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Design of a Study: Sampling, Surveys, and Experiments Review Problems for AP Statistics

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By — McGraw-Hill Professional
Updated on Feb 5, 2011

Review the following concepts if necessary:

Problems

1. The five-number summary for a set of data is [52, 55, 60, 63, 85]. Is the mean most likely to be less than or greater than the median?
2. Pamela selects a random sample of 15 of her classmates and computes the mean and standard deviation of their pulse rates. She then uses these values to predict the mean and standard deviation of the pulse rates for the entire school. Which of these measures are parameters and which are statistics?
3. Consider the following set of values for a dataset: 15, 18, 23, 25, 25, 27, 28, 29, 35, 46, 55. Does this dataset have any outliers if we use an outlier rule that
1. is based on the median?
2. is based on the mean?
4. For the dataset of problem #3 above, what is z55?
5. A study examining factors that contributes to a strong college GPA finds that 62% of the variation in college GPA can be explained by SAT score. What name is given to this statistic and what is the correlation (r) between SAT score and college GPA?

Solutions

1. The dataset has an outlier at 85. Because the mean is not resistant to extreme values, it tends to be pulled in the direction of an outlier. Hence, we would expect the mean to be larger than the median.
2. Parameters are values that describe populations, and statistics are values that describe samples. Hence, the mean and standard deviation of the pulse rates of Pamela's sample are statistics, and the predicted mean and standard deviation for the entire school are parameters.
3. Putting the numbers in the calculator and doing 1-Var Stats, we find that – = 29.64, s = 11.78, Q1 = 23, Med = 27, and Q3 = 35.
1. The interquartile range (IQR) = 35 – 23 = 12, 1.5(IQR) = 1.5(12) = 18. So the boundaries beyond which we find outliers are Q1 – 1.5(IQR) = 23 – 18 = 5 and Q3 + 1.5(IQR) = 35 + 18 = 53. Because 55 is beyond the boundary value of 53, it is an outlier, and it is the only outlier.
2. The usual rule for outliers based on the mean is – ± 3s. – ± 3s = 29.64 ± 3(11.78) = (–57,64.98). Using this rule there are no outliers since there are no values less than –5.7 or greater than 64.98. Sometimes – ± 2s is used to determine outliers. In this case, – ± 2s = 29.64 ± 2 (11.78) = (6,08,53.2) Using this rule, 55 would be an outlier.
4. For the given data, – = 29.64 and s = 11.78. Hence,
5. Note that in doing problem #3, we could have computed this z-score and observed that because it is larger than 2, it represents an outlier by the x– ± 2s rule that is sometimes used.

6. The problem is referring to the coefficient of determination—the proportion of variation in one variable that can be explained by the regression of that variable on another.

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