4.4 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? The point estimate is 171.1 and the median is 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 of the heights of active individuals is 9.4, the IQR is 14.

c)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. This person is not considered unusually tall, since they are within one standard deviation of the mean.

  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.

I wouldn’t expect the next sample’s mean to be the exact same, since it will be a new group of people, but I would expect the mean to be similar to this one.

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

The calculations would be 9.4/((507)^.5) = 0.417.

4.14 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. FALSE. We know everything about the 436 American adults. So we can 100% be sure that the average is between the range.

  2. This confidence interval is not valid since the distribution of spending in the sample is right skewed. False, snce the sample size is large enough.

  3. 95% of random samples have a sample mean between $80.31 and $89.11.

This is true, since the confidence interval is caluclated of the distribution of sample means, so 95% of the sample means should be between $80.31 and $89.11.

  1. We are 95% confident that the average spending of all American adults is between $80.31 and $89.11. This is true, this is the definition of the confidence interval.

  2. A 90% confidence interval would be narrower than the 95% confidence interval since we don’t need to be as sure about our estimate. This is true, because the z-score is lower. Also, intuitively, as we increase our span of a guess we can increase our confidence in that guess and vice versa.

  3. 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. This is false, we would need to decrease it by 9, since the calculation of standard deviation is divided by the square root of n.

  4. The margin of error is 4.4. We would divide the interval by 2, to get (89.11-80.31)/2=4.4, so yes the margin of error is 4.4

4.24 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.

  1. Are conditions for inference satisfied? yes, every event is independent and the sample size in greater than 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.

Ho: μ = 32 H1: μ =/= 32 x = 30.69 sdev = 4.31 SE = sdev/sqrt(36)

(30.69-32)/(4.31/6) = -1.82. The p-value is .0344 which is less than our significance level of .1, so we to reject Ho. The true mean for gifted children could be something different than 32.

  1. Interpret the p-value in context of the hypothesis test and the data. The p-value is .0344 which is less than our significance level of .1, so we to reject Ho. There is evidence that the true mean for gifted children could be something different than 32.

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

30.69+(1.64)(4.31/6) = 31.87 30.69-(1.64)(4.31/6) = 29.51 The CI is (29.51, 31.97)

  1. Do your results from the hypothesis test and the confidence interval agree? Explain. Yes, since 32 is outside of both intervals, the CI are saying 32 is an unlikely mean for this population.

4.26 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.

  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 significance level of 0.10. Ho: μ = 100 H1: μ =/= 100 x = 118.2 sdev = 6.5 SE= sdev/sqrt(36)

(118.2-100)/(6.5/6)= 16.8, the associated p-value is basically 0, which is less than our significance level of 0.10. Therefore, we reject the null-hypothesis. There is sufficient evidence to say the average IQ of mothers of gifted children is different than the average IQ for the population at large.

  1. Calculate a 90% confidence interval for the average IQ of mothers of gifted children. 118.2+(1.64)(6.5/6) = 119.980 118.2-(1.64)(6.5/6) = 116.423 The CI is (116.423, 119.980)

  2. Do your results from the hypothesis test and the confidence interval agree? Explain. Yes, the lower boundary of the CI is far higher than the population mean of 100.

4.34 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 the means of various samples from the population. The distribution should be more normal and have a smaller spread than the originial distribution.

4.40 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? (10500-9000)/(1000)=1.5, the associated p-value is .066807. sot eh chance that a randomly chosen light bulb lasts more than 10,500 hours is 6.6807%

  2. Describe the distribution of the mean lifespan of 15 light bulbs. The mean should be around 9000, the standard deviation should be about 11000/sqrt(15)=258.1989 hours and it should be an almost normal distribution.

  3. What is the probability that the mean lifespan of 15 randomly chosen light bulbs is more than 10,500 hours? 1-pnorm(10500,9000,258) (10500-9000)/(1000/(15^.5))=5.809, the associated p-value is practically zero, so teh probability that the mean lifespan of 15 randomly chosen light bulbs is more than 10,500 hours is almost 0.

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

m <- seq(9000 - (4 * 1000), 9000 + (4 * 1000), length=15)
norm1 <- dnorm(m,9000,1000)
n <- seq(9000 - (4 * (1000/sqrt(15))), 9000 + (4 * (1000/sqrt(15))), length=15)
norm2 <- dnorm(n,9000,(1000/sqrt(15)))

plot(m, norm1, type="l",col="green",
     xlab=" Population vs Sample", 
     main=" CFLBs",
     ylim=c(0,0.002))
lines(n, norm2, col="pink")

  1. Could you estimate the probabilities from parts (a) and (c) if the lifespans of light bulbs had a skewed distribution? no, because the sample size is small.

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

Since the sample size is in the denominator of the formula for the z-score, as the sample size increase, the p-value will inevitably decrease.