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
- What is the point estimate for the average height of active individuals? What about the median?
The point estimate for the average height of active individuals is 171.1. The median is 170.3
- What is the point estimate for the standard deviation of the heights of active individuals? What about the IQR?
The SD is 9.4. The IQR is 177.8 - 163.8 = 14
- 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.
x <- 180
mu <- 171.1
sigma <- 9.4
z <- (x - mu)/sigma
z## [1] 0.9468085x <- 155
mu <- 171.1
sigma <- 9.4
z <- (x - mu)/sigma
z## [1] -1.712766They are both within 2 SD, so not ununsual heights.
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 same values listed in the histogram. Since they are a random sample, the values will somewhat change everytime.
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 SDx¯ = ????? n)? Computethis quantity using the data from the original sample under the condition that the data are a simple random sample
n <- 507
SE <- sigma/sqrt(n)
SE## [1] 0.41746874.14 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.
- We are 95% confident that the average spending of these 436 American adults is between $80.31 and $89.11.
False. 100% that the average spending costs of the sampled adults is between $80.31 and $89.11. The point estimate is always in the confidence interval if it was obtained from a population, in this case our population is 436 American adults.
- This confidence interval is not valid since the distribution of spending in the sample is right skewed.
False. We already know n=436 and n>=30. The skew doesn’t play an important role.
- 95% of random samples have a sample mean between $80.31 and $89.11.
False.
- We are 95% confident that the average spending of all American adults is between $80.31 and $89.11.
True.
- 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 we don’t need to be as accurate, lowering the confidence interval will make it less exact.
- 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. We would have to increase our sample size at least 9 times.
- The margin of error is 4.4.
True.
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.
- Are conditions for inference satisfied?
Yes. We can assume the sample data is independent, n is large enough (over 30).
- 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 first count to 10 successfully is less than the general average of 32 months. Use a significance level of 0.10.
x <- 32
n <- 36
min <- 21
mu <- 30.69
sigma <- 4.31
max <- 39
SignificanceLevel <- 0.10
SE <- sigma/sqrt(n)
Z <- (mu - x)/(SE)
p <- pnorm(Z, mean = 0, sd = 1) * 2
p## [1] 0.0682026- Interpret the p-value in context of the hypothesis test and the data.
Since the p value is lower than the significance level, it might suggest that gifted children count to 10 faster than a normal child
- Calculate a 90% confidence interval for the average age at which gifted children first count to 10 successfully.
LowerTail <- round(mu - 1.645 * SE,2)
UpperTail <- round(mu + 1.645 * SE,2)
LowerTail## [1] 29.51UpperTail## [1] 31.87- Do your results from the hypothesis test and the confidence interval agree? Explain. Yes, since we reject the Null hypothesis by taking u = 32.
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.
- 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.
x <- 100
n <- 36
min <- 101
mu <- 118.2
sigma <- 6.5
max <- 131
SignificanceLevel <- 0.10
SE <- sigma/sqrt(n)
Z <- (mu - x)/(SE)
p <- (1 - pnorm(Z, mean = 0, sd = 1)) * 2
p## [1] 0Null hypothesis (H0): The average of mother’s IQ of gifted children = population’s IQ average.
Alternate hypothesis (HA): The average of mother’s IQ of gifted children ??? population’s IQ average.
We reject the Null Hypotehsis and accept the Alternate.
- Calculate a 90% confidence interval for the average IQ of mothers of gifted children.
LowerTail <- round(mu - 1.645 * SE,2)
UpperTail <- round(mu + 1.645 * SE,2)
LowerTail## [1] 116.42UpperTail## [1] 119.98- Do your results from the hypothesis test and the confidence interval agree? Explain.
The results agree since we reject the null hypotehsis.
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.
The sampling distribution of the mean is the mean of the population. A sampling distribution of sample means is a theoretical distribution of the values that the mean of a sample takes on in all of the possible samples of a specific size that can be made from a given population.
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
- What is the probability that a randomly chosen light bulb lasts more than 10,500 hours?
mu <- 9000
sd <- 1000
x <- 10500
p <- round(1 - pnorm(x, mean = mu, sd = sd),4)
p## [1] 0.0668Describe the distribution of the mean lifespan of 15 light bulbs.
What is the probability that the mean lifespan of 15 randomly chosen light bulbs is more than 10,500 hours?
n <- 15
x <- 10500
mu <- 9000
sd <- 1000
SE15 <- sd/sqrt(n)
p15 <- round((1 - pnorm(x, mean = mu, sd = sd))*100, 4)
p15## [1] 6.6807- Sketch the two distributions (population and sampling) on the same scale.
- Could you estimate the probabilities from parts (a) and (c) if the lifespans of light bulbs had a skewed distribution? If you have a skewed distribution, you cannot estimate the probabilities sinc ethey are not normal distributions
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.
The P value will change depending on the sample size. The p value will decrease, since we have more values in our sample.