Simulado AP Statistics: Unidades 1 a 5

• A. The mean is larger than the median, and the median is the better summary of a typical household
• B. The mean is smaller than the median, and the median is the better summary of a typical household
• C. The mean equals the median, so either one summarizes the center equally well
• D. The mean is larger than the median, and the mean is the better summary because it uses every value

Resposta: A. The mean is larger than the median, and the median is the better summary of a typical household

In a right-skewed distribution the long tail of high values pulls the mean up toward the tail, while the median stays near the bulk of the data. So the mean is larger than the median. Because the mean is sensitive to those extreme values, the median describes a typical household better. Using every value is a weakness here, not a strength. The mean and median are about equal only for roughly symmetric distributions.

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• A. The median
• B. The interquartile range
• C. The standard deviation
• D. The range
• E. The mean

Resposta: A. The median, B. The interquartile range

The median is the fourth value, 8, and it does not move. The first quartile is the median of 3, 5, 6, which is 5, and the third quartile is the median of the upper three values, 11 before and after the change, so the IQR is 6 both times. The mean jumps from 54/7 (about 7.7) to 102/7 (about 14.6), the range grows from 9 to 57, and the standard deviation grows a lot. Statistics built from the middle of the data are resistant to outliers, and statistics built from every value or from the extremes are not.

• A. Cluster sampling
• B. Stratified random sampling
• C. Simple random sampling
• D. Systematic sampling

Resposta: A. Cluster sampling

The population is divided into natural groups (the schools), a random sample of whole groups is chosen, and every member of the chosen groups is surveyed. That is cluster sampling. In stratified sampling you would take some students from every group, which would mean visiting all 12 schools. A simple random sample would pick individual students from the whole county, and a systematic sample picks every kth name from a list.

• A. Put every sunny plant on fertilizer A and every shady plant on fertilizer B
• B. Within the sunny plants and within the shady plants separately, randomly assign half to each fertilizer
• C. Let the gardener pick which fertilizer each plant gets
• D. Randomly assign fertilizers to all 20 plants and ignore the light difference

Resposta: B. Within the sunny plants and within the shady plants separately, randomly assign half to each fertilizer

This is a randomized block design. Light is a known source of variation, so the plants are blocked by sun and shade, and the fertilizers are randomly assigned within each block. That way each fertilizer is tried equally in both light conditions and the effect of light is separated from the effect of fertilizer. The first option confounds fertilizer with light, the third invites bias, and the last one is valid but wastes the chance to remove the light variation.

• A. 0.16
• B. 0.31
• C. 0.52
• D. 0.57

Resposta: D. 0.57

The condition is that the student plays an instrument, so the sample space shrinks to the 40 + 30 = 70 instrument players. Of those, 40 also play a sport, so the probability is 40/70, about 0.57. The answer 0.31 is the reverse, 40/130, the chance that a sport player also plays an instrument. The answer 0.16 is 40/250, the chance of doing both, and 0.52 is 130/250, the chance of playing a sport at all.

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• A. A and B are independent
• B. A and B are mutually exclusive
• C. P(A or B) = 0.7
• D. P(A given B) = 0.4
• E. P(A or B) = 0.9

Resposta: A. A and B are independent, C. P(A or B) = 0.7, D. P(A given B) = 0.4

Independence means P(A and B) = P(A) times P(B), and 0.4 times 0.5 is 0.2, so A and B are independent. Equivalently, P(A given B) = 0.2/0.5 = 0.4, which equals P(A). For the union, P(A or B) = 0.4 + 0.5 - 0.2 = 0.7. Mutually exclusive events cannot happen together, so P(A and B) would be 0, but here it is 0.2. The value 0.9 forgets to subtract the overlap, which is counted twice.

A basketball player makes 70 percent of free throws, and attempts are independent. The number of made shots in 10 attempts follows a binomial distribution with mean _____ and standard deviation about _____.

Opções para a lacuna 1: 3, 7, 7.7, 10

Opções para a lacuna 2: 2.1, 1.45, 2.65, 7

Resposta: 7, 1.45

For a binomial random variable the mean is n times p = 10 times 0.7 = 7. The standard deviation is the square root of n times p times (1 - p) = sqrt(10 times 0.7 times 0.3) = sqrt(2.1), about 1.45. The value 2.1 is the variance, not the standard deviation, and 2.65 is sqrt(np), which forgets the (1 - p) factor.

• A. We are 95% confident that the proportion of all residents who support the bus line is between 0.501 and 0.599
• B. 95% of the residents in the city support the bus line
• C. There is a 95% probability that a new sample proportion will fall between 0.501 and 0.599
• D. 95% of all possible samples have a sample proportion between 0.501 and 0.599

Resposta: A. We are 95% confident that the proportion of all residents who support the bus line is between 0.501 and 0.599

A confidence interval is a statement about the population parameter, here the proportion of all residents. The 95% describes the method: if you repeated the sampling many times, about 95% of the intervals built this way would capture the true proportion. The interval says nothing about what percent of residents hold a view, and it is not a range for future sample proportions. Those would vary around the true value, not around this one sample's 0.55.

In a random sample of 250 students, 90 own a bicycle. The sample proportion is _____, and the standard error of the sample proportion is about _____.

Opções para a lacuna 1: 0.64, 0.36, 0.28, 90

Opções para a lacuna 2: 0.48, 0.0009, 0.0304, 0.0608

Resposta: 0.36, 0.0304

The sample proportion is p-hat = 90/250 = 0.36. The standard error is sqrt(p-hat times (1 - p-hat) / n) = sqrt(0.36 times 0.64 / 250) = sqrt(0.0009216), about 0.0304. The value 0.48 comes from forgetting to divide by n, 0.0009 is the variance, and 0.0608 is twice the standard error, roughly the margin of error for 95% confidence.

• A. Chi-square test for independence, df = 6
• B. Chi-square test for homogeneity, df = 12
• C. Chi-square test for homogeneity, df = 6
• D. Chi-square test for independence, df = 11

Resposta: C. Chi-square test for homogeneity, df = 6

Separate random samples were taken from each of several populations (the three schools) and one categorical variable was recorded, and the question is whether the distributions match. That is a test for homogeneity. Independence would apply to one random sample from one population with two categorical variables recorded. The table has 3 rows and 4 columns, so df = (3 - 1)(4 - 1) = 6. The number 12 is the count of cells, and 11 is one less than that.

• A. Mean 50, standard deviation 12
• B. Mean 50, standard deviation 2
• C. Mean 50, standard deviation 0.33
• D. Mean 8.33, standard deviation 2

Resposta: B. Mean 50, standard deviation 2

The sampling distribution of x-bar is centered at the population mean, 50, because the sample mean is an unbiased estimator. Its standard deviation is the population standard deviation divided by the square root of the sample size: 12/sqrt(36) = 12/6 = 2. The answer 0.33 divides by n instead of the square root of n. With n = 36 the shape is also close to normal even if the population is not, by the central limit theorem.

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• A. Zero is a plausible value for the difference in population means
• B. A two-sided test of H0: mu_A = mu_B at the 5% significance level would not reject H0
• C. There is a 95% probability that the difference in sample means is between -1.2 and 4.6
• D. The data prove that the two population means are equal
• E. The interval shows mean A is larger than mean B

Resposta: A. Zero is a plausible value for the difference in population means, B. A two-sided test of H0: mu_A = mu_B at the 5% significance level would not reject H0

Because 0 lies inside the interval, a difference of 0 is a plausible value, and a two-sided test at the matching 5% level would not reject the null hypothesis of no difference. That is not the same as proving the means are equal. The data are simply not convincing evidence of a difference. The sample difference is the center of the interval and is known, so there is no probability statement to make about it. And since the interval includes negative values, it does not show that mean A is the larger one.

• A. A student who studies 0 hours is predicted to score 6.5
• B. Every student who studies one more hour scores exactly 6.5 points higher
• C. 6.5 percent of the variation in exam scores is explained by hours studied
• D. For each additional hour studied, the predicted exam score increases by about 6.5 points

Resposta: D. For each additional hour studied, the predicted exam score increases by about 6.5 points

The slope is the change in the predicted response for each one-unit increase in the explanatory variable, so each extra hour goes with about 6.5 more points on average. It describes the model's prediction, not a guarantee for each student, which rules out the second option. The intercept, 52, is the predicted score at 0 hours. The percent of variation explained is r squared, a different quantity.

A linear model has a correlation of r = -0.80. About _____ percent of the variation in the response is explained by its linear relationship with the explanatory variable, and the slope of the least-squares line is _____.

Opções para a lacuna 1: 8, 64, 80, 20

Opções para a lacuna 2: positive, zero, negative, equal to -0.80

Resposta: 64, negative

The proportion of variation explained is r squared: (-0.80) squared is 0.64, so about 64 percent. The slope always has the same sign as r, so a negative correlation means a negative slope. The slope is not equal to r in general, because it also depends on the ratio of the two standard deviations: slope = r times (s_y / s_x).

• A. 73 kg
• B. 5 kg
• C. -5 kg
• D. 68 kg

Resposta: C. -5 kg

The predicted weight is -12 + 0.5(170) = 73 kg. A residual is the actual value minus the predicted value, so it is 68 - 73 = -5 kg. The negative sign means the line overestimates this person's weight. The value 73 is the prediction itself, and 5 kg has the sign reversed.