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$hgt)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   147.2   163.8   170.3   171.1   177.8   198.1
sd(bdims$hgt)
## [1] 9.407205
(177.8-170.3)
## [1] 7.5

  1. What is the point estimate for the average height of active individuals? What about the median? The point estimate = 171.1 while the median = 170.3.

  2. What is the point estimate for the standard deviation of the heights of active individuals? What about the IQR? The point estimate for the standard deviation is 9.4 while the IQR is 7.5.

  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. Yes above 180cm is unusually tall and 155cm is unusually short because they are beyond the IQR meaning they are beyond the middle 50%.

  4. 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. The mean and standard deviation would almost always vary from the ones given above. For a randomly picked sample it is usually going to provide different means and standard deviations because of the different observations however it is possible to get the same values but it is unlikely.

  5. 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. To measure the standard deviation of a sampling distribution or an estimate of that standard deviation we use standard error.

sd(bdims$hgt) / sqrt(nrow(bdims))
## [1] 0.4177887

The standard error of the original sample is 0.42.

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. This is true because the 95% confidence interval for this sample is between those two values.

  2. This confidence interval is not valid since the distribution of spending in the sample is right skewed. False. Although the distribution is skewed our sample size is relatively large (n >> 30).

  3. 95% of random samples have a sample mean between $80.31 and $89.11. False, the % of random samples with a mean inbetween those values will always change, however the more random samples provided the closer the proportion will be to 0.95.

  4. We are 95% confident that the average spending of all American adults is between $80.31 and $89.11. False, this only holds true for our sample size. Statistically speaking this sample cannot represent the population.

  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, if the CI decreases then we will notice a decrease in our critical value which narrows the range of estimations.

  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 because margin of error is a factor of (n)^.5 we would need a sample size 9x larger.

  7. The margin of error is 4.4. True.

1.96*sd(tgSpending$spending)/sqrt(436)
## [1] 4.405038

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?

These conditions are satisfied only when making assumptions about gifted children from schools in large cities. The distribution is approxametly normal and observations are independent with the sample size above thirty.

  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.

Null Hypoth:The average age of gifted children to successfully count to 10 is 32 months. Alt Hypoth:The average age of gifted children to successfully count to 10 is less than 32 months.

z <- (32 - 30.69) / (4.31/(36^.5))
pnorm(-abs(z))
## [1] 0.0341013

Our p-value is less than .10 so it would be very unlikely to ovserve the dada if the null hypothesis were true, so we can successfully reject our null hypothesis.

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

The probability of our null hypothesis being true (i.e. the age of a gifted child to successfully count to 10 at age 32 months) is very unlikely, in fact it has a 3.4% chance, therefore gifted children are more likely to count to 10 before 32 months.

  1. Calculate a 90% confidence interval for the average age at which gifted children first count to 10 successfully.
se <- 4.31 / sqrt(36)
lower <- 30.69 - 1.65 * se
upper <- 30.69 + 1.65 * se
c(lower, upper)
## [1] 29.50475 31.87525
  1. Do your results from the hypothesis test and the confidence interval agree? Explain.

Yes, 32 months falls just outside the range of our interval.

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 mothers and fathers 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.
x <- gifted$motheriq
t.test(x, mu = 100)
## 
##  One Sample t-test
## 
## data:  x
## t = 16.756, df = 35, p-value < 2.2e-16
## alternative hypothesis: true mean is not equal to 100
## 95 percent confidence interval:
##  115.9657 120.3676
## sample estimates:
## mean of x 
##  118.1667
  1. Calculate a 90% confidence interval for the average IQ of mothers of gifted children.
se <- 6.5 / sqrt(36)
lower <- 118.2 - 1.65 * se
upper <- 118.2 + 1.65 * se
c(lower, upper)
## [1] 116.4125 119.9875
  1. Do your results from the hypothesis test and the confidence interval agree? Explain.

Yes, 100 falls far outside of the interval which makes sense when comparing this to our p-value.

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 of the mean is the process of taking random, independent observations from a population preferably above 30 and calculating the mean of this sample group. As sample size increases we notice the shape to be more symmetric, center is about the same and the spread is smaller.

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

The probability is 0.0668 of choosing a light bulb that lasts over 10,500 hours.

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

The distribution would be centered around 9000 hours, would have a large spread and it would be asymmetric.

  1. What is the probability that the mean lifespan of 15 randomly chosen light bulbs is more than 10,500 hours?
SE <- 1000 / (15^.5)
z <- (10500 - 9000) /  SE
pnorm(-abs(z))
## [1] 3.133452e-09

The data shows that the probability of the sample mean of 15 light bulbs being more than 10,500 is almost zero.

  1. Sketch the two distributions (population and sampling) on the same scale.

The population distribution would be bell shaped, a narrow spread and centered around 9000. We would notice on the right tail the value of 10,500. For the sampling distribution we would notice a wider spread, asymmetric look yet still a center around 9000.

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

For part (a) we could measure the probability as long as the skew is not extreme because of the large size of the population. For a sampling distribution of size 15 we cannot accurately determine a probability because the variance in data is too large.

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.

By increasing population size you increase the z-score. Increasing your z-score will decrease the probability of your event occurring. A decreased probability correlates to a lower p-value.