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?
summary(bdims$hgt)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   147.2   163.8   170.3   171.1   177.8   198.1
  1. What is the point estimate for the standard deviation of the heights of active individuals? What about the IQR?
sd_hgt <- sd(bdims$hgt)
# iqr is Q3 -Q1
iqr_hgt <- 177.8 - 163.8
iqr_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.
z180 <- (180-171.1) / 9.41
z155 <- (155-171.1) / 9.41
cat(z180, z155)
## 0.9458023 -1.710946

The height of 180cm is within 1 standard deviation so it is not unusually tall. The height of 155cm is within 2 standard deviations so it is not unusually short either.

  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, I would not expect the same mean and standard deviation for another random sample. They would vary among samples since different sample would have randomly selected observations.

  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.

We use sample error to quantify the variability of point estimates.

se_hgt <- sd_hgt/sqrt(507)
se_hgt
## [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.

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

False. It is not the definition of 95% confidence interval. 95% confidence interval refers that we are 95% confident that true mean of population would lie within this interval for different samples.

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

False. If the sample is large enough (n=436) and sample observations are independent, we can have confidence interval.

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

False. It is not the definition of 95% confidence interval. 95% confidence interval refers that we are 95% confident that true mean of population would lie within this interval for different samples.

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

True. This is similar to I explained in answer (a).

  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. Lower confidence interval would be narrower or lesser wide.

  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 margin of error to a third, we would need a sample of size \(3^2=9\)

  1. The margin of error is 4.4.

True. Margin of error is 4.4

moe <- (89.11 - 80.31)/2
moe
## [1] 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?

The sample is random and the sample size is over 30. It doesnt show any strong skewness in the population. So I would say the conditions for inference seems satisfied.

  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.

\(H_0: mean = 32\) (average age at which gifted children fist count to 10 successfully equals the general average of 32 months.) \(H_A: mean < 32\) (average age at which gifted children fist count to 10 successfully is less than the general average of 32 months.)

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

se <- sd / sqrt(n)
z_score <- (30.69-32)/se
pnorm(z_score)
## [1] 0.0341013

Since p-value is less then 0.10, we reject the null hypothesis \(H_0\) in favor of \(H_A\).

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

p-value of 0.034 is less then the given significance level of 0.10. So we reject the null hypothesis \(H_0\) in favor of \(H_A\)

  1. Calculate a 90% confidence interval for the average age at which gifted children first count to 10 successfully.
c(mean-1.65*se, mean + 1.65*se)
## [1] 29.50475 31.87525

90% confidence interval is (29.50, 31.88)

  1. Do your results from the hypothesis test and the confidence interval agree? Explain.

Yes, the results from the hypothesis test and confidence interval are aligned. The upper bound of 90% confidence interval is 31.88 which is lower then the average age.

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.

\(H_0: mean = 100\) (average IQ of mothers of gifted children equals the average IQ for the population at large.) \(H_A: mean \neq 100\) (average IQ of mothers of gifted children is different than the average IQ for the population at large.)

n_m <- 36
mean_m <- 118.2
sd_m <- 6.5

se_m <- sd_m / sqrt(n_m)
z_score_m <- (118.2-100)/se_m
z_score_m
## [1] 16.8

With z score as 16.8, p-value will come close to 0 ehich is less then the significance 0.10. So we reject the null hypothesis \(H_0\) in favor of \(H_A\)

  1. Calculate a 90% confidence interval for the average IQ of mothers of gifted children.
c(mean_m-1.65*se_m, mean_m + 1.65*se_m)
## [1] 116.4125 119.9875

90% confidence interval is (116.41, 119.99)

  1. Do your results from the hypothesis test and the confidence interval agree? Explain.

Results from the hypothesis test and the confidence interval seems aligned. We are 90% confident that the average IQ of mothers of gifted children is between 116.4 and 119.99 which is significantly above the population average of 100.

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.

The “sampling distribution” of the mean is generated by first taking several samples of a particular size from the population, and then taking the mean of each sample. It is the distribution of sample means. Per Central Limit Theorem, it can be approximated by a normal model. As the sample size increases, the shape of the distribution becomes more normal and the center gets closer to the true 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.

  1. What is the probability that a randomly chosen light bulb lasts more than 10,500 hours?
1 - pnorm(q=10500, mean = 9000, sd = 1000)
## [1] 0.0668072
  1. Describe the distribution of the mean lifespan of 15 light bulbs.
sd <- 1000
mean <- 9000

sd_sample <- sd/sqrt(15)
sd_sample
## [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?
1 - pnorm(q=10500, mean = mean, sd = sd_sample)
## [1] 3.133452e-09
  1. Sketch the two distributions (population and sampling) on the same scale.
s <- seq(3000,15000,0.01)
plot(s, dnorm(s,mean = mean, sd = sd), type="l", ylim = c(0,0.002), ylab = "", xlab = "Lifespan (in hours)")
lines(s, dnorm(s,mean = mean, sd = sd_sample), col="red")

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

We could not estimate part a probability without a nearly normal distribution. Also could not estimate part c since 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.

**Z_score = (value-mean)/(sd/sqrt(n)) = (value-mean)*sqrt(n)/sd**

p-value decreases in this case. The p-value depends on the z-score. Since the denominator of the z-value calculation is the standard error, an increase in sample size results a decrease in the standard error and an increase in the z-value. That would decrease the p-value