Graded Exercises

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

library("png")
pp <- readPNG("HgtAdults.png")
plot.new() 
rasterImage(pp,0,0,1,1)

  1. What is the point estimate for the average height of active individuals? What about the median?

Mean = 171.1; Median = 170.3

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

SD = 9.4; IQR = 14 (Q3-Q1)

  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.

No, both are within 2 SDs from the mean height (171.1) which should account for 95% of the population

  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.

No, the new sample values will be similar but not identical as random samples pick up different population values

  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 = pn)? Compute this quantity using the data from the original sample under the condition that the data are a simple random sample.

Standard Error (SD/sqrt(n)) = 0.4174687

4.14 Thanksgiving spending, Part I.

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% con???dence interval based on this sample is ($80.31, $89.11). Determine whether the following statements are true or false, and explain your reasoning

library("png")
pp <- readPNG("ThxGivSpending.png")
plot.new() 
rasterImage(pp,0,0,1,1)

  1. We are 95% con???dent that the average spending of these 436 American adults is between $80.31 and $89.11.

False, for the sample average spending, we are 100%, based on the calculations, we are 95% confident for the whole population

  1. This confidence interval is not valid since the distribution of spending in the sample is right skewed.

False, even though the distribution is right-skewed, the sample size is large enough (436) to be lenient with the skew

  1. 95% of random samples have a sample mean between $80.31 and $89.11.

False,the confidence interval only try to campture the population parameter, not individual observations

  1. We are 95% con???dent that the average spending of all American adults is between $80.31 and $89.11.

True

  1. A 90% confidence interval would be narrower than the 95% con???dence interval since we don’t need to be as sure about our estimate.

True, 90% interval would be narrower than the 95% one

  1. In order to decrease the margin of error of a 95% con???dence interval to a third of what it is now, we would need to use a sample 3 times larger.

False, we would need to use a sample 9 times larger as per the SE formula of SD/sqrt(n)

  1. The margin of error is 4.4.

True, difference between avg=$84.71 and the confidence intervals ($80.31, $89.11)

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 identi???ed 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 ???rst counted to 10 successfully. Also provided are some sample statistics

library("png")
pp <- readPNG("Giftedch.png")
plot.new() 
rasterImage(pp,0,0,1,1)

  1. Are conditions for inference satis???ed?

No, even thoug sample size (n) is >=30 and that the observations are independent, the distribution is not normal (Bimodal, left-skewed) and the sample is small (less than 100) to be lenient on the skewness

  1. Suppose you read online that children ???rst 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 ???st count to 10 successfully is less than the general average of 32 months. Use a signi???cance level of 0.10.

H0:Average age = 32 months, HA: Average age <32 months, Significance level is 10% (meaning 90% confidence level)

Confidence Interval between 29.50475 and 31.87525. At 90% confidence level

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

Zscore -1.8236659. p-Value (probability of the Zscore) = 0.0341013, because the p-value is less than the significance level of 0.10, we reject the Null hypothesis

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

Confidence Interval between 29.50475 and 31.87525

  1. Do your results from the hypothesis test and the con???dence interval agree? Explain.

Yes both agree, p-Value compared to the significance level of 0.10 and the confidence interval at a 90% confidence level

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.

library("png")
pp <- readPNG("MothersIQ.png")
plot.new() 
rasterImage(pp,0,0,1,1)

  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 signi???cance level of 0.10.

H0: IQ=100, HA: IQ!=100, Significance level is 10%

Zscore 16.8. p-Value (probability of the Zscore) = 2.440440110^{-63}, because the p-value is much lower than the significance level of 0.10, we reject the Null hypothesis

  1. Calculate a 90% con???dence interval for the average IQ of mothers of gifted children.

Confidence Interval between 116.4125 and 119.9875

  1. Do your results from the hypothesis test and the con???dence interval agree? Explain.

Yes both agree, p-Value is very smal compared to the significance level of 0.10 (two-sided hypothesis, that’s why we multiply by 2 as the normal model is symmetric) and the confidence interval at a 90% confidence level does not include the H0: IQ=100

4.34 CLT.

De???ne 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 represents the distribution of the point estimates based on samples of a ???xed size from a certain population. If a sample consists of at least 30 independent observations and the data are not strongly skewed, then the distribution of the sample mean is well approximated by a normal model. As the sample size increases so the frequency at the center and the spread gets reduced

4.40 CFLBs.

A manufacturer of compact ???uorescent 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?

p-Value (probability of the Zscore > 10,500) = 0.0668072

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

Normal distribution with mean=9000 and SE=1000/Sqrt(15)

  1. What is the probability that the mean lifespan of 15 randomly chosen light bulbs is more than 10,500 hours?

p-Value (probability of the Zscore > 10,500) = 3.133452210^{-9}

  1. Sketch the two distributions (population and sampling) on the same scale.
scalenum <- 7000:12000
pop <- rnorm(scalenum,mean=9000,sd=1000)
sample <- rnorm(scalenum,mean=9000,sd=(1000/sqrt(15)))

hist(pop)

hist(sample)

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

Not with a normal distribution approach as the sample size is small (less than 30 observations) in order to be lenient with the skewed distribution

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

p-Value will decrease as the SE decreases due to the larger sample size. Meaning, Z score value increases as wel as its corresponding probability or area under the curve (complement to p-Value)