DATA 606 - Homework

Exercise 4.4

Mean = 171.1 cm
Median = 170.3 cm
Std dev = SD = 9.4 cm
IQR = Q3 - Q1 = 14 cm
The sample distribution is approximately normal, so we can use Z scores.
- \(Z_{180} = 0.95\), so 180 is within 1 SD of the mean, which is not that unusual (~68% probability of observations within 1 SD of the mean).
- \(Z_{155} = -1.71\), which is within 2 SD of the mean, which suggests this is not that unusual. However the probability of 155 cm or less is \(P(Z \leq Z_{155}) = 4.3\%\), so 155 is not that common either.
```
m <- 171.1
sd <- 9.4
(z180 <- (180 - m) / sd)
```
```
## [1] 0.9468085
```
```
(z155 <- (155 - m) / sd)
```
```
## [1] -1.712766
```
```
pnorm(z155)
```
```
## [1] 0.0433778
```
No, the sample mean and sample standard deviation are only point estimates, and as we’ve seen, they have their own sampling distributions. From sample to sample, we expect random variability in these point estimates.
We use the standard error: \(SE_{\bar{x}} = s_{\bar{x}} / \sqrt{n} = 0.42\). Assuming that all the conditions for a normal distribution are met, the sampling means \(\bar{x}\) are expected to be normally distributed with a mean equal to the population mean \(\mu\), and a standard deviation equal to the standard error \(SE_{\bar{x}}\).
```
(se <- sd / sqrt(507))
```
```
## [1] 0.4174687
```

Exercise 4.14

False; the confidence interval by definition includes the sample mean (so we are 100% confident!).
False: most likely all the conditions are met for a normal sampling distribution:
- Independent sample observations: random sample, and < 10% of the population
- Large enough sample size: 436 >> 30
- Distribution not strongly skewed: this can be relaxed somewhat because of the large sample size of 436.
False: the confidence interval for one sample says nothing about the sample means for other samples.
True: the confidence interval has a 95% probability of containing the true population mean.
True: a confidence interval with a lower level of confidence is constructed using a smaller critical value \(z^{*}\), which means the confidence interval will be narrower.
False: the margin of error \(= z^{*} SE_{\bar{x}}\), and the standard error \(SE_{\bar{x}} = s_{\bar{x}} / \sqrt{n}\). So in order to decrease the margin of error by a factor of 1/3, we need to increase the sample size n by a factor of 9.
True: the margin of error is half the width of the confidence interval, which is 4.4.
```
(89.11 - 80.31) / 2
```
```
## [1] 4.4
```

Exercise 4.24

The conditions for using the normal model in inference are mostly met:
- Sample observations are independent: children are randomly selected, and the sample size is <10% of the population
- Sample size is sufficiently large: 36 > 30
- Data are not strongly skewed: this is the condition that worries me, since there is skew evident in the right tail. However, the sample size is >30, and we’ll assume the slight skew is not a problem.
\(H_{0}\): \(\mu_{gifted} = 32\)
\(H_{A}\): \(\mu_{gifted} < 32\)
\(\alpha = 0.10\)
Note that this is a 1-tailed test.
- Standard error \(SE_{\bar{x}} = s / \sqrt{n} = 0.718\)
- Z-score \(Z = -1.82\)
- 1-tailed p-value = 0.034
- The p-value \(\lt \alpha = 0.10\), so we reject \(H_0\) and accept \(H_A\), concluding that gifted children on average first count to 10 earlier than the general population of children.
```
mu = 32
n <- 36
m <- 30.69
s <- 4.31
(se <- s / sqrt(n))
```
```
## [1] 0.7183333
```
```
(z <- (m - mu) / se)
```
```
## [1] -1.823666
```
```
pnorm(z)
```
```
## [1] 0.0341013
```
The 1-tailed p-value of 0.034 indicates that, if the null hypothesis is true, then there is only a 3.4% chance that the sample mean would be 30.69 months or less. Since this is less than the signficance level of 10%, we reject the null hypothesis in favor of the alternative hypothesis.
The critical value \(z^* = 1.645\) for a two-tailed 90% confidence interval. In this case, the confidence interval is [29.51, 31.87]. Note that the population mean \(\mu = 32\) barely falls outside the confidence interval; on the surface, this would suggest that the average age of gifted children when they first count to 10 may not be statistically significantly different than 32, at a 90% confidence level.

However, it’s important to keep in mind that the hypothesis test was done as a 1-tailed test, so the 2-tailed confidence interval is not directly comparable. Consider how we would construct a 1-tailed confidence interval about the sample mean \(m = 30.69\), with 10% of the probability in the right tail. Then \(z^* = 1.282\), and the interval would span from 0 to \(m + z^* SE_{\bar{x}} = 31.61\). This more strongly suggests that the population average is outside the 90% confidence interval of [0, 31.61], compared to the 2-tailed confidence interval.
```
(z <- qnorm(0.05))
```
```
## [1] -1.644854
```
```
qnorm(0.95)
```
```
## [1] 1.644854
```
```
m + z * se
```
```
## [1] 29.50845
```
```
m - z * se
```
```
## [1] 31.87155
```
```
(z1 <- qnorm(0.9))
```
```
## [1] 1.281552
```
```
m + z1 * se
```
```
## [1] 31.61058
```
At first they appeared to almost disagree, since the hypothesis test was a 1-tailed test and the usual confidence interval is 2-tailed. However, after adjusting the confidence interval to be 1-tailed, they agree.

Exercise 4.26

\(H_0: \mu_{\text{gifted}} = 100\)
\(H_A: \mu_{\text{gifted}} \neq 100\)
\(\alpha = 0.10\)
- The sample distribution looks approximately normal, and we assume that all conditions are met for applying the normal model for inference.
- Note this is a 2-tailed test, since we’re only testing whether the average IQ of mothers of gifted children is “different than” the average IQ for the population at large.
- \(Z = (m - \mu) / SE_\bar{x} = 16.8\)
- The p-value is extraordinarily small, and \(\ll \alpha\), in which case we reject \(H_0\) and conclude that mothers of gifted children have a different IQ on average than the general population.
```
mu <- 100
n <- 36
m <- 118.2
s <- 6.5
(se <- s / sqrt(n))
```
```
## [1] 1.083333
```
```
(z <- (m - mu) / se)
```
```
## [1] 16.8
```
```
# 2-tailed p-value
pnorm(-z)*2
```
```
## [1] 2.44044e-63
```
For a 90% confidence interval around the sample mean, the critical value is \(z^* = 1.645\), in which case the confidence interval is [116.4, 120.0]. Note that the population average IQ of 100 is dramatically far from the confidence interval.
```
# 2-tailed critical value at alpha = 90%
(zstar <- qnorm(0.95))
```
```
## [1] 1.644854
```
```
m - zstar * se
```
```
## [1] 116.4181
```
```
m + zstar * se
```
```
## [1] 119.9819
```
Yes the results agree, and both indicate that at a 10% significance level, mothers of gifted children have a different (higher) IQ on average than the general population. In fact, this would be true even at a much lower significance level, say 0.01.

Exercise 4.34

The “sampling distribution” of the mean refers to the distribution that arises when we (a) select an infinite number of independent random samples (each having the same sample size) from the same population, and (b) then compute the mean for each sample and plot the means on a histrogram.

As the sample size increases, the sampling distribution approaches a normal distribution with mean equal to the population mean and standard deviation equal to the standard error. In other words, as the sample size increases:

The distribution shape more closely resembles a normal distribution
The center of the distribution approaches the population mean \(\mu\)
The spread of the distribution, measured in terms of standard deviation, approaches the standard error \(SE_\bar{x}\).

Exercise 4.40

Normal distribution with \(\mu = 9,000\) and \(\sigma = 1,000\)

\(P(x > 10,500) = 6.68\%\)

mu = 9000
sigma = 1000
pnorm(10500, mean = mu, sd = sigma, lower.tail = FALSE)

## [1] 0.0668072

normalPlot(mean = mu, sd = sigma, bounds = c(10500, Inf))

Assuming the 15 light bulbs are independently chosen at random from the overall population, which is approximated by a normal distribuion, then the sampling distribution should also be approximately normal. The sampling distribution should be centered at \(\mu = 9000\) and have standard deviation \(SE_\bar{x} = \sigma / \sqrt{n} = 258.2\). We can see this in the graphic below.
```
n <- 15
(se <- sigma / sqrt(n))
```
```
## [1] 258.1989
```
```
# generate samples of size 15    
sample_means <- rep(NA, 5000)
for (i in 1:5000) {
    samp <- rnorm(n, mean = mu, sd = sigma)
    sample_means[i] <- mean(samp)
}

hist(sample_means, breaks = 50, probability = TRUE)
x <- seq(7000, 11000, 0.5)
y <- dnorm(x = x, mean = mu, sd = se)
lines(x = x, y = y, col = "blue")
```
The Z-score of 10,500 from the sampling distribution is \(Z_{10500} = 5.81\), which corresponds to a p-value of 3 parts in a billion, e.g., almost 0. This much smaller than the probability of a single bulb lasting longer than 10,500 hours.
```
(z <- (10500 - mu) / se)
```
```
## [1] 5.809475
```
```
pnorm(z, lower.tail = FALSE)
```
```
## [1] 3.133452e-09
```
```
# alternatively, just computed directly
pnorm(10500, mu, se, lower.tail = FALSE)
```
```
## [1] 3.133452e-09
```
See the graphic below; the blue line graphs the population distribuion, while the black line graphs the sampling distribution of the mean.
```
curve(dnorm(x, mu, se), 7000, 11000)
x <- seq(7000, 11000, 1)
y <- dnorm(x, mu, sigma)
lines(x = x, y = y, col = "blue")
```
No. If the distribution were skewed, then we couldn’t have used the normal distribution to calculate part (a). And for part (c), the sample size of 15 is \(\ll\) 30, so not large enough to rely on the Central Limit Theorem to provide for a normally distributed sampling distribution.

Exercise 4.48

The wording in this question is ambiguous, as there seem to be two ways to interpret the question:

First, we hold the sample observation constant (from the question heading “Same observation, different sample size”), which means the sample mean \(\bar{x}\) and sample standard deviation \(s_{\bar{x}}\) remain constant and we just change the sample size \(n\). This corresponds to the case where the sample size was in fact 500 and the observations were made, and in the last step, a sample size of 50 was mistakenly used to arrive at the p-value. In this case, the sample size increases by a factor of 10, and other quantities change as follows:

\[n = 50 \rightarrow 500 = 10n\] \[SE_{\bar{x}} = s_{\bar{x}} / \sqrt{n} \rightarrow SE_{\bar{x}} / \sqrt{10} = 0.316 SE_{\bar{x}}\] \[Z = (\bar{x} - \mu_0) / SE_{\bar{x}} \rightarrow \sqrt{10} Z = 3.16 Z\]

So we have a smaller standard error (by a factor of 0.316) and a larger Z-score (by a factor of 3.16), which means that the p-value will decrease substantially.

For instance, if we assume a two-tailed hypothesis test, then the p-value of 0.08 correspondes to a Z-score of +1.75 or -1.75. Following the logic above, the Z-score would increase to 5.54 (3.16 * 1.75), in which case the p-value would decrease to 3 parts in 100 million.
```
# 2-tailed
(z <- qnorm(0.04, lower.tail = FALSE))
```
```
## [1] 1.750686
```
```
(z1 <- z * sqrt(10))
```
```
## [1] 5.536155
```
```
pnorm(-z1) * 2
```
```
## [1] 3.091832e-08
```
On the other hand, this question could be interpreted as follows: We assume that the sample was done for 50 observations, and the sample mean and sample standard deviation were computed correctly for this sample of size 50. In this case, the sample would have to be redone, this time randomly selecting 500 observations. Now the sample mean and sample standard deviation will change, and we can’t infer in general what the new values will be. However given that the sample size is increasing by a factor of 10, it would be expected that the standard error should decrease, the z-score should increase, and the p-value should decrease. Of course, this all assumes that the sample of size 500 has a sample mean and sample standard deviation having the same order of magnitude as the first sample of size 50.

DATA 606 - Homework - Chapter 4

Kevin Benson

October 18, 2018