Sampling Distributions

Show your work for each problem.

1.
A survey of 200 randomly chosen voters finds that 54% support a new law. In this situation, the 54% is a...
1. a)standard error
2. b)parameter
3. c)statistic
4. d)population
Answer: statistic
The 54% was computed from a sample of 200 voters, so it is a statistic ( $\hat{p} = 0.54$ ). The share among ALL voters would be the parameter $p$ .
2.
A population has standard deviation $\sigma = 18$ . What is the standard error of the sample mean for samples of size $n = 36$ ?
Answer: 3
$\sigma_{\bar{x}} = \dfrac{18}{\sqrt{36}} = \dfrac{18}{6} = 3$ .
3.
A population has mean $\mu = 40$ and standard deviation $\sigma = 8$ . What is the mean of the sampling distribution of $\bar{x}$ for samples of size $n = 25$ ?
1. a)8
2. b)40
3. c)200
4. d)1.6
Answer: 40
$\mu_{\bar{x}} = \mu = 40$ , whatever the sample size. The sample size only changes the spread: the standard error is $8 / \sqrt{25} = 1.6$ .
4.
Bolts from a machine have mean length $\mu = 50$ mm and $\sigma = 2$ mm. A worker measures random samples of 16 bolts. Find the mean and the standard error of $\bar{x}$ .
Answer: 0.5
1. What is the mean of the sampling distribution, $\mu_{\bar{x}}$ (in mm)? 50
2. What is $\sqrt{16}$ ? 4
3. What is the standard error $2 / 4$ (in mm)? 0.5
5.
IQ scores have $\mu = 100$ and $\sigma = 10$ in a test group. What is the standard error of the mean IQ of random samples of $n = 100$ people?
Answer: 1
$\sigma_{\bar{x}} = \dfrac{10}{\sqrt{100}} = \dfrac{10}{10} = 1$ . The mean $\mu = 100$ plays no role in the standard error.
6.
20% of the apps in a store are games ( $p = 0.2$ ). You pick random samples of $n = 100$ apps. Check the conditions and find the standard error of $\hat{p}$ .
Answer: 0.04
1. What is $np = 100 \cdot 0.2$ ? 20
2. What is $n(1-p) = 100 \cdot 0.8$ ? 80
3. Both are at least 10. What is $\sqrt{\dfrac{0.2 \cdot 0.8}{100}}$ ? 0.04
7.
House prices in a region are strongly skewed to the right. For which sample size is the sampling distribution of the mean price closest to normal?
1. a)n = 5
2. b)n = 10
3. c)n = 40
4. d)n = 15
Answer: n = 40
The population is skewed, so only a large sample makes $\bar{x}$ approximately normal (central limit theorem, rule of thumb $n \geq 30$ ). Of the options, $n = 40$ is the only one that qualifies.
8.
A population has $\mu = 80$ and $\sigma = 12$ . A random sample of $n = 16$ has mean $\bar{x} = 86$ . What is the z-score of this sample mean?
Answer: 2
Standard error $= 12 / 4 = 3$ , so $z = \dfrac{86 - 80}{3} = \dfrac{6}{3} = 2$ . The sample mean is 2 standard errors above $\mu$ .
9.
A population has $\sigma = 20$ . Compare the standard error of $\bar{x}$ for $n = 25$ and for $n = 100$ .
Answer: 0.5
1. What is the standard error for $n = 25$ ? 4
2. What is the standard error for $n = 100$ ? 2
3. What is (new standard error) $\div$ (old standard error)? 0.5
10.
A coin is fair ( $p = 0.5$ ). You flip it $n = 100$ times and record the proportion of heads $\hat{p}$ . What is the standard error of $\hat{p}$ ?
1. a)0.0025
2. b)0.5
3. c)0.05
4. d)0.25
Answer: 0.05
$\sqrt{\dfrac{0.5 \cdot 0.5}{100}} = \sqrt{0.0025} = 0.05$ . The value 0.0025 is the variance (forgot the square root), 0.25 is $p(1-p)$ and 0.5 is $p$ itself.
11.
Exam scores are normally distributed with $\mu = 70$ and $\sigma = 15$ . For a random class of $n = 25$ students, find $P(\bar{x} < 67)$ . Use $P(Z < -1) = 0.1587$ .
Answer: 0.1587
1. What is $\sqrt{25}$ ? 5
2. What is the standard error $15 / 5$ ? 3
3. What is $z = \dfrac{67 - 70}{3}$ ? -1
4. What is $P(\bar{x} < 67)$ ? 0.1587
12.
Phone batteries last $\mu = 50$ hours with $\sigma = 12$ hours. For a random sample of $n = 36$ batteries, find $P(\bar{x} > 54)$ . Use $P(Z < 2) = 0.9772$ and give the answer as a decimal.
Answer: 0.0228
$n = 36 \geq 30$ , so $\bar{x}$ is approximately normal. Standard error $= 12 / 6 = 2$ , $z = \dfrac{54 - 50}{2} = 2$ , $P(\bar{x} > 54) = 1 - 0.9772 = 0.0228$ .
13.
About 5% of a factory's light bulbs are faulty ( $p = 0.05$ ). An inspector takes random samples of $n = 120$ bulbs. Can she use the normal model for $\hat{p}$ ? If not, how large must $n$ be?
Answer: 200
1. What is $np = 120 \cdot 0.05$ ? 6
2. What is $n(1-p) = 120 \cdot 0.95$ ? 114
3. Is the normal model OK? No
4. What is the smallest $n$ with $n \cdot 0.05 \geq 10$ ? 200
14.
Apple weights are normally distributed with $\mu = 180$ g and $\sigma = 20$ g. Find the probability that ONE apple weighs more than 185 g, and that the mean of a bag of 16 apples is more than 185 g. Use $P(Z < 0.25) = 0.5987$ and $P(Z < 1) = 0.8413$ .
Answer: 0.1587
1. z-score of one apple of 185 g: $\dfrac{185 - 180}{20}$ ? 0.25
2. $P$ (one apple $> 185$ g)? 0.4013
3. Standard error of the bag mean, $20 / \sqrt{16}$ ? 5
4. z-score of a bag mean of 185 g? 1
5. $P$ (bag mean $> 185$ g)? 0.1587
15.
Which statement about the central limit theorem is correct?
1. a)The standard error grows as the sample size grows
2. b)For large n, the population itself becomes normal
3. c)It only works when the population is already normal
4. d)For large n, the sample mean is approximately normal whatever the population shape
Answer: For large n, the sample mean is approximately normal whatever the population shape
The central limit theorem says that for large $n$ the distribution of the sample MEAN is approximately normal, whatever the population shape. It does not change the population, it also works for non-normal populations, and the standard error $\sigma / \sqrt{n}$ gets smaller, not larger, as $n$ grows.
16.
Delivery times have $\sigma = 15$ minutes. A company wants the standard error of the mean delivery time to be 1.5 minutes. How many deliveries must it sample? How many for a standard error of 0.75 minutes?
Answer: 400
1. From $15 / \sqrt{n} = 1.5$ : what is $\sqrt{n}$ ? 10
2. So what is $n$ ? 100
3. From $15 / \sqrt{n} = 0.75$ : what is $\sqrt{n}$ ? 20
4. So what is $n$ now? 400
17.
40% of a city's households have a pet ( $p = 0.4$ ). In a random sample of $n = 150$ households, find $P(\hat{p} > 0.46)$ . Use $P(Z < 1.5) = 0.9332$ and give the answer as a decimal.
Answer: 0.0668
Conditions: $np = 60$ , $n(1-p) = 90$ , both at least 10. Standard error $= \sqrt{0.24 / 150} = \sqrt{0.0016} = 0.04$ . $z = \dfrac{0.46 - 0.4}{0.04} = 1.5$ , so $P(\hat{p} > 0.46) = 1 - 0.9332 = 0.0668$ .
18.
LED lamps last $\mu = 1000$ hours with $\sigma = 80$ hours. For a random sample of $n = 64$ lamps, find $P(990 < \bar{x} < 1020)$ . Use $P(Z < -1) = 0.1587$ and $P(Z < 2) = 0.9772$ .
Answer: 0.8185
1. What is the standard error $80 / \sqrt{64}$ ? 10
2. What is the z-score of $\bar{x} = 990$ ? -1
3. What is the z-score of $\bar{x} = 1020$ ? 2
4. What is $P(990 < \bar{x} < 1020)$ ? 0.8185