Heights of adults. (7.7, p. 260) 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 = 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?

Standard deviation = 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.

180 cm is taller than the average person likewise 155 cm is shorter than the average. If the definition of unusual is greater than one standard deviation then both of these individuals are unusually tall/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.

I expected the mean and standard deviation to be close to the one we already have.

  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 \(SD_x = \frac{\sigma}{\sqrt{n}}\))? Compute this quantity using the data from the original sample under the condition that the data are a simple random sample.

Use the standard error

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

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.

\(\mu=84.71 \\\) 95% confidence (80.31, 89.11)

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

False, the range is a population parameter so we are 100% sure the average spending of these 436 American adults 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.

Since n=436 which is > 30 we know by the Central limit theorem the distribution will be normal, this sample is right skewed.

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

False, based on the information provided here we can’t tell. However if the actual mean is 84.71 then we can say that 95% of random samples with have a mean between $80.31 and $89.11.

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

True for this sample of 436.

  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 lower confidence interval is narrower

  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, No you would need to sample 9 times more since margin of error is calculated by sqrt(p ( (1-p) / n)), the square root causes the non-linear requirement

  1. The margin of error is 4.4. True, p = (89.11+80.31)/2=84.71 which is 1 margin of error from the upper and lower bound

Gifted children, Part I. Researchers investigating characteristics of gifted children col- lected 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 dis- tribution 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? The requirement for inference is n>30 and 36 just meets the requirement

  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 fist count to 10 successfully is less than the general average of 32 months. Use a significance level of 0.10.

\(H_o: \mu = 32 \\\) \(H_a: \mu < 32 \\\)

For significance level 0.1 -> z = 1.645

std = 4.31
n = 36
df = n - 1
X_bar = 32
mu = 30.69
se <- std / sqrt(n)
t <- (X_bar - mu) / se
t_crit <- 1.69

Since t (1.823) > t_crit(1.69) we reject the null hypothesis

  1. Interpret the p-value in context of the hypothesis test and the data.
p_val <- 1 - pnorm(t, mean=0, sd=1)

The p_val is 0.034 so there is a 3.4% chance that the difference between those two are due to chance

  1. Calculate a 90% confidence interval for the average age at which gifted children first count to 10 successfully.
c(30.69 - t_crit * se, 30.69 + t_crit * se)
## [1] 29.47602 31.90398
  1. Do your results from the hypothesis test and the confidence interval agree? Explain.

The confidence interval of 10% produced a range of 29.5 - 31.9 months at which a child who counts to 10 can be considered gifted. The hypothesis test of 32 is outside the range so it was correct to reject it.

Gifted children, Part II. Exercise above 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 the 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.
se = 6.5 / sqrt(36)
mu = 118.2
t = (mu - 100) / se

It looks like t=16.8 >> t_crit=1.69 therefore we fail to reject the null hypothesis

  1. Calculate a 90% confidence interval for the average IQ of mothers of gifted children.
c(mu - t_crit * se, mu + t_crit * se)
## [1] 116.3692 120.0308
  1. Do your results from the hypothesis test and the confidence interval agree? Explain.

The confidence interval of 10% produced an IQ range of 116-119 for mothers of gifted children. The hypothesis test of 100 is outside the range so it was correct to fail to reject it.

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.

As sample size (n) increases the shape of the curve becomes more narrow and skewness less apparent. The shape is more normal and the center p is a closer approximation of the true population mean.

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.

$ H_o: = 9000, =1000$ $ H_a: $

  1. What is the probability that a randomly chosen light bulb lasts more than 10,500 hours?
X = 10500
mu = 9000
std = 1000
z = (X - mu) / std
1 - pnorm(z)
## [1] 0.0668072

There is a 6.7% chance that a a randomly chosen light bulb lasts more than 10,500 hours

  1. Describe the distribution of the mean lifespan of 15 light bulbs.
n = 15
se = std/sqrt(n)

The problem statement states the distribution is nearly normal; however since n<30 the standard error will be large, here it’s 258.2

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

\(P(X > \mu | n=15) = 1 - P(X <= \mu | n=15)\)

n = 15
X_bar = 10500
1 - pnorm(X_bar, mean=mu, sd = se)
## [1] 3.133452e-09

It looks like there is a near 0 chance that 15 randomly chosen light bulbs will last more than 10,500 hours.

  1. Sketch the two distributions (population and sampling) on the same scale.
library(ggplot2)
library(dplyr)
## 
## Attaching package: 'dplyr'
## The following objects are masked from 'package:stats':
## 
##     filter, lag
## The following objects are masked from 'package:base':
## 
##     intersect, setdiff, setequal, union
population = 100000
mu = 9000
std = 1000
n = 15
dist = rnorm(population, mu, std)
sample_pop15 = sample(dist, size=n)

data <- rbind(data.frame(x=dist, label=c('Population')), data.frame(x=sample_pop15,             label=c('Sampling')))

data %>% ggplot(aes(x=x,color=label)) + geom_density()

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

I believe we can always “estimate” the probabilities; however, the accuracy of those estimates will be low with a skewed distribution especially since n is so small (<30)

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.

n=50 p_val=0.08

Noting that our initial n > 30 and our true n >> 30 ie 500, I would first expect the stand error to go down since due to it’s sensitivity to n. A lower standard error will give us results closer to the true population parameters and thus lowering our p_value.