4.4 Heights of adults. 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 average height is 171.1 and the median is 170.3

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

    The standard deviation is 9.4 and the IQR is 14

  3. 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.

iqr = 14
#upper outliers
177.8 + 1.5 * iqr 
## [1] 198.8
#lower outliers
163.8 - 1.5 * iqr
## [1] 142.8
Values greater than 198.8 are upper outliers and values less than 142.8 are lower outliers. Thus, a person who is 180 cm is not considered unusually tall, and someone who is 155 cm is not considered unusually short. In addition, 180 is within 1 standard deviation of the mean and 155 is almost 2 standard deviations below the mean.

  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.

    I would not expect the mean and the standard deviation of this new sample to be the ones given above because the new sample with have different point estimates.

  2. 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? Compute this quantity using the data from the original sample under the condition that the data are a simple random sample.

#standard error
9.4/sqrt(507)
## [1] 0.4174687

4.14 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 statement is about the sample mean not population mean

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

    False: Since the sample size is greater than 30, according to the Central Limit Theorem the sample mean is approximately normal

  3. 95% of random samples have a sample mean between $80.31 and $89.11.

    False: Confidence intervals quantify the level of confidence that the parameter is in the interval

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

    True: We are 95% confident that the interval conatains the population parameter

  5. 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: A 90% confidence interval would be narrower because we are less certain it contains the population parameter

  6. 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: 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 9 times larger

  7. The margin of error is 4.4.

    True: calculation below

    round((89.11-80.3)/2, 1)
    ## [1] 4.4

4.24 Gifted children, Part I. Researchers investigating characteristics of gifted children collected 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 distribution 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, conditions for inference satisfied.
    1. Random - random sample
    2. Normal - sample size larger than 30
    3. Independence - sample is less than 10% of city’s population

  2. 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.

    Null hypothesis: The average age at which gifted children fist count to 10 is less than 32 months

    Alternative hypothesis: The average age at which gifted children fist count to 10 is greater than or equal to 32 months

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

zscore <- (30.69-32)/(4.31/sqrt(36)) 
p <- round(pnorm(zscore),2)
p
## [1] 0.03

The p-value is less than 0.10, so there is strong evidence against the null hypothesis and we reject the null hypothesis in favor of the alternative.

  1. Calculate a 90% confidence interval for the average age at which gifted children first count to 10 successfully.
upper <- 30.69 + 1.28*4.31 / sqrt(36)
upper
## [1] 31.60947
lower <- 30.69 - 1.28*4.31 / sqrt(36)
lower
## [1] 29.77053
#90% confidence interval: (29.77053, 31.60947)

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

    Yes, results from the hypothesis test and the confidence interval agree because the confidence interval does not contain 32 and we rejected the the null hypothesis in the hypothesis test.

4.26 Gifted children, Part II. Exercise 4.24 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.

    Null hypothesis: The average IQ of mothers of gifted children is equal to 100

    Alternative hypothesis: The average IQ of mothers of gifted children is not equal to 100

zscore <- (118.2-100)/6.5*sqrt(36)
p <- 1- round(pnorm(zscore),2)
p
## [1] 0
#p-value is less than 0.10 so we reject the null hypothesis in favor of the alternative
  1. Calculate a 90% confidence interval for the average IQ of mothers of gifted children.
upper <- 118.2 + 1.65 * (6.5/sqrt(36))
upper
## [1] 119.9875
lower <- 118.2 - 1.65 * (6.5/sqrt(36))
lower
## [1] 116.4125
#90% confidence interval: (116.4125, 119.9875)

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

    Yes, results from the hypothesis test and the confidence interval agree because we rejected the the null hypothesis in the hypothesis test and the confidence interval does not contain 100.

4.34 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 probability distribution of a statistic obtained through a large number of samples drawn from a population. According to the Central Limit Theorem, the shape of a sampling distribution is normal if sample size is greater than 30. The larger the sample size, the smaller the spread and the closer we converge towards the true mean.

4.40 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(10500, 9000, 1000)
## [1] 0.0668072

  1. Describe the distribution of the mean lifespan of 15 light bulbs.

    Since the population is normal, the sample is also normal.

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

sd<- 1000/sqrt(15)
1-pnorm(10500, 9000, sd)
## [1] 3.133452e-09
  1. Sketch the two distributions (population and sampling) on the same scale.
par(mfrow = c(2,1))
pop <- rnorm(10000, 9000, 1000)
hist(pop, xlim = c(5000, 12000))
sample <- rnorm(15, 9000, 1000)
hist(sample, xlim = c(5000, 12000))

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

    No, we could not estimate the probabilities from parts (a) and (c) if the lifespans of light bulbs had a skewed distribution because it must be normal.

4.48 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.

Increasing sample size will decrease the p-value. As we saw in problem 4.14 part g, as sample size increases, margin of error decreases, and z value increases.