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.38

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

Point Estimate = Sample Mean = 171.1

Sample Median = 170.3

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

Point Estimate Standard Deviation = 9.4

IQR = Q3 - Q1 = 177.8 - 163.8 = 14

  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.

Someone 180cm tall is considered average, not unusually tall because they are within the inter-quartile range of the distribution.

  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.

One would expect the mean and standard deviation to be approximately similar because we have a large distribution and a large enough sample size to determine popultion estimates from a sample.

  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?

SDx = σ/√n SDx = 9.4/√507

9.4/(sqrt(507))
## [1] 0.4174687

4.14 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.

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

FALSE: We know the average spending of these 436 adults, the 95% confidence interval speaks for average consumer spending for all American Adults.

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

FALSE: The sample size is large enough for us not to be concerned with skewedness.

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

TRUE: Values within the random sample will follow our confidence interval

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

TRUE: This is the correct interpretation of a confidence interval

  1. 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: As our confidence increases, we must also broaden our confidence interval. We make up for a lack of confidence with a larger buffer

  1. 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: The standard error is computed as sample standard deviation/sqrt(n). If n is multiplied by 3, the standard error will be reduced by sqrt(3). Therefore, we would need a sample 9 times as large.

  1. The margin of error is 4.4.

TRUE: It is the difference between the mean and the upper and lower bounds of the confidence interval.

89.11 - 84.71
## [1] 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.

  1. Are conditions for inference satisfied?

Yes. Since this is a simple random sample from less than 10% of all children, we can assume the observations are independent. The sample size is large because it is greater than 30. The sample appears roughly normal.

  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 first count to 10 successfully is less than the general average of 32 months. Use a significance level of 0.10.

Ho = Average age gifted children count to 10 successfully = 32 months Ha = Average age gifted children count to 10 successfully < 32 months This is a one-sided hypothesis test with a significance level of .1

Z = xbar - null value / SEx

SE_gifted = 4.31/(sqrt(36))
mean_gifted <- 30.69
mean_kids <- 32

Z_hypothesis <- (mean_gifted - mean_kids)/SE_gifted
pnorm(-abs(Z_hypothesis))
## [1] 0.0341013
  1. Interpret the p-value in context of the hypothesis test and the data.

The probability of observing a sample mean as small as 30.69 for a sample of 36 students is 3.4%. Since this is smaller than our p-value of .1, we can reject the null hypothesis. We can then reject the null hypothesis that gifted students can counted to 10 at 32 months and accept the alternative hypothesis that gifted students can count to 10 at a rate less than average.

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

point esimate +- Z * SE

Z_value <- qnorm(.95)

Upper_Confidence <- mean_gifted + Z_value*SE_gifted
Upper_Confidence
## [1] 31.87155
Lower_Confidence <- mean_gifted - Z_value*SE_gifted
Lower_Confidence
## [1] 29.50845
  1. Do your results from the hypothesis test and the confidence interval agree? Explain.

These results match the hypothesis test because we are 90% confident that the mean is between 29.5 and 31.8 months. This is lower than our hypothesis test, which compared against the average of 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.

  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.

We can assume indepdence. Normality is confirmed from the distribution. Sample size is large enough.

Ho = Average mother IQ of gifted children = 100 Ha = Average mother IQ of gifted children != 100 This is a two-sided test.

mean_g_mothers = 118.2
mean_mothers = 100
SE_mothers = 6.5/(sqrt(36))

Z_hypothesis_mothers <- (mean_g_mothers - mean_mothers)/SE_mothers
2*pnorm(-abs(Z_hypothesis_mothers))
## [1] 2.44044e-63

The p-value is nearly zero. We can accept the alternative hypothesis that the average IQ of mothers with gifted children is different than the average.

  1. Calculate a 90% confidence interval for the average IQ of mothers of gifted children.
Upper_Confidence_mothers <- mean_g_mothers + Z_value*SE_mothers
Upper_Confidence_mothers
## [1] 119.9819
Lower_Confidence_mothers <- mean_g_mothers - Z_value*SE_mothers
Lower_Confidence_mothers
## [1] 116.4181
  1. Do your results from the hypothesis test and the confidence interval agree? Explain.

Yes, they agree. We are 90% confident that mothers with gifted children have have an IQ between 116.42 and 119.98, far different than the average of 100.

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.

Sampling distribution represents the distribution of the point estimates based on sample of a fixed size from a certain population. As sample size increases, standard error decreases. Therefore, the spread becomes narrower, the shape files more towards the mean and becomes more normal. The center should not change if our sample was adequate.

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.

  1. What is the probability that a randomly chosen light bulb lasts more than 10,500 hours?
pnorm(10500, mean=9000, sd=1000, lower.tail = FALSE)
## [1] 0.0668072
  1. Describe the distribution of the mean lifespan of 15 light bulbs.
# sd/sqrt(n)
bulbs15 <- 1000/sqrt(15)
bulbs15
## [1] 258.1989

The standard deviation is 258 and the mean is 1000.

  1. What is the probability that the mean lifespan of 15 randomly chosen light bulbs is more than 10,500 hours?
pnorm(10500, mean=9000, sd=bulbs15, lower.tail = FALSE)
## [1] 3.133452e-09
  1. Sketch the two distributions (population and sampling) on the same scale.

Using the below website for assistance https://steemit.com/programming/@dkmathstats/plotting-probability-distribution-functions-in-r-using-ggplot2

suppressMessages(library(ggplot2))

x_lower_norm <- 6000
x_upper_norm <- 12000

mean <- 9000
pop_sd <- 1000
sam_sd <- 258

ggplot(data.frame(x = c(x_lower_norm , x_upper_norm)), aes(x = x)) + 
  xlim(c(x_lower_norm , x_upper_norm)) + 
  stat_function(fun = dnorm, args = list(mean, pop_sd), aes(color = "Population")) + 
  stat_function(fun = dnorm, args = list(mean, sam_sd), aes(color = "Sample"))

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

It would be difficult if it were skewed, we would have to use a different probability function.

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 Z-value is determined by Z = xbar - null value / SE. Since SE decreases when sample sizes increase, the overall SE decreases. If SE decreases, the Z increases and the p-value decreases.