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.

library(openintro)
data(bdims)
par(mar=c(3.7,2.5,0.5,0.5), las=1, mgp=c(2.5,0.7,0), cex.lab = 1.5)
histPlot(bdims$hgt, col = COL[1], xlab = "Height", ylab = "")

  1. What is the point estimate for the average height of active individuals? What about the median?
head(bdims$hgt)
## [1] 174.0 175.3 193.5 186.5 187.2 181.5
mean_h <- mean(bdims$hgt)
mean_h 
## [1] 171.1438
median_h <- median((bdims$hgt))
median_h
## [1] 170.3
  1. What is the point estimate for the standard deviation of the heights of active individuals? What about the IQR?
sd_h <- sd(bdims$hgt)
sd_h
## [1] 9.407205
IQR(bdims$hgt, na.rm = T)
## [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.
pos <- mean_h  + 2*sd_h 
neg <- mean_h  - 2*sd_h 

pos
## [1] 189.9582
neg
## [1] 152.3294
pos2 <- mean_h + 1*sd_h  
neg2 <- mean_h - 1*sd_h  

pos2
## [1] 180.551
neg2
## [1] 161.7366
  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.
hist(bdims$hgt, probability = TRUE)
x <- 140:200
y <- dnorm(x = x, mean = mean_h, sd = sd_h)
lines(x = x, y = y, col = "blue")

I would expect the mean and the standard deviation of the new sample to be similar to the ones given above, because the bars are distributed the curve steadly and the normal plot shows regularity between , nothing unusual.

  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.
x <- sd_h/sqrt(nrow(bdims))

x
## [1] 0.4177887

std error is 0.417

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.

library(openintro)
data(data(thanksgiving_spend))
par(mar=c(3.7,2.2,0.5,0.5), las=1, mgp=c(2.5,0.7,0), cex.lab = 1.5)
histPlot(thanksgiving_spend$spending, col = COL[1], xlab = "Spending", ylab = "")

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

True. We are 95% confident that the stated range contains the mean for the population at large and also for the sampled population.

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

False. The confidence interval is not affected by this. The data is skewed beyond what looks to be within normal parameters required for our confidence interval.

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

False. This is not what the confidence interval shows.

  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.

True

  1. The margin of error is 4.4.

True

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.

library(openintro)
data(gifted)
par(mar=c(3.7,2.2,0.5,0.5), las=1, mgp=c(2.5,0.7,0), cex.lab = 1.5)
histPlot(gifted$count, col = COL[1], 
         xlab = "Age child first counted to 10 (in months)", ylab = "", 
         axes = FALSE)
axis(1)
axis(2, at = c(0,3,6))

  1. Are conditions for inference satisfied?

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

  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.

HN: mean = 32 HA: mean < 32

mean <- 30.69
sd <- 4.31
n <- 36
se <- sd/sqrt(n)
Z_s <- (mean-32)/se
pvalue <- pnorm(Z_s)
pvalue
## [1] 0.0341013
  1. Interpret the p-value in context of the hypothesis test and the data.

We reject the null hypothesis since there is sufficient evidence to reject HNULL in favor of HAlternative.

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

c(l, u)
## [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. It means that results from the hypothesis test and the confidence interval agree.

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.

library(openintro)
data(gifted)
par(mar=c(3.7,2.2,0.5,0.5), las=1, mgp=c(2.5,0.7,0), cex.lab = 1.5)
histPlot(gifted$motheriq, col = COL[1], 
         xlab = "Mother's IQ", ylab = "", axes = FALSE)
axis(1)
axis(2, at = c(0,4,8,12))

  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.
Z_s <- (118.2 - 100) / (6.5 / sqrt(36))
Z_s
## [1] 16.8
pvalue <- 2 * (pnorm(Z_s, 0, 1, lower.tail = FALSE))
pvalue
## [1] 2.44044e-63
  1. Calculate a 90% confidence interval for the average IQ of mothers of gifted children.
l <- 118.2 - (qnorm(0.95) * (6.5 / sqrt(36)))
u <- 118.2 + (qnorm(0.95) * (6.5 / sqrt(36)))
c(l,u)
## [1] 116.4181 119.9819
  1. Do your results from the hypothesis test and the confidence interval agree? Explain.

Yes, the interval is well above the mean of HNULL. The analyses agree.

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.

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?
z <- (10500 - 9000) / 1000
p <- 1 - pnorm(z)
p
## [1] 0.0668072
  1. Describe the distribution of the mean lifespan of 15 light bulbs.
sd<-1000
mean<-9000
n<-15
  1. What is the probability that the mean lifespan of 15 randomly chosen light bulbs is more than 10,500 hours?
s <- sd / sqrt(n)
z_b <-(10500-mean)/s
pro <- pnorm(z_b, mean, sd)
pro
## [1] 1.189897e-19
  1. Sketch the two distributions (population and sampling) on the same scale.
ns <- seq(mean - (4 * sd), mean + (4 * sd), length=15)
rs<- seq(mean - (4 * s), mean + (4 * s), length=15)
hn <- dnorm(ns,mean,sd)
hr<- dnorm(rs,mean,s)

plot(ns, hn, type="l",col="green",
xlab="Population vs Sampling",
ylab="",main="Distribution", ylim=c(0,0.002))
lines(rs, hr, col="orange")

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

A normal distribution is required to use the Z-score, also the sample size needs to be greater use the z-score.

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.

The p value should decrease as the sample size increases. The standard error decreases, the z-score increases, and therefore the p value will decrease.