Chapter 4 Foundations for Inference
Graded: 4.4, 4.14, 4.24, 4.26, 4.34, 4.40, 4.48
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? (See the next page for parts (b)-(e).)
No unsual heights because both are within 2 SD
- 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 SD¯ x = pn)? Compute this 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] 0.4174687
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.
- We are 95% confident that the average spending of these 436 American adults is between $80.31 and $89.11.
False, based on the data above
- This confidence interval is not valid since the distribution of spending in the sample is right skewed.
False. From the data provided we already know n=436 and n > = 30. The skew does not 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. If we wanted to increase it up to 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 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
- Are conditions for inference satis???ed?
Yes, the sample data is considered independent, n is over 30
- 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 ???rst count to 10 successfully is less than the general average of 32 months. Use a signi???cance 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.
The p-value is lower than the signi???cance level, we can conclude that gifted children count to 10 quicker than a normal child.
- Calculate a 90% con???dence interval for the average age at which gifted children ???rst count to 10 successfully.
LowTail <- round(mu-1.645 * SE, 2)
UpTail <- round(mu+1.645* SE, 2 )
LowTail
## [1] 29.51
UpTail
## [1] 31.87
- Do your results from the hypothesis test and the con???dence interval agree? Explain.
Yes, by rejecting the NULL hypotheis from 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 signi???cance level of 0.10.
x <- 100
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] 0
Null hypothesis (H0): The average of mother’s IQ of gifted children will = population’s IQ average.
Alternate hypothesis (HA): The average of the mother’s IQ of gifted children population’s IQ average.
We reject the Null Hypotehsis and accept the Alternate.
- Calculate a 90% con???dence interval for the average IQ of mothers of gifted children.
LowTail <- round(mu - 1.645 *SE, 2)
UpTail <- round(mu + 1.645* SE, 2)
LowTail
## [1] 116.42
UpTail
## [1] 119.98
- Do your results from the hypothesis test and the con???dence interval agree? Explain
The results were both NULL hypothesis and rejected.
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 of the mean is the mean of the population of which the scores were sampled. Therefore, if a population has a mean ??, then the mean of the sampling distribution of the mean is also ??.
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.
- 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] 0.0668
Describe 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, di???erent 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.