Chapter 4 Foundations for Inference Graded: 4.4, 4.14, 4.24, 4.26, 4.34, 4.40, 4.48

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?

Mean: 171.1 Median: 170.3

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

SD: 9.4 IQR: Q3- Q1 = 177.8 - 163.8 = 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.

Yes, a person who is 180cm tall is considered unsually tall because this value is above the third quartile, ie this person taller than over 75% of this group. A person who is 155cm tall is below the average but taller than over 25% of the group.

  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.

The mean and stadard deviation of this new sample would probably be different from these values because they are randomly sampled and there is a lot of variance among “physically active individuals”.

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

We use the standard error to quantify the variability of such an estimate. Here the standard error is:

9.4/sqrt(507)
## [1] 0.4174687

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 population mean is between these values.

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

True. The distribution of the sample should be normal in order to extrapolate a confidence interval.

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

False. Random samples of different sizes will have different means that may not fall in this range.

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

True. This is how we interpret confidence intervals.

  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.

False. It would be narrower because we would have to be more sure about our estimate.

  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.

The standard error is sd/sqrt(n). Let’s say the standard error is 1, we would have 1 = sd/sqrt(n). To get a standard error with a third of the size, we would need a proportion of 1:sd/sqrt(n) = 1/3:sd/sqrt(m). So, sqrt(n)/sd = sqrt(m)/3sd which becomes sqrt(n) = sqrt(m)/3 which becomes 3sqrt(n) = sqrt(m). Squaring both sides, we have 9n = m. So, in order to get a standard error that is a third of what it is now, we would need to use a sample size 9 times larger than what we have now.

  1. The margin of error is 4.4.

The margin of error is

89.11-84.71
## [1] 4.4

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 because we havea sample size of 36 > 30 and the distribution of the sample appears normal.

  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.

H0: mean = 32 Ha: mean < 32

Our p-value is 0.03.

mean <- 30.69
sd <- 4.31
n <- 36
se <- sd/sqrt(n)
z <- (mean - 32)/se
p <- pnorm(z)
p
## [1] 0.0341013
  1. Our p-value is 0.03 < 0.10, so we reject the null hypothesis that the mean age when children countot 10 is 32 and conclude the alternative, that is, that the mean age is less than 32.
  1. Calculate a 90% confidence interval for the average age at which gifted children first count to 10 successfully.
margin <- se * 1.64
upper <- mean + margin
lower <- mean - margin
lower
## [1] 29.51193
upper
## [1] 31.86807
  1. Do your results from the hypothesis test and the confidence interval agree? Explain.

Yes, our results agree as we are 95% confidence that the true population mean lies between 29.5 and 31.9. 32 is not in this range.

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 different than the average IQ for the population at large, which is 100. Use a significance level of 0.10.

H0: mean = 100 Ha: mean > 100

Our p = 0 < 0.10 so we reject the null hypothesis and conclude that the average IQ of mothers of gifted children is greater than 100.

mean <- 118.2
sd <- 6.5
n <- 36
se <- sd/sqrt(n) 
z <- (mean - 100)/se
p <- 1 - pnorm(z)
p
## [1] 0
  1. Calculate a 90% confidence interval for the average IQ of mothers of gifted children.
margin <- se * 1.64
lower <- mean - margin
upper <- mean + margin
lower
## [1] 116.4233
upper
## [1] 119.9767
  1. Do your results from the hypothesis test and the confidence interval agree? Explain.

Our results from our hypothesis test agree because we are 95% confident that the true population mean IQ of mothers of gifted children is between 116.4 and 119.98, and 100 does not fall in this range.

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 from a set of samples that we take from the population. This distribution becomes more normal as the number of samples taken increases. It also becomes centered on the true population mean and shows less variance as the sample size increases.

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?
mean <- 9000
sd <- 1000
z <- (10500 - mean)/sd
1-pnorm(z)
## [1] 0.0668072
  1. Describe the distribution of the mean lifespan of 15 light bulbs.

This would be a normal distribution with a mean center of 9000.

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

The probability of this is extremely small, close to 0.

se <- sd/sqrt(15)
z <- (10500 - mean)/se
p <- pnorm(z, mean, sd)
p
## [1] 1.189897e-19
  1. Sketch the two distributions (population and sampling) on the same scale.
pop <- seq(mean - (4 * sd), mean + (4 * sd), length=15)
sample<- seq(mean - (4 * se), mean + (4 * se), length=15)
y1 <- dnorm(pop,mean,sd)
y2 <- dnorm(sample,mean,se)
plot(pop, y1, type="l",col="blue", ylim=c(0,0.0015))
lines(sample, y2, col="red")

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

No because we would need a normal distribution to estimate the probabilities with these methods.

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.

If the sample size increases, the standard error decreases and our z score increases. If our z score increases, our p-value will decrease.