Description

The following is the graded problems for the Foundations for Inference chapter of Open Intro to Statistics. My answers to the questions are bolded. The headers correspond to a question.

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.

  1. What is the point estimate for the average height of active individuals? 171.1 cm What about the median? 170.3 cm

  2. What is the point estimate for the standard deviation of the heights of active individuals? 9.4 cm What about the IQR? 177.8 cm - 163.8 cm = 14 cm

  3. 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. A person who is 180 cm is pretty tall but I wouldn’t call it unusually tall b/c it’s within 1 standard deviation of the mean. By the same token, I would consider someone who is 155 cm small because they are less than one standard deviation from the mean.

  4. 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 expect both the mean and standard deviation to be different. There is variation within the population which would translate into a different sample mean and standard deviations.

  5. 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 \(SD_\bar{x} = \frac{\sigma}{\sqrt{n}}\))? The standard error Compute this quantity using the data from the original sample under the condition that the data are a simple random sample. 0.4174687 \(SD_\bar{x} = \frac{\sigma}{\sqrt{n}} = \frac{9.4}{\sqrt{507}} = 0.4174687\)

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. The 95% applies to the population parameter not the sample.

  2. This confidence interval is not valid since the distribution of spending in the sample is right skewed. True. This is one of the conditions for a confidence interval.

  3. 95% of random samples have a sample mean between $80.31 and $89.11. False. Cofidence interval does not contstrain other sample means.

  4. We are 95% confident that the average spending of all American adults is between $80.31 and $89.11. True. This is assuming that the 436 people in the sample were a renadom draw.

  5. A 90% confidence interval would be narrower than the 95% confidence interval since we don’t need to be as sure about our estimate. False. We need to be just as sure about the estimate. The 90% means that the population parameter is within the CI 90% of the time or that we are less sure that our sample captures the population parameter.

  6. 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 had a large sample size trippling it would not decrease the margin of error.

  7. The margin of error is 4.4. False. That is the difference between the lower bound of the CI and the mean, not the margin of error.

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? The N > 30 rule of thumb is satisfied, the children were randomly selected, and the histogram appears to be normally distributed. I think all conditions have been met.

  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 first count to 10 successfully is less than the general average of 32 months. Use a significance level of 0.10. \(H_0\) is that x_bar >= 32 and \(H_a\) is x_bar < 32

confidence_level <- 0.10
n <- 36
x_bar <- 30.69
sigma <- 4.31
mu <- 32

se_x_bar <- sigma / sqrt(n)

z <- (x_bar - mu) / se_x_bar

z_score_for_confidence_level <- 1.645

p_value <- pnorm(z, mean=0, sd=1) * 2

if(z < z_score_for_confidence_level){
  message(paste(round(z, 3), "is less than", z_score_for_confidence_level, "so reject the null hypothesis (p-value:", round(p_value, 3), ")"))
} else {
  message(paste(round(z, 3), "is not less than", z_score_for_confidence_level, "so do not reject the null hypothesis (p-value:", round(p_value, 3), ")"))
}
## -1.824 is less than 1.645 so reject the null hypothesis (p-value: 0.068 )

  1. Interpret the p-value in context of the hypothesis test and the data. There is a 0.068 probability that the difference between these two means was due to chance.

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

# Same method as http://onlinestatbook.com/2/calculators/inverse_normal_dist.html
zinv <- function(p) {
    if (p >= 1 || p <= 0) {
        return(FALSE)
    }
    if (p < .5) {
        t <- p
        theSign <- -1
    }
    else {
        t <- 1 - p
        theSign <- 1
    }
    
    v <- sqrt(-2.0 * log(t))
    x <- 2.515517 + (v * (0.802853 + v * 0.010328))
    y <- 1 + (v * (1.432788 + v * (0.189269  + v * 0.001308)))
    Q <- theSign * (v - (x / y))
    return(Q)
}

p <- 1 - confidence_level
x1 <- x_bar + sigma * zinv(p)
p2 <- p / 2
x1 <- zinv(.5 - p2)

error <- sigma * -x1
ci_low <- x_bar - error
ci_high <- x_bar + error

Confidence interval: (23.5991387, 37.7808613) or 30.69 \(\pm\) 7.0908613

  1. Do your results from the hypothesis test and the confidence interval agree? Explain. No. Since 32 is inside of the confidence interval you would not reject the null hypotheses. It is the opposite conclusion you reach with the hypothesis test, which indicates to me that I did something wrong.

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.
confidence_level <- 0.10
n <- 36
x_bar <- 118.2
sigma <- 6.5
mu <- 100

se_x_bar <- sigma / sqrt(n)

z <- (x_bar - mu) / se_x_bar

z_score_for_confidence_level <- 1.645

p_value <- pnorm(z, mean=0, sd=1) 

if(z < z_score_for_confidence_level){
  message(paste(round(z, 3), "is less than", z_score_for_confidence_level, "so reject the null hypothesis (p-value:", round(p_value, 3), ")"))
} else {
  message(paste(round(z, 3), "is not less than", z_score_for_confidence_level, "so do not reject the null hypothesis (p-value:", round(p_value, 3), ")"))
}
## 16.8 is not less than 1.645 so do not reject the null hypothesis (p-value: 1 )
  1. Calculate a 90% confidence interval for the average IQ of mothers of gifted children.
p <- 1 - confidence_level
x1 <- x_bar + sigma * zinv(p)
p2 <- p / 2
x1 <- zinv(.5 - p2)

error <- sigma * -x1
ci_low <- x_bar - error
ci_high <- x_bar + error

Confidence interval: (107.5061256, 128.8938744) or 118.2 \(\pm\) 10.6938744

  1. Do your results from the hypothesis test and the confidence interval agree? Explain. No. 100 lies outside the CI which would suggest that we should reject the null hypothesis, but the hypothesis test disagrees. Again it would appear there is something wrong with my understanding as both should lead you to the same conclusion.

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. This is the distributions of the mean value of the samples. It is normally distributed. The mean approximates the population mean. As the sample size increases the sample mean approaches the population mean, and the spread decreases.

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? 0.067
mu <- 9000 
sd <- 1000
x <- 10500
1 - pnorm(x, mu, sd)
## [1] 0.0668072

  1. Describe the distribution of the mean lifespan of 15 light bulbs. Normal and centered around 9,000 hours.

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

n <- 15
se <- sd / sqrt(n)
1 - pnorm(x, mu, se)
## [1] 3.133452e-09

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

  2. Could you estimate the probabilities from parts (a) and (c) if the lifespans of light bulbs had a skewed distribution? Not using the above method.

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. Increase. Because as you increase the N the standard error would decrease and the p-value would increase.