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

JR Answer: the average height 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?
sd(bdims$hgt)
## [1] 9.407205
IQR <- 177.8-163.8
IQR
## [1] 14

JR Answer: The point estimate is 9.41 and the point estimate for 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.
tall2sd <- mean(bdims$hgt) + 2*(sd(bdims$hgt)) 
tall2sd
## [1] 189.9582
short2sd <- mean(bdims$hgt) - 2*(sd(bdims$hgt))
short2sd
## [1] 152.3294

Both are not considered to0 tall or too short as they fall within the 95% conf. intervals. If he/she was taller then 189.96 they would be tall. If he/she was shorter then 152.33 they would be considered too short.

  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.

JR Answer: I would expect that mean and SD to be be similiar but not exact as this is a random sample. It would really depend on if more outliers were to be part of the sample if it would deviate much from the total 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.
n <- 507
SE_hgt <- sd(bdims$hgt)/sqrt(n)
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. JR Answer: False - needs to be entire population

  2. This confidence interval is not valid since the distribution of spending in the sample is right skewed. JR Answer: False as long as random and sample is large enough

  3. 95% of random samples have a sample mean between $80.31 and $89.11. JR Answer: False CI refers to population mean not sample mean

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

  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. JR Answer: True - smaller confidence level is narrower.

  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. JR Answer: False. Standard of Error is the standard deviation * the square root of the sample size

  7. The margin of error is 4.4. JR Answer:

Linterval <- 80.31
Uinterval <- 89.11
SE <- (Uinterval - Linterval)/2
SE 
## [1] 4.4

True. CI of 95% is a 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? JR Answer: Yes - Random and independant and sample >30

  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. JR Answer:

mean <- 30.69
n <- 36
sd <- 4.31
x <- 32
se <- sd / sqrt(n)
Z_score <- (mean - x) / se
p <- pnorm(Z_score, mean = 0, sd = 1)
p
## [1] 0.0341013

  1. Interpret the p-value in context of the hypothesis test and the data. JR Answer: Since P value is .03 which is less then the sign. level of .10 , the hypothesis in not correct that the average age of Gifted children first count to 10 is the same as general average.

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

lower <- mean - 1.645 * se
upper <- mean + 1.645 * se
c(lower, upper)
## [1] 29.50834 31.87166
  1. Do your results from the hypothesis test and the confidence interval agree? Explain. JR Answer:

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.
  2. Calculate a 90% confidence interval for the average IQ of mothers of gifted children.
  3. Do your results from the hypothesis test and the confidence interval agree? Explain.

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.

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?
  2. Describe the distribution of the mean lifespan of 15 light bulbs.
  3. What is the probability that the mean lifespan of 15 randomly chosen light bulbs is more than 10,500 hours?
  4. Sketch the two distributions (population and sampling) on the same scale.
  5. Could you estimate the probabilities from parts (a) and (c) if the lifespans of light bulbs had 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.