Question 4.4: Heights of adults.

  1. What is the point estimate for the average height of active individuals? What about the median?
  1. What is the point estimate for the standard deviation of the heights of active individuals? What about the IQR?
  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. 180cm
(180 - 171.1) / 9.4
## [1] 0.9468085

NO. Falls within 2 Standard deviations of the mean.

155 cm

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

NO. Falls within 2 Standard deviations 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.

No. Point estimates that are based on samples only approximate the population parameter, and they vary from one sample to another.

  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 \(SD_\bar{x} = \frac{\sigma}{\sqrt{n}}\). Compute this quantity using the data from the original sample under the condition that the data are a simple random sample.
9.4 / sqrt(507)
## [1] 0.4174687

Question 4.14: Thanksgiving spending, Part I.

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

FALSE: Inference is made on the population parameter and not on the point estimate.

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

FALSE: The sample is random and >=30 so we can be lenient with the skew.

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

FALSE: The confidence interval is not about the mean.

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

TRUE: The sample is random which can be used to represent the general population.

  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: Sample \(2^3 = 8\) times the number of people in the initial sample.

  1. The margin of error is 4.4.

TRUE: The margin of error is half the width of the interval.

Question 4.24: Gifted children, Part I.

  1. Are conditions for inference satisfied? YES:
    • Randomly sampled
    • Size >= 30
    • Slightly skewed but because the sample size is > 30 we can be lenient with the skew.
  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 first count to 10 successfully is less than the general average of 32 months. Use a significance level of 0.10.

\(H_0 = 30.69\)

\(H_A > 30.69\)

\(Z = \frac{(\bar{x} - H_0)}{SE_\bar{x}}\)

SE <- 4.31 / sqrt(36)
SE
## [1] 0.7183333
Z <- (32 - 30.69) / SE
Z
## [1] 1.823666
normalPlot(bounds = c(Z, Inf));

  1. Interpret the p-value in context of the hypothesis test and the data.
p_value <- 1 - pnorm(Z)
p_value
## [1] 0.0341013

p_value is less than 0.10 so we reject the null hypothesis in favor of the alternative. It takes more than 30 months for the average child to count from 1 to 10 successfully.

  1. Calculate a 90% confidence interval for the average age at which gifted children ???rst count to 10 successfully.
lower_level <- 32 - (1.69 * SE)
upper_level <- 32 + (1.69 * SE)

c(lower_level, upper_level)
## [1] 30.78602 33.21398
  1. Do your results from the hypothesis test and the con???dence interval agree? Explain YES. The null value 30.69 is not in the confidence interval and is therefore implausible.

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

\(H_0 = 100\)
\(H_A \neq 100\)

SE <- 6.5 / sqrt(36)
Z <- (118 - 100) / SE
Z
## [1] 16.61538
normalPlot(bounds = c(-Z, Z))

The p value > 0.10 so we failed to reject the null hypthesis

  1. Calculate a 90% confidence interval for the average IQ of mothers of gifted children.
lower_level <- 100 - (1.69 * SE)
upper_level <- 100 + (1.69 * SE)

c(lower_level, upper_level)
## [1]  98.16917 101.83083
  1. Do your results from the hypothesis test and the confidence interval agree? Explain.
    YES, the alternative does not fall within the confidence interval range.

Question 4.34: CLT.

De???ne 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.

Under the Central Limit Theorem, the sampling distribution of the mean approaches a normal distribution when the sample size increases. Sampling distributions help us understand variability so as the size increase, the spread gets smaller, the center moves closer to the population mean and the shape becomes more normal.

Question 4.40: CFLBs.

A manufacturer of compact ???uorescent 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?

\(Z > 10,500\)

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

The population standard deviation is known and the data are nearly normal, so the sample mean will be nearly normal with distribution \(N(\mu = 9000, SE = 258.199)\)

  1. What is the probability that the mean lifespan of 15 randomly chosen light bulbs is more than 10,500 hours?
Z <- (10500 - 9000) / SE
1-pnorm(Z)
## [1] 3.133452e-09
1 - 1
## [1] 0
  1. Sketch the two distributions (population and sampling) on the same scale.
samp <- 10500
pop_mean <- 9000
sd <- 1000
n <- 15

bulbs <- rnorm(n, samp, sd)
options(scipen = 999)
hist(bulbs,  probability = T, xlim = range(5000:15000))

x <- 3000:15000
y <- dnorm(x = x, mean = pop_mean, sd = sd)
lines(x = x, y = y, col = "blue")

y2 <- dnorm(x = x, mean = samp, sd = SE)
lines(x = x, y = 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 the sample size is less than 30 which is not sufficient enough to yield a nearly normal sampling distribution if the population distribution is not nearly normal.

Question 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.

With other factors remaining constant, increasing the sample size makes the standard error smaller which in turns decreases the p value. This is evidence why the althernative is chosen when the p value gets smaller because the sample size deviates from the null hypothesis.