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?

Answer

summary(bdims$hgt)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   147.2   163.8   170.3   171.1   177.8   198.1

The mean is 171.1 and the median is 170.3

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

Answer

sd(bdims$hgt)
## [1] 9.407205
IQR(bdims$hgt)
## [1] 14

The standrad deviation of the heights of active individuals is 9.40 and the IQR 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.

Answer

Since we have the average is 171.1 and SD is 9.4, a person who is 180 cm would not be considered unusually tall because if we calculate the Z-score, it will be 0.95 away from the mean which is not unusual.

On the other hand, a person who is 155cm will be considered short because there is a big difference in between average which is 1.71.

  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.

Answer

I believe the standard deviation will not be the same due to the variability of the population.

  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.

Answer

#Standard Error = Standard Deviation/sqrt(n)

SE <- (9.4)/sqrt(507)
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.

Answer

FALSE, because the confidence interval refers to the population not the sample.

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

Answer

FALSE, the sample is enough and the confidence interval is valid when the sample distribution is skewed.

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

Answer

FALSE, because 95% confidence interval does not mean 95% of the sample.

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

Answer

TRUE, because the confidence refers to the acual 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.

Answer

FALSE, 90% confidence interval would be wider.

  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.

Answer

FALSE, because we will need larger sample.

  1. The margin of error is 4.4.

Answer

TRUE. (89.11- 80.31)/2 = 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?

Answer

Yes, since we have good sample size and sample was randomly selected.

  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.

Answer

H0 : μ = 32 , HA : μ < 32

#Calculate the Standard Error:

x <- 32
mean <- 30.69
sd <- 4.31
n <- 36

SE <- sd/sqrt(n) 

#Calculate Z-score:

Z_Score <- round((mean - x) / SE,2)
Z_Score
## [1] -1.82

Next, we calculate the p-value

P_value <- pnorm(Z_Score, mean=0, sd = 1)

P_value
## [1] 0.0343795

Since the P_value is lower than 0.10.

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

Answer

The P_value is 0.034 which is lower than 0.10. This reject the hypothesis of H0.

  1. Calculate a 90% confidence interval for the average age at which gifted children first count to 10 successfully.

Answer

#Calculate the upper limits:

Upper <- mean - 1.64 * SE

Upper
## [1] 29.51193
#Calculate lower limits.

LOWER<- mean+1.64 * SE
LOWER
## [1] 31.86807
  1. Do your results from the hypothesis test and the confidence interval agree? Explain.

Answer

The results from the hypothesis test and the confidence interval agree because the confidence interval says that 90% chance the true mean for gifted children is between 29.51 and 31.87 months.

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. Performahypothesistesttoevaluateifthesedataprovideconvincingevidencethattheaverage 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.
#Given 

x <- 100
n <- 36
mean <- 118.2
sd <- 6.5

#Calculate Standard Error

SE <- sd/sqrt(n)

#Calculate Z score

z_score <- (mean - 100) / SE
z_score
## [1] 16.8
pnorm(z_score)
## [1] 1
  1. Calculate a 90% confidence interval for the average IQ of mothers of gifted children.

Answer

#Calculate the upper:

#for .90 confidence interval, the z value is 1.645

upper <- mean + (1.645 * SE)

upper
## [1] 119.9821
#Calculate lower.

lower<- mean- 1.64 * SE
lower
## [1] 116.4233
  1. Do your results from the hypothesis test and the confidence interval agree? Explain.

The results agree since we rejected the null and the 90% CI range is 116.4 - 119.98

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.

Answer

The sampling distribution of the mean describe the mean of the population and how it is distributed. The central limit theorem is the assumption of normality inherent within a relatively non-skewed distribution, having more than 30 independent samples. As the sample size increases, the shape of sampling distribution becomes close to normal distribution, the center will have higher value, and the spread become narrow.

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?

Answer

#mean = 9000
#sd= 1000
#q= 10500
# calculating the probability

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

The population of the sample has normal distribution with mean = 9000 and SD =1000/15 = 258.2

Answer

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

Answer

#mean = 9000
#sd= 258.2
#q= 10500
# calculating the probability

Prob2 <-1-pnorm(q=10500, mean=9000, sd=258.2)
Prob2
## [1] 3.13392e-09
  1. Sketch the two distributions (population and sampling) on the same scale.
seq <- seq(5000,12000,100)
Population<- dnorm(seq, 9000,1000)
Sample<- dnorm(seq, 9000, 258)

#Population
plot(seq, Population, type="l", main="Population")

#Sample
plot(seq, Sample, type="l", main="Sample")

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

Answer

We can not estimate the probabilities with a skewed 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.

Answer

The p-value will decrease as the sample increase. The SE for sample 1 is sd/sqrt(50) = 0.141

The SE for sample 2 which is increased is sd/sqrt(500) = .045

We can see that the SE decrease as we increase the sample.