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?

The point estimate for the average height is approximately 171.1cm. The median is also approximately 170.3cm

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

The standard deviation is 9.4cm

Q1 <- 163.8
Q3 <- 177.8
IQR <- Q3-Q1
paste("The interquartile range is ", IQR)
## [1] "The interquartile range is  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.
lower_limit <- Q1 - 1.5*IQR
upper_limit <- Q3 + 1.5*IQR

paste0("lower limit: ", lower_limit) 
## [1] "lower limit: 142.8"
paste0("Upper limit: ", upper_limit)
## [1] "Upper limit: 198.8"

A person that is 1m 55cm (155cm) is not considered unsually short and a person that is 1m 80cm (180 cm) is not considered unusually tall in the height distribution above because both (155cm) and (180cm) are within the range of the lower and upper limit of the distribution. Both heights seem not to represent outliers.

  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.

The mean and standard deviation may not be exactly the same. But they will be close or approximately the same if the the number of sample is the same.

  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.

The quantity is estimated using standard error (SE) population standard deviation = sample standard deviation = 9.4

pop_std <- 9.4
n<-507
SE<-9.4/sqrt(n)
SE
## [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.

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

False. This is because confidence interval is an estimation of population parameter. In this case, the 95% confidence interval ought to be the an estimation of population mean, not the sample mean.

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

False. The confidence interval is valid because the sample size is large enough (i.e n>=30) according to the central limit theorem.

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

False. The confidence interval measures the range of population parameter, mean. It does not focus on the sample parameter.

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

True. The confidence interval generalizes on the population not the sample

  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. The lower the confidence interval, the less the range of the lower and upper limit of the distribution

  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. To decrease the 95% confidence interval to a third of what it is now, we would need to increase the sample size 9x because the standard deviation of sampling distribution has an inverse relationship with the sample size

  1. The margin of error is 4.4.

Lower limit(lm) = $80.31 Upper limit(ul) = $89.11

#Margin of error
lm <- 80.31
ul<-89.11
ME<- (ul-lm)/2
paste0("The margin of error is ", ME)
## [1] "The margin of error is 4.4"

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?

Yes. For the fact that the sample is randomly selected from a target population and the summary statistics are determined including a sample size that is above 30, with a distribution devoid of extreme skewness

  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.
sm<-30.69
pm<-32
significance <- 0.10
n<-36
df<-35
sd<-4.31
se <-sd/sqrt(n)
z_score <- (sm-pm)/se
z_score
## [1] -1.823666
pnorm(z_score)
## [1] 0.0341013

The p-value is less then 0.10, so, we reject the null hypothesis H0 in favor of the alternative hypothesis HA

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

The p-value is less then 0.10. So, we reject the null hypothesis H0 in favor of the alternative hypothesis HA. In other words, there is not enough evidence to proof that the average gifted children count to 10 is at 32 months

  1. Calculate a 90% confidence interval for the average age at which gifted children first count to 10 successfully.
CI_90 <- c(sm-1.65*se, sm + 1.65*se)
CI_90
## [1] 29.50475 31.87525
  1. Do your results from the hypothesis test and the confidence interval agree? Explain.

Yes. The hypothesis test favored the alternative hypothesis (HA) which argued that the average month talented children count to 10 is less than 32 months and the 90% confidence interval supported the claim as the it shows that the average month is between 29.50 to 31.87

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 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:μ=100

HA:μ≠100

sm<-118.2
pm<-100
n <- 36
sdm <- 6.5

sem <- sdm / sqrt(n)
z_score_m <- (sm-pm)/sem
z_score_m
## [1] 16.8

The p-value that corresponds to the z score of 16.8 at α=0.10 for 35 degree of freedom is approaching 0.So, null hypothesis is rejected in favor of the alternative hypothesis.

  1. Calculate a 90% confidence interval for the average IQ of mothers of gifted children.
#z-score at 90% confident interval
z90 <- (-1)*qnorm(0.05)

#IQ Range
CI_90_m <- c((118.2 - z90*sem),(118.2 + z90*sem))
CI_90_m
## [1] 116.4181 119.9819
  1. Do your results from the hypothesis test and the confidence interval agree? Explain.

Yes. The hypothesis test favored the alternative hypothesis (HA) which argued that the average IQ of mothers that birthed gifted children is not 100 and the 90% confidence interval supported the claim as the it shows that the average IQ is between 116.41 to 119.98 approximately.

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.

A sampling distribution is a data distribution that contains only the means or averages of selected samples from a target population. The shape of a sampling distribution is bell-shaped. The mean of a sampling distribution approximates the population mean. The higher the sample size, the closer the sample mean is to the population mean and vice versa. Also, the higher the sample size, the more symmetric the distribution or closer(less standard deviation) and vice versa.

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?
prob <- (1 - pnorm(q=10500, mean = 9000, sd = 1000))

paste0("The probability that a randomly chosen light bulb lasts more than 10,500 hours is ",prob)
## [1] "The probability that a randomly chosen light bulb lasts more than 10,500 hours is 0.0668072012688581"
  1. Describe the distribution of the mean lifespan of 15 light bulbs.
mean <- 9000
sd <- 1000
n <- 15

#Standard Error
SE = sd/sqrt(n)
SE
## [1] 258.1989

The distribution of the mean lifespan of 15 light bulbs is 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?
prob15 <- (1 - pnorm(q=10500, mean = 9000, sd = SE))

paste0("The probability that the mean lifespan of 15 randomly chosen bulbs is more than 10,500 hours is ",prob15)
## [1] "The probability that the mean lifespan of 15 randomly chosen bulbs is more than 10,500 hours is 3.13345216440553e-09"
  1. Sketch the two distributions (population and sampling) on the same scale.
x<-seq(3000,15000,0.01)
population <- dnorm(x, mean = 9000, sd = sd)
sample <- dnorm(x, mean = 9000, sd = SE)
plot(x, population, type = "l", xlab = "Lifespan", ylab = "Probabilities", col = "tomato", ylim = c(0, 0.0017))
lines(x, sample, col = "green")
legend(11300, 0.0012, legend = c("Population", "Sample"), fill = c("tomato", "green"))

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

No. This is because the sample size is small and not enough to reflect a nearly normal distribution.

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.

Hint:

SE <- σ/sqrt(n)

z <- (x – μ) / SE

μ = population mean

σ = population standard deviation

SE <- Standard Error

The value of the z-values depends on the standard error which decreases as the number of sample size increases. Invariably, the z-value increases as the standard error is decreased s a result of increased sample size