# AP Statistics Unit 5 Practice Test: Sampling Distributions

Test 5 for AP Stats.

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 Question 1

### For which of the following situations would this question be an appropriate question: “Does the amount of homework correlate to students’ grades?”

Question 1 Explanation:
When asking questions about different samples, the samples need to be taken from the same population. Two different grade levels are not considered the same population for this topic, because there are other factors that change from 3rd to 10th grade. Collecting grade data from a single class, while it would be the same population, wouldn’t provide comparison. Students from different schools are not necessarily the same population, as there could be other factors besides homework assigned. The closest way to have different sample data within the same population is to have students from the same school randomly assigned to classes within that school. Answer choice D is correct.
 Question 2

### A potter is working on creating clay pots. The mean height of the pots thrown is 10 inches with a standard deviation of 1.5 inches. What is the probability that a randomly selected pot is between 10.5 and 10.75 inches? A 6.22% B 9.87% C 12.93% D 10.15%
Question 2 Explanation:
The Z-table shows the probability from the mean to the given proportion of the standard deviation away from the mean. 10.5 is approximately 0.33 standard deviations from the mean, corresponding to 0.1293. 10.75 is 0.5 standard deviations from the mean, corresponding to 0.1915. 0.1915 - 0.1293 = 0.0622 or 6.22% and answer choice A is correct.
 Question 3

### The finishing times for a track and field race had an mean of 26.3 seconds with a standard deviation of 0.25 seconds. Between what two values will the middle 33% of the data fall? A 25.9667 to 26.6333 B 26.1926 to 26.4075 C 26.2175 to 26.3825 D 26.2013 to 26.3987
Question 3 Explanation:
The middle 33% of data will have approximately 16.6% of data on each side of the mean. This corresponds most closely with 0.43 standard deviations from the mean. With a standard deviation of 0.25 seconds, 0.25*0.43 = 0.1075. Adding this number to 26.3 and subtracting this number from 26.3 gives us the interval 26.1926 to 26.4075.
 Question 4

### What is an unbiased estimator?

 A As the sample size increases, the estimates converge to the true value of the parameter. B The estimates account for confounding variables. C The mean of the sampling distribution of the estimator is the expected value of the sampling distribution of the parameter. D None of the above
Question 4 Explanation:
An unbiased estimator will, on average, give the same value as the population parameter. Answer choice C is correct.
 Question 5

### A random sample of 100 high school students is taken from the population of students in a given state, for which the overall proportion of students with a driver’s license is 0.45. There is a 68% chance that the sample proportion falls between what two values?

 A 0.14 and 0.76 B 0.35 and 0.55 C 0.35 and 0.45 D 0.40 and 0.50
Question 5 Explanation:
The mean is 0.45 and the standard deviation is $\sqrt {\dfrac{0.45(1-0.45)}{100}≈0.05}.np = 100(0.45) = 45$ and both are greater than 10. The Standard Deviation rule tells us that 68% of the data will fall within 1 standard deviation of the mean, or $0.45 + 0.5$ and $0.45 - 0.5$, which gives us an interval of $0.4 and 0.5$. Answer choice D is correct.
 Question 6

### A video game company is looking at data regarding how many people play its flagship game, and comparing that data between two countries. In country A, 60% of 7,000 people play the game; in country B, 52% of 9,000 people play. The survey will ask 200 people, chosen at random, whether or not they play the game. What are the mean and standard deviation between the proportions of people who play the game in each sample?

 A Mean = 0.08, deviation = 0.004 B Mean = 0.08, deviation = 0.049 C Mean = 0.56, deviation = 0.004 D Mean = 0.56, deviation = 0.049
Question 6 Explanation:
The mean is is the difference of the population proportions $(0.60-0.52 = 0.08)$.

The standard deviation is $\sqrt {\dfrac {0.6(1-0.6)}{200}} + \dfrac {0.52(1-0.52)}{200} =$
$\sqrt {\dfrac {0.24}{200} + \dfrac{0.2496}{200}} = \sqrt{0.0012+0.001248} ≈0.049$.

 Question 7

### For a given population of students, foot length is normally distributed with a mean of 8 inches and a standard deviation of 0.25 inches. A researcher is taking samples of 40 students. What are the mean and standard deviation of the sample mean?

 A Mean 8, deviation 0.00625 B Mean 8, deviation 0.0395 C Mean 8, deviation 0.25 D Mean 8, deviation 0.05
Question 7 Explanation:
The mean will be the same as the population sample, 8. The standard deviation will be $\dfrac{0.25}{\sqrt{40}}≈0.0395$. Answer B is correct.
 Question 8

### A new species has been discovered and researchers are comparing two separate populations on different continents. The height of population A is 20 inches with a standard deviation of 0.8 inches. The height of population B is 22 inches with a standard deviation of 1.2 inches. What are the mean and standard deviation for the difference in sample means with samples of size 100?

 A Mean = 2, deviation = 0.0208 B Mean = 2, deviation = 0.144 C Mean = 2, deviation = 0.456 D Mean = 2, deviation = 1.0
Question 8 Explanation:
The mean = $22-20 = 2$. The standard deviation = $\sqrt {\dfrac{0.8^2}{100} + \dfrac{1.2^2}{100}} = \sqrt{0.0064+0.0144} ≈ 0.144$ Answer B is correct.
 Question 9

4. ### The sampling distribution of a statistic can be simulated by generating repeated random samples from a population.

 A I and IV B I and II C III and IV D I, III, and IV
Question 9 Explanation:
The Central Limit Theorem states that when the sample size is sufficiently large, a sampling distribution of the mean of a random variable will be approximately normally distributed. It requires independent samples and can be simulated by generating repeated random samples from a population. Answer A is correct.
 Question 10

### Data is collected on a sample of 20 students with a population proportion of 0.4 student-athletes. Can the sampling distribution be described as approximately normal?

 A No, because $np$ is too small. B No, because $n(1-p)$ is too small. C No, because $np$ and $n(1-p)$ are too small. D Yes, because $np$ and $n(1-p)$ are large enough.
Question 10 Explanation:
In this sample, $np = 20*0.4 = 5$ and $n(1-p) = 20*0.6 = 12$. Because $np<10$, the value is too small to say the data is approximately normal. Answer choice A is correct.
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