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)
## Please visit openintro.org for free statistics materials
## 
## Attaching package: 'openintro'
## The following objects are masked from 'package:datasets':
## 
##     cars, trees
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?

The average height is 171.1. The median is 170.3.

summary(bdims$hgt)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   147.2   163.8   170.3   171.1   177.8   198.1
  1. What is the point estimate for the standard deviation of the heights of active individuals? What about the IQR?

The standard deviation is 9.407205. The IQR is 14.

sd(bdims$hgt)
## [1] 9.407205
IQR(bdims$hgt)
## [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.

If the average is 171.1, 180 cm will be considered tall and 155 cm will considered short.

mean<-mean(bdims$hgt)
n<-180
if (n>mean){
print("Tall")
}else if(n<mean){
print("Short")
}
## [1] "Tall"
n<-155
if (n>mean){
print("Tall")
}else if(n<mean){
print("Short")
}
## [1] "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 because another random sample will give us a new set of mean and standard deviation.

  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.
# $SD_x = \frac{\sigma}{\sqrt{n}}$)
n<-507
hgt_sd<-sd(bdims$hgt)
hgt_mean<-mean(bdims$hgt)
hgt_sd / sqrt(n)
## [1] 0.4177887

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(tgSpending)
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(tgSpending$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.

[1] 80.30173 89.11180 False. We are close though.

n<-436
tg_mean<-mean(tgSpending$spending)
tg_sd<-sd(tgSpending$spending)
lower_vector <- tg_mean - 1.96 * tg_sd / sqrt(n)
upper_vector <- tg_mean + 1.96 * tg_sd / sqrt(n)

c(lower_vector, upper_vector)
## [1] 80.30173 89.11180
  1. This confidence interval is not valid since the distribution of spending in the sample is right skewed.

False. Based on the Central Limit Theorem, if the sample size selected at a random and is greater than 30, the sample mean is normally distributed and its confidence interval is valid.

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

True.

n<-436
tg_mean<-mean(tgSpending$spending)
tg_sd<-sd(tgSpending$spending)
lower_vector <- tg_mean - 1.96 * tg_sd / sqrt(n)
upper_vector <- tg_mean + 1.96 * tg_sd / sqrt(n)

c(lower_vector, upper_vector)
## [1] 80.30173 89.11180
  1. We are 95% confident that the average spending of all American adults is between $80.31 and $89.11.

True, 95% CI covers up to 2 standard deviations of the American adults.

  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.

n<-436
tg_mean<-mean(tgSpending$spending)
tg_sd<-sd(tgSpending$spending)
lower_vector <- tg_mean - 1.65 * tg_sd / sqrt(n)
upper_vector <- tg_mean + 1.65 * tg_sd / sqrt(n)

c(lower_vector, upper_vector)
## [1] 80.99844 88.41509
  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.

n=436
1.96*(tg_sd / sqrt(n))
## [1] 4.405038
m=436*3
1.96*(tg_sd / sqrt(m))
## [1] 2.54325
  1. The margin of error is 4.4.

True.

n=436
1.96*(tg_sd / sqrt(n))
## [1] 4.405038

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?

Yes, sample is randomly selected and have more than 30 sample size.

  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.

Since the significance level is 0.10, the confidence interval will be 0.90. So we will be using +/- 1.645.

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

lower_vector<- mean - 1.645 * sd / sqrt(n) 
upper_vector<- mean + 1.645 * sd / sqrt(n)

c(lower_vector, upper_vector)
## [1] 29.50834 31.87166

Our data shows that on 90% confidence that the general average counts to 10 between ages 29.50834 to 31.87166 (months).

  1. Interpret the p-value in context of the hypothesis test and the data.
Z_score <- (30.69 - 32) / (4.31 / sqrt(36)); Z_score
## [1] -1.823666
p_value <- pnorm(Z_score); p_value
## [1] 0.0341013

The p_value is less than 0.10 so we can reject the hypothesis.

  1. Calculate a 90% confidence interval for the average age at which gifted children first count to 10 successfully.
n<-36
mean<-30.69
sd<-4.31

lower_vector<- mean - 1.645 * sd / sqrt(n) 
upper_vector<- mean + 1.645 * sd / sqrt(n)

c(lower_vector, upper_vector)
## [1] 29.50834 31.87166
  1. Do your results from the hypothesis test and the confidence interval agree? Explain.

Yes, because confidence interval and significance levels correlate. They both total up to 1. For example, if CI= 0.90, 1 minus 0.90= 0.10 for the significance levels.

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.
n=36
sd=6.5
H0=118.2
HA=100

Z_score <- (H0 - HA) / (sd/ sqrt(n)); 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.
n<-36
mean<-118.2
sd<-6.5

lower_vector<- mean - 1.645 * sd / sqrt(n) 
upper_vector<- mean + 1.645 * sd / sqrt(n)

c(lower_vector, upper_vector)
## [1] 116.4179 119.9821
  1. Do your results from the hypothesis test and the confidence interval agree? Explain.

Yes they agree. The p value is small so we reject the null hypothesis. In the confidence interval, it does not include the average of 100 IQ for the population.

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.

Sampling distribution is the randomly selection of samples from a population and calculating the mean, and then creating a distribution from the results of the mean values. As the sample size increases, the spread gets smaller and the shape will get taller and more sharp. The center will stay relatively the same because that is the 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.

lb_mean<-9000 lb_sd<-1000

H0<-9000 HA<-10500 (a) What is the probability that a randomly chosen light bulb lasts more than 10,500 hours?

pnorm(10500, 9000, 1000, lower.tail = FALSE)
## [1] 0.0668072
  1. Describe the distribution of the mean lifespan of 15 light bulbs.
sd <- 1000
mean <- 9000

sample_sd <- sd/sqrt(15)
sample_sd
## [1] 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.
par(mfrow = c(2, 1))

xpop <- 7000:12000
ypop <- dnorm(xpop,mean=9000,sd=1000)

xsample <- 7000:12000
ysample <- dnorm(xsample,mean=9000,sd=1000/sqrt(15))

plot(xpop,ypop)
plot(xsample, ysample)

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

It will be hard to estime the probabilities because the sample size is less than 30, it would’nt be a normal distribution based the Central Limit Theorem.

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 will decrease as the sample size increases. Sample size increases cause the standard of error to decrease, which lowers the standard deviations from the population mean which lowers the p-value.