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?

Point estimate for average height = 147.12, Median = 170.3

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

Point estimate for st. dev. = 9.4

IQR = 177.8 - 163.8
IQR
## [1] 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.
Mean = 171.7
SD = 9.4
Mean - 2* SD
## [1] 152.9
Mean + 2* SD
## [1] 190.5
# 180 cm not considered unusually tall, 155 cm not considered unusually tall either. They are both within 2 SD of the mean


  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.

Point estimates based on samples approximate the population so the mean and st dev. of a new sample may 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 SDx ̄ = pn )? Compute this quantity using the data from the original sample under the condition that the data are a simple random sample.

We use SE:

SE = 9.4 / 507^.5
SE
## [1] 0.4174687


4.14 Thanksgiving spending, Part I.

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

False. The point estimate is always in the confidence interval. Inference is made on the population parameter.

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

We can be lenient with the skew given the size of the sample.

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

False, the confidence interval does not make inferences on sample means.

  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, we would need to use a sample 9 time larger.

  1. The margin of error is 4.4.
# True
(89.11-80.31) / 2
## [1] 4.4


4.24 Gifted children, Part I.

  1. Are conditions for inference satisfied?

Sample observations are independent since it is a random sample and number of children, 36, is very likely to be less than 10 percent of the population.

The sample is greater than 30.

The population distribution is not strongly skewed.

Conditions seem to be satisfied for inference.

  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

mean <- 30.69
sd <- 4.31
n <- 36

se <- sd/sqrt(n)

Z_score <- (mean-32)/se

pvalue <- pnorm(Z_score)
pvalue
## [1] 0.0341013


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

P-value of 0.034 is smaller than the significance level of 0.10. We reject the null hypothesis since there is sufficient evidence to reject H0 in favor of HA.

  1. Calculate a 90% confidence interval for the average age at which gifted children first count to 10 successfully.
low <- 30.69 - (qnorm(0.95) * (4.31 / sqrt(36)))

up <- 30.69 + (qnorm(0.95) * (4.31 / sqrt(36)))

c(low, up)
## [1] 29.50845 31.87155


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

While the hypothesis test rejected the null hypothesis, the average of 32 months is also not within the confidence interval. So results from the hypothesis test and the confidence interval agree.


4.26 Gifted children, Part II.

  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 = 100 IQ HA ≠ 100 α = 0.10

Z_score <- (118.2 - 100) / (6.5 / sqrt(36))
Z_score
## [1] 16.8
p_value <- 2 * (pnorm(Z_score, 0, 1, lower.tail = FALSE))
p_value
## [1] 2.44044e-63


  1. Calculate a 90% confidence interval for the average IQ of mothers of gifted children.
low <- 118.2 - (qnorm(0.95) * (6.5 / sqrt(36)))
up <- 118.2 + (qnorm(0.95) * (6.5 / sqrt(36)))
c(low,up)
## [1] 116.4181 119.9819


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

Yes they do. The p-value is much lower than the significance level of 0.10. The confidence interval does not include the average IQ of 100 for 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.

The distribution of sample means, called the sampling distributio The sampling distribution is the distribution of sample means. As the sample size increases, the shape of the distribution becomes more normal, the center is closer to the true mean, and the spread of the samples are smaller and vice versa.


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?
1 - pnorm(q = 10500, mean = 9000, sd = 1000)
## [1] 0.0668072


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

The sample will approximate normal distribution since the population has normal distribution.

se <- 1000/sqrt(15)
se
## [1] 258.1989

N(9000, 258.1989)


  1. What is the probability that the mean lifespan of 15 randomly chosen light bulbs is more than 10,500 hours?
pnorm(10500, 9000, 1000/sqrt(15), lower.tail = FALSE)
## [1] 3.133452e-09


  1. Sketch the two distributions (population and sampling) on the same scale.
sd<-1000
mean<-9000
se <- sd / sqrt(15)

norm <- seq(mean - (4 * sd), mean + (4 * sd), length=15)
random<- seq(mean - (4 * se), mean + (4 * se), length=15)
norm2 <- dnorm(norm, mean, sd)
random2 <- dnorm(random, mean, se)

plot(norm, norm2, type="l",col="red",
  xlab="Lightbulb Population vs Sampling",
  main="Distribution", ylim=c(0,0.0020))
lines(random, random2, col="blue")


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

The parameters are based on the population so We can’t estimate (a) without a normal distribution. We can’t estimate (c) either because the sample size is less than 30.


4.48 Same observation, di↵erent 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 depends on the z-value. Since the denominator of the z-value calculation is the standard error, an increase in sample size would result in a decrease in the standard error and an increase the z-value determined. That would result in a decrease in p-value.