Advanced Placement Test On Statistics! Trivia Questions Quiz

1. Moving times (in minutes) and weights (in pounds) were recorded for a random sample of 20 moving jobs requiring three-man crews, and the results of the regression analysis are shown below. The equation for the least-squares regression line is:

Predicted weight = 21.84 + 0.037(Time)

Predicted time = 21.84 + 0.037(Weight)

Predicted weight = 25.54 + 0.003(Time)

Predicted time = 25.54 + 0.003(Weight)

Predicted time = 0.037 + 21.84(Weight)

The correct answer is "predicted time = 21.84 + 0.037(Weight)". This equation represents the least-squares regression line for predicting the time required for a moving job based on the weight of the items being moved. The constant term 21.84 represents the estimated time when the weight is 0, and the coefficient 0.037 indicates that for every 1-pound increase in weight, the predicted time increases by 0.037 minutes.

Explanation

The correct answer is "predicted time = 21.84 + 0.037(Weight)". This equation represents the least-squares regression line for predicting the time required for a moving job based on the weight of the items being moved. The constant term 21.84 represents the estimated time when the weight is 0, and the coefficient 0.037 indicates that for every 1-pound increase in weight, the predicted time increases by 0.037 minutes.

2. For Hospital A, the average waiting time (time between walking in the door and seeing a doctor) in the emergency room is 135 minutes with a standard deviation of 45 minutes. For Hospital B, the average waiting time in the emergency room is 90 minutes with a standard deviation of 22.5 minutes. In which hospital are you more likely to wait less than 45 minutes? Assume the distribution of waiting times are normal.

Hospital B, because the average wait-time is only 90 minutes, rather than 135 minutes in Hospital A.

Hospital A, because with twice the standard deviation of Hospital B, it has twice the spread.

Neither, because for both hospitals, the probability of waiting less than 45 minutes is 2.275%.

Neither, because for both hospitals, the probability of waiting less than 45 minutes is 97.725%.

The correct answer is neither, because for both hospitals, the probability of waiting less than 45 minutes is 2.275%. This is because the question states that the waiting times follow a normal distribution, and the average waiting time and standard deviation are provided for each hospital. To determine the probability of waiting less than 45 minutes, we can use the z-score formula and standard normal distribution table. By calculating the z-score for 45 minutes using the average and standard deviation for each hospital, we find that the probability is the same for both hospitals, which is 2.275%. Therefore, neither hospital is more likely to have a waiting time less than 45 minutes.

Explanation

The correct answer is neither, because for both hospitals, the probability of waiting less than 45 minutes is 2.275%. This is because the question states that the waiting times follow a normal distribution, and the average waiting time and standard deviation are provided for each hospital. To determine the probability of waiting less than 45 minutes, we can use the z-score formula and standard normal distribution table. By calculating the z-score for 45 minutes using the average and standard deviation for each hospital, we find that the probability is the same for both hospitals, which is 2.275%. Therefore, neither hospital is more likely to have a waiting time less than 45 minutes.

3. The cause of death and the age of the deceased are recorded for 454 patients from a hospital. Use these values to estimate the probability that a person at this hospital died as a result of an accident if it is known the person was between the ages of 45 and 54.

0.0264

0.0976

0.1322

0.2000

0.4878

The probability that a person at this hospital died as a result of an accident if it is known the person was between the ages of 45 and 54 is estimated to be 0.0976. This estimation is based on the recorded cause of death and age of the deceased for 454 patients from the hospital.

Explanation

The probability that a person at this hospital died as a result of an accident if it is known the person was between the ages of 45 and 54 is estimated to be 0.0976. This estimation is based on the recorded cause of death and age of the deceased for 454 patients from the hospital.

4. A regression equation is given as log(y-hat) = 0.214 - 1.28(x). What is the (approximate) predicted value for y when x = 2?

- 2.346

- 0.171

0.005

0.167

Cannot be determined.

The given regression equation is in the form of log(y-hat) = 0.214 - 1.28(x). To find the predicted value for y when x = 2, we substitute x = 2 into the equation. Therefore, log(y-hat) = 0.214 - 1.28(2) = 0.214 - 2.56 = -2.346. To find the actual value of y, we need to take the antilog of -2.346, which is approximately 0.005. Therefore, the approximate predicted value for y when x = 2 is 0.005.

Explanation

The given regression equation is in the form of log(y-hat) = 0.214 - 1.28(x). To find the predicted value for y when x = 2, we substitute x = 2 into the equation. Therefore, log(y-hat) = 0.214 - 1.28(2) = 0.214 - 2.56 = -2.346. To find the actual value of y, we need to take the antilog of -2.346, which is approximately 0.005. Therefore, the approximate predicted value for y when x = 2 is 0.005.

5. The median of a distribution is 150, and the interquartile range is 50. Identify the statement(s) that must be true. (I). Apprx. 50% of the data are between 125 and 175. (II). Apprx. 50% of the data are less than or equal to 150. (III). Apprx. 75% of the data are greater than 125.

I only

II only

I and III only

II and III only

I, II, and III

The correct answer is II only because the median represents the middle value of a distribution, meaning that approximately 50% of the data are less than or equal to the median. The interquartile range is a measure of the spread of the middle 50% of the data, so it does not provide information about the percentage of data between specific values such as 125 and 175. The statement III cannot be determined based on the information given.

Explanation

The correct answer is II only because the median represents the middle value of a distribution, meaning that approximately 50% of the data are less than or equal to the median. The interquartile range is a measure of the spread of the middle 50% of the data, so it does not provide information about the percentage of data between specific values such as 125 and 175. The statement III cannot be determined based on the information given.