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.

a. Mean and median  of the sample distribution 171.1 and 170.3, respectively.

b. The Standard deveiation is 9.4  and the IQR is Q3-Q1 = (177.8 - 163.9) 0r 14.

c. Heights of 180cm and 155cm are both within two sd from the mean and therefore not that unusual. Relatively speaking 155cm is furthr from the mean than 180cm.
Z180 <- (180 - 171.1) / 9.4
Z155 <- (155 - 171.1) / 9.4
Z180
## [1] 0.9468085
Z155
## [1] -1.712766
d.  No, per the question this is another randon sample so I would not expect the point estimates to be excactly like the original. I would, however, expect that estimates to be similar to the origional estimates because the were sampled from the same population and assuming the sample sizes are the same.

e. The variability of the estimate is measured by the Standard Error, S = (sd / sqrt(n)) = 0.4174687.

4.14 Thanksgiving spending, Part 1

The 2009 holiday retail season, which kicked o??? 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.

a. FALSE - We draw inferences on the Population and the sample distribution.  The point estimate will always fall within the confidence interval by definition.

b. FALSE - The sample observations were independent, the sample size was sufficently large and the population was not "strongly" skewed. Additionaly the larger the sample size the more forgiving you can be with the sample's skew.

c. FALSE - The 95% confidence interval refers to our confidence that the Population mean falls withing the interval not the sampling mean. 

d. TRUE - This is true because the confidence interval is refering to the population mean.

e. TRUE - This is correct, as a 90% confidence interval would only capture 90% vs 95% of the values.

f. FALSE - You would have to increase it by 3^2 or 9 to decrease by 3. Remember, SE = (mean / sqrt(n)).

g. TRUE - The margin of error equals (z* x SE) where z* = 1.96 and SE = (89.11 - 84.71) / 1.96, thus the margin of error  =  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.

a. Yes, condidtions appear to be met since the samples is random / independent, the sample    size is greater than 30 and the histogram does not indicate a significant skew.

b. Ho: mu = 32 Ha: mu <32; 
   SE = (4.31/sqrt(36)) = 0.7183333  
   Z = (30.69-32)/ 0.72 = -1.8194444
   p-val = 0.0344 < .10 
   We would reject Ho and accept Ha.
   
c. The p-value of 0.03 indicates that if the null hypothesis, Ho =32, were true the   probability of an observed value of 30.69 or lower is only 3%.

d. The 90% confidence interval is (30.69  +/- (1.65 ∗ SE)) = (29.502, 31.878).

e. Yes, the hypothesis test and confidence interval appear to agree.  We are 90% confident the average age gifted children count fomr 0 to 10 falls between (29.5, 31.9) this fall short of Ho = 32 month - Ho rejected.

4.26 Gifted Children, P 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.

a. Ho: mu = 100 and Ha: mu != 100
   SE = (6.5 / sqrt(36)) = 1.0833333.
   Z = (118.2 - 100)/1.08 = 16.8518519.
   p-value = (1-pnorm(q=118.2, mean = 100, sd=1.0833)) = 0 
   Given p-value = 0, we would reject Ho and accept Ha.
   
b. The 90% confidence interval is (118.2  +/- (1.65 ∗ 1.08333)) = (116.4125055, 119.9874945).

c. Results seem to agree, We are 90% confident that the average IQs of mothers of gifted kids is falls between 116.8 and 119.9, which is inconsistent with Ho = 100.

4.34 Central Limit Theorem

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 the distribution of sample means of multiple samples (ideally greater than 30). The Central Limit Theorem tell us, the sample distribution of the mean will approximate a normal distribution. As the sample size, n, increases the distribution becomes more normal and the spread decreases.

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.

a. p = 1-pnorm(q=10500, mean=9000, sd = 1000) = 0.0668072
   The probability = 6.68%.
   
b. If the light bulbs are randomly sampled and independent the distribution should approximate the random distribution with mu = 9000 and sd = 258.1988897.

c. p = 1-pnorm(q=10500, mean=9000, sd = 258.20) = 3.1339198\times 10^{-9}
   The probability = 0%, with Z-score of ((10500-9000)/258.20) = 5.81.

d. See plot below; with population = black and sample = blue.
s <- seq(5000,13000,0.01)
plot(s, dnorm(s,9000, 1000), type="l", ylim = c(0,0.002), ylab = "", xlab = "Lifespan (hours)")
lines(s, dnorm(s,9000, 258.1989), col="blue")

e.  No, neither part a or c could be estimated without a nearly normal distribution.  The skew violates the requirements.

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.

Z = [(x - null value) / (sd / sqrt(n))], therefore:
  a. z-score negative - p-value decrease - left tail 
  b. z-score positive - p-value increase - right tail
  
This is because the sd/spread decreases as n get larger.