportional to \(1 / \sqrt{n}\).
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.
(a) What is the probability that a randomly chosen light bulb lasts more than 10,500 hours? (b) Describe the distribution of the mean lifespan of 15 light bulbs. (c) What is the probability that the mean lifespan of 15 randomly chosen light bulbs is more than 10,500 hours? (d) Sketch the two distributions (population and sampling) on the same scale. (e) Could you estimate the probabilities from parts (a) and (c) if the lifespans of light bulbs had a skewed distribution?
  1. Let \(X = \text{the lifespan of a lightbulb}\) \(P(X>10,500)\):
1 - pnorm(10500, mean = 9000, sd = 1000)
## [1] 0.0668072

  1. Because the population distribution is nearly normal, the sampling distribution for mean lifespan will also be nearly normal, even though the sample size is small. The mean of the sampling distribution will be the population mean, 1000 hours. The standard error for the sampling distribution is \(1000/\sqrt{15} = 258.20\).

  2. Let \(X = \text{the mean lifespan of 15 randomly chosen lightbulbs}\) \(P(X>10,500)\):

1 - pnorm(10500, mean = 9000, sd = 1000/sqrt(15))
## [1] 3.133452e-09
normal_plot(mean = 10500, sd = 1000, cv = c(0,0), limits = c(7500,13500))
## Warning in max(ids, na.rm = TRUE): no non-missing arguments to max; returning -
## Inf

normal_plot(mean = 10500, sd = 1000/sqrt(15), cv = c(0,0), limits = c(7500, 13500))
## Warning in max(ids, na.rm = TRUE): no non-missing arguments to max; returning -
## Inf

  1. If the distribution of the lifespans of the lightbulbs was skewed, then we would need more information to be able to estimate the probability that a randomly chosen light bulb lasts more than 10,500 hours. If we were taking large samples, n > 30, then we would still be able to estimate the probability that any particular sample had a mean greater than 10,500 hours. That’s because the sampling distribution for means even for a skewed population distribution is itself normal. In this case, \(n\) is small, at 15. So I don’t think we could confidently make a claim about the mean lifespan of 15 randomly chosen light bulbs if the population were skewed.

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.

The \(p\)-value of a hypothesis test is the total area under the distribution to the extremes of the test statistic. The test statistics for hypothesis tests of proportions and means both increase with \(n\). So a larger value of \(n\) means that test statistics would be greater, and \(p\) values would be lower. These lower \(p\) values indicate a greater confidence in the conclusion of the hypothesis tests.