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.

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 average height of active individuals? What about the median?

Solution: The point estimate for the average height of active individuals is 171.1 Median is 170.3

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

Solution: Point estimate for standard deviation = 9.4 IQR = 177.8 - 163.8 = 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.

Solution: The heights are not unusual since both Z scores are within 2 stardard deviations as indicated below.

Z_180 <- (180 - 171.1) / 9.4
Z_180
## [1] 0.9468085
Z_155 <- (155 - 171.1) / 9.4
Z_155
## [1] -1.712766
  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.

Solution: The point estimates based on samples approximate the population so the mean and standard Deviation of a different sample may not be 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.

Solution: The standard error (SE) from the sample is 0.4174687 as indicated below

SE = 9.4 / 507^.5
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. Solution: FALSE - Inference is measured on the population parameter. The CI should not be a representation of the sample but the population.

  2. This confidence interval is not valid since the distribution of spending in the sample is right skewed. Solution: False - The data is only slightly skewed.

  3. 95% of random samples have a sample mean between $80.31 and $89.11. Solution: False - CI for the mean of a sample is not representative of other sample means.

  4. We are 95% confident that the average spending of all American adults is between $80.31 and $89.11. Solution: TRUE-The population parameter is estimated by the Point Estimate and the CI respectively.

  5. A 90% confidence interval would be narrower than the 95% confidence interval since we do not need to be as sure about our estimate. Solution: TRUE -With a small percentage of CI, the interval becomes narrow

  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. Solution: False - The sample should 9 times larger

  7. The margin of error is 4.4. Solution: TRUE - The margin of error is half the CI

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? Solution: The conditions for inference are satisfied because the sample random, sample size is large and the data is not skewed.

  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. Solution: H0:??=32 and HA:??<32;

Sig level = 0.10

  1. Interpret the p-value in context of the hypothesis test and the data. Solution: z = (30.69-32)/4.31 pnorm(-.3) = 0.38 p-value = 0.38

We fail to reject H0 since p-value of 0.38 > 0.1

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

The 90% CI is 30.69 ? 1.65???SE where SE = 4.3136???36 =0.72 CI = 29.502 or CI = 31.878

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

Yes, the results from the hypothesis test and the confidence interval seem to agree, the CI of 90% is between 29.50 and 31.88 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 and father IQ of the 36 randomly sampled gifted children. The histogram below shows the distribution of mother 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. Solution: H0:??=100 and HA:?????100
Mean = 118.2
n = 36
SD = 6.5
SE = SD/sqrt(n)

z_score = (Mean -100)/SE
pnorm(z_score)
## [1] 1

If the null hypothesis is true, the probability of sample mean different at significance level of ?? = 0.10, we reject the null hypothesis H0 in favor of HA.

  1. Calculate a 90% confidence interval for the average IQ of mothers of gifted children. Solution:
low = Mean - 1.645*SE
high =  Mean +1.645*SE
low
## [1] 116.4179
high
## [1] 119.9821

The CI is (116.42,119.98)

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

We are 90% confident that the average IQ of mothers of gifted children is between 116.42 and 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.

Solution:

The sampling distribution of the mean is the distribution of sample means of multiple samples. As the sample approaches the size of the population, the sampling distribution becomes a normal distribution. The distribution takes a bell shape.

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? Solution: The probability is 6.68% as indicated below.
mean = 9000
sd = 1000
z_score = (10500 - 9000)/sd
prob = 1-pnorm(z_score)
prob
## [1] 0.0668072
normalPlot(bounds = c(z_score,10000000))

  1. Describe the distribution of the mean lifespan of 15 light bulbs. Solution: This is normal as indicated below.
SE_sample = 1000/sqrt(15)
SE_sample
## [1] 258.1989
  1. What is the probability that the mean lifespan of 15 randomly chosen light bulbs is more than 10,500 hours? Solution:
z_score = (10500 - 9000)/258.2
probability = 1 - pnorm(z_score)
probability
## [1] 3.13392e-09
  1. Sketch the two distributions (population and sampling) on the same scale. Solution: The sampling distribution is the blue below while the population distribution is black.
s <- seq(5000,13000,0.01)
plot(s, dnorm(s,9000, 1000), type="l", ylim = c(0,0.002),)
lines(s, dnorm(s,9000, 258.1989), col="blue")

  1. Could you estimate the probabilities from parts (a) and (c) if the lifespans of light bulbs had a skewed distribution? Solution: We cannot 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.

Solution: It should be expected that the p-value will decrease with a larger sample size as the spead of the distribution will narrow causing the standard deviation to decrease with increase in n.