4.4

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.

Question4.4 Image

Question4.4 Image

  1. What is the point estimate for the average height of active individuals? What about the median?

Average of 171.1 cm Median of 170.3 cm

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

SD is 9.4 cm IQR is 14 (i.e. 177.8 - 163.8)

  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.

180 cm tall would have a z-score of (180 - 171.1)/9.4 = 0.947

We usually consider measurements beyond 2 SD away from the mean to be unusual. This is less than one SD away from the mean, so not unusually tall.

155 cm tall would have a z-score of (155 - 171.1)/9.4 = -1.71

This is still less than 2 SD away from the mean, so not unusually short.

  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.

No, they would likely not be the exact values as above. A random sample of people may have similar measurements, but would rarely, if ever, be exactly the same.

  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 of the sample mean = sigma/sqrt(n))?

Compute this quantity using the data from the original sample under the condition that the data are a simple random sample.

We use Standard Error to quantify the variability of sample means.

In this case the Standard Error would be: 9.4/sqrt(507) = 0.4174687

4.14

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.

Question4.14 Image

Question4.14 Image

  1. We are 95% confident that the average spending of these 436 American adults is between $80.31 and $89.11.

False. We know the average spending of this sample for certain. It’s generalizing the the whole population that requires a confidence interval.

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

False. While the sample is right skewed, we are using a large sample size (>100), so we can assume a normal distribution and a valid confidence interval calculation

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

False. We are 95% confident that the POPULATION mean is contained within the interval, not the SAMPLE means

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

True. That is the definition of this confidence interval

  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. A 90% confidence interval would have a narrower range, since it only needs to cover 90% of possible sample means

  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, since we use the square root of ‘n’ in our calculation of SE, we would need to use a sample that is 9x as large in order to reduce error by 3x

  1. The margin of error is 4.4.

True. Our mean 84.71, and our confidence interval is +/- 4.4.

4.24

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.

Question 4.24 Image

Question 4.24 Image

  1. Are conditions for inference satisfied?

Yes. n > 30, random sample, and roughly symmetrical.

  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.
n <- 36
s <- 4.31
x_mean <- 30.69
a <- .10
null_hyp <- 32

SE = s/sqrt(n)

z <- (x_mean - null_hyp)/SE

p <- pnorm(z)

if(p>=a){print("Accept the Null Hypothesis")}else{print("Reject the Null Hypothesis")}
## [1] "Reject the Null Hypothesis"
  1. Interpret the p-value in context of the hypothesis test and the data.

The p-value is the probability that children count to 10 in at most 30.69 months.

  1. Calculate a 90% confidence interval for the average age at which gifted children first count to 10 successfully.
z <- qnorm(.95)

u_lim <- x_mean + z*SE
l_lim <- x_mean - z*SE

CI <- c(l_lim,u_lim)

CI
## [1] 29.50845 31.87155
  1. Do your results from the hypothesis test and the confidence interval agree? Explain.

Yes, a 90% confidence interval will be contained in all larger confidence intervals (i.e. 95%, etc).

4.26

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 mothers and fathers IQ of the 36 randomly sampled gifted children. The histogram below shows the distribution of mothers IQ. Also provided are some sample statistics.

Question4.26 Image

Question4.26 Image

  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.
n <- 36
s <- 6.5
x_mean <- 118.2
a <- .10
null_hyp <- 100

SE = s/sqrt(n)

z <- (x_mean - null_hyp)/SE

p <- 1-pnorm(z)*2

if(p>=a){print("Accept the Null Hypothesis")}else{print("Reject the Null Hypothesis")}
## [1] "Reject the Null Hypothesis"

The sample mean is far outside the significance level (z value of nearly 17!). Therefore we can state with a high level of confidence that the average IQ of the mothers is higher than the average IQ of the population at large.

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

Since this is 2 sided, we wil compute z values for 95%:

z <- qnorm(.95)
l_lim <- x_mean - z * SE
u_lim <- x_mean + z * SE

CI <- c(l_lim,u_lim)

CI
## [1] 116.4181 119.9819
  1. Do your results from the hypothesis test and the confidence interval agree? Explain.

Yes, the confidence interval for the mothers of gifted children does no contain the population mean, hence why we rejected the null hypoethesis that the mothers IQs were the same as the population.

4.34

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.

Sampling distribution of the mean is the distribution of values (means of a sample) if many different samples were taken from the same population.

The higher the sample size, the closer the shape comes to a normal/symmetric distribution.

The center of this distribution tends to hover around the true population mean, but it will get closer and closer as the sample size increases

The spread of the distribution will decrease with higher sample sizes. As more sample observations are included, the effect of outliers is reduced, hence reducing the spread

4.40

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.

mu <- 9000
sigma <- 1000
  1. What is the probability that a randomly chosen light bulb lasts more than 10,500 hours?
z <- (10500 - mu)/sigma
p <- 1-pnorm(z)

p
## [1] 0.0668072
  1. Describe the distribution of the mean lifespan of 15 light bulbs.

The sample of 15 comes from a population that is normally distributed, so it will roughly follow a normal distribution that has the same mean as the population, and the standard deviation as the Standard Error, i.e. sigma/sqrt(n)

n <- 15
SE <- sigma/sqrt(n)

sample_1 <- rnorm(n = n, mean = mu, sd = SE)
hist(sample_1)

  1. What is the probability that the mean lifespan of 15 randomly chosen light bulbs is more than 10,500 hours?
z <- (10500 - mu)/SE
1 - pnorm(z)
## [1] 3.133452e-09
  1. Sketch the two distributions (population and sampling) on the same scale.
#Define  the x-axes based on 4 standard devs or standard errors above/below the mean, in 15 chunks.

pop_x <- seq(mu - 4 * sigma, mu + 4 * sigma, length = 15)
sam_x <- seq(mu - 4 * SE, mu + 4 * SE, length = 15)

hpop <- dnorm(pop_x,mu,sigma)
hsam <- dnorm(sam_x,mu,SE)


plot(pop_x, hpop, type = "l", col = "red",ylim=c(0,0.002), main = "Population vs. Sampling Distriubtion", xlab="Population (Red), Sample (Blue)", ylab = "")
lines(sam_x,hsam, col = "blue")

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

No, because the sample size is too small to assume a normal disribution.

4.48

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.

As sample size increases, standard error would decrease. As standard error decreases, z values would increase. Increases z value would lead to a decreased p-value

Therefore, p-value would decrease as n increases