4.4 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 of the average is the mean of this sample, 171.1cm. The point estimate of the median is the median of this sample, 170.3cm.

  1. What is the point estimate for the standard deviation of the heights of active individuals? What about the IQR?

The point estimate of the standard deviation is the standard deviation of this sample, 9.4cm. The point estimate of the IQR is the IQR of this sample, calculated from the difference between Q3 and Q1, 177.8 - 163.8 = 14cm.

  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.

Calculating the z-scores of these two individuals shows that a person who is 180cm tall is less than one standard deviation away from the mean of this sample. The person is tall, but not incredibly so. The z-score of the person who is 155cm is 1.71 standard deviations below the mean, and therefore more unusual than the taller person.

(180-171.1)/9.4
## [1] 0.9468085
(155-171.1)/9.4
## [1] -1.712766
  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.

Since the new sample would be taken randomly, I would not expect the mean and the standard deviation to be the same as the first sample. Due to the variation in the population, the point estimates for each random sample will be different.

  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 = ! pn )? Compute this quantity using the data from the original sample under the condition that the data are a simple random sample.

The measure used to quantify the variability is standard error (SE).

\[SE = \frac{s}{\sqrt{n}} = \frac{9.4}{\sqrt{507}} \approx 0.4175\]

4.14 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 population 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.

FALSE - With 436 observations in the sample, the sample size is large enough to accommodate the skew in the data.

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

FALSE - the confidence interval refers to the population mean, not the sample mean.

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

TRUE

  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

  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

The margin of error is the difference between the upper bound of the confidence interval and the mean:

\[89.11 - 84.71 = 4.4\]

From the margin of error, we can compute the standard error:

\[ \begin{aligned} 4.4 &= 84.71 + 1.96SE \\ SE &= 2.24 \end{aligned} \]

We use this standard error to find the standard deviation of the sample:

\[ \begin{aligned} SE &= \frac{s}{\sqrt{n}} \\ 2.24 &= \frac{s}{\sqrt{436}} \\ s &= 46.77 \end{aligned} \]

To reduce the margin of error to a third of its current value, the standard error would also need to be reduced by to a third:

\[\frac{1}{3}(2.24) = 0.747\]

We can use this new standard error to find the needed sample size:

\[ \begin{aligned} SE &= \frac{s}{\sqrt{n}} \\ 0.747 &= \frac{46.77}{\sqrt{n}} \\ n &= 3920 \end{aligned} \]

The sample size is approximately 9 times larger to reduce the margin of error to a third of its size.

  1. The margin of error is 4.4.

TRUE - see part (f)

4.24 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 sampling was performed randomly, the sample size is over 30, and the distribution is not skewed.

  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.

We will perform a one-tailed hypothesis test to determine whether the average age is less than 32 months.

\[H_{0}: \mu = 32\] \[H_{A}: \mu < 32\]

From the given information, we can calculate the standard error (\(SE\)):

\[SE = \frac{s}{\sqrt{n}} = \frac{4.31}{\sqrt{36}} \approx 0.7183\]

We use the standard error to calculate the \(Z\)-score and use to provide the probability:

\[Z_{30.69} = \frac{30.69-32}{0.7183} \approx -1.82\]

\[P(Z < -1.82) = 0.034\]

normalPlot(0, 1, c(-Inf, -1.82))

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

Since our \(\alpha = 0.10\), we see that the \(p\)-value is less than the significance level, and we reject the null hypothesis in favor of the alternate hypothesis

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

\[CI_{upper} = 30.69 + 1.645(0.7183) \approx 31.87\] \[CI_{lower} = 30.69 - 1.645(0.7183) \approx 29.51\]

  1. Do your results from the hypothesis test and the confidence interval agree? Explain.

The results do agree - the \(p\)-value instructs us to reject the null hypothesis that the average age is 32 months, and the confidence interval does not include 32 months.

4.26 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 di???erent than the average IQ for the population at large, which is 100. Use a significance level of 0.10.

We will perform a one-tailed hypothesis test to determine whether the IQ of the mothers is greater than the average IQ.

\[H_{0}: \mu = 100\] \[H_{A}: \mu > 100\]

From the given information, we can calculate the standard error (\(SE\)):

\[SE = \frac{s}{\sqrt{n}} = \frac{6.5}{\sqrt{36}} \approx 1.083\]

We use the standard error to calculate the \(Z\)-score and use to provide the probability:

\[Z_{118.2} = \frac{118.2-100}{1.083} \approx 16.80\]

\[P(Z > 16.80) = 1 - P(Z < 16.80) = 1 - 0 = 1\]

normalPlot(0, 1, c(16.80, Inf))

We reject the null hypothesis.

  1. Calculate a 90% confidence interval for the average IQ of mothers of gifted children.

\[CI_{upper} = 118.2 + 1.645(1.083) \approx 119.98\] \[CI_{lower} = 118.2 - 1.645(1.083) \approx 116.42\]

  1. Do your results from the hypothesis test and the confidence interval agree? Explain.

The results do agree - the \(p\)-value instructs us to reject the null hypothesis that the average IQ of the mothers is 100, and the confidence interval does not include 100.

4.34 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 mean taken from repeated samples of the population. As the sample size increases, the shape becomes more normal, the center approaches the true population mean, and the spread decreases.

4.40 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?

Let \(X\) be the number of hours a lightbulb lasts. Since \(X\) is normally distributed, we can calculate the \(Z\)-score for 10,500 hours as:

\[Z_{10500} = \frac{10500-9000}{1000} = 1.5\]

Therefore, \[P(X > 10500) = P(Z > 1.5) = 1 - P(Z < 1.5) = 1 - 0.9332 = 0.0668\]

normalPlot(0, 1, c(1.5, Inf))

  1. Describe the distribution of the mean lifespan of 15 light bulbs.

Knowing the population standard deviation, the distribution of the sample mean of 15 light bulbs is \(N(9000, \frac{1000}{\sqrt{15}}) = N(9000, 258.20)\).

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

\[Z_{10500} = \frac{10500-9000}{258.20} \approx 5.81\] \[P(Z > 5.81) \approx 0\]

  1. Sketch the two distributions (population and sampling) on the same scale.
x <- 4000:14000
y1 <- dnorm(x, 9000, 1000)
y2 <- dnorm(x, 9000, 258)
plot(x,y1,type="l",col="red")
lines(x,y2,col="green")

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

You could not estimate the probabilities for (a) and (c). For part (a), a normal distribution is required to use the \(Z\)-score. For part (c), the sample size is too small to represent a normal distribution.

4.48 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.

Because the number of observations (\(n\)) is in the denominator of the standard error (\(SE\)), the \(SE\) decreases as \(n\) increases. With a smaller \(SE\), the \(Z\)-score increases in magnitude (i.e., moves farther from the mean), and the rejection region becomes smaller. Since the \(p\)-value reflects the area of the rejection region, the \(p\)-value decreases.