Our statistics students, as noted were asked to rate their admiration of Hillary Rodham Clinton on a scale of 1 to 7. They also were asked to rate their admiration of Jennifer Lopez and Venus Williams on a scale of 1 to 7. As noted earlier, the mean rating of Clinton was 4.06, with a standard deviation of 1.70.The mean rating of Lopez was 3.72, with a standard deviation of 1.90. The mean rating of Williams was 4.58,with a standard deviation of 1.46. One of our students rated her admiration of Clinton and Williams at 5 and her admiration of Lopez at 4.
a. What is her z score for her admiration rating of Clinton?
b. What is her z score for her admiration rating of Williams?
c. What is her z score for her admiration rating of Lopez?
d. Compared to the other statistics students in our sample, which celebrity does this student most admire?(We can tell by her raw scores that she prefers Clinton and Williams to Lopez, but when we take into account the general perception of these celebrities, how does this student feel about them?)
To calculate the z-scores for each admiration rating, we'll use the formula:
z = (x - μ) / σ
Where:
- z is the z-score
- x is the raw score (in this case, the admiration rating of a student)
- μ is the mean of the population (in this case, the mean admiration rating of the respective celebrity)
- σ is the standard deviation of the population (in this case, the standard deviation of the respective celebrity)
Now, let's calculate the z-scores for each admiration rating:
a. Z-Score for admiration rating of Clinton:
Given: x = 5, μ(Clinton) = 4.06, σ(Clinton) = 1.70
Using the formula:
z(Clinton) = (5 - 4.06) / 1.70
z(Clinton) ≈ 0.553
Therefore, the z-score for the admiration rating of Clinton is approximately 0.553.
b. Z-Score for admiration rating of Williams:
Given: x = 5, μ(Williams) = 4.58, σ(Williams) = 1.46
Using the formula:
z(Williams) = (5 - 4.58) / 1.46
z(Williams) ≈ 0.288
Therefore, the z-score for the admiration rating of Williams is approximately 0.288.
c. Z-Score for admiration rating of Lopez:
Given: x = 4, μ(Lopez) = 3.72, σ(Lopez) = 1.90
Using the formula:
z(Lopez) = (4 - 3.72) / 1.90
z(Lopez) ≈ 0.147
Therefore, the z-score for the admiration rating of Lopez is approximately 0.147.
d. To compare the z-scores, we look at the magnitude of each z-score. The larger the magnitude, the farther away the observation is from the mean in terms of standard deviation units.
In this case, the absolute value of the z-score for Clinton (0.553) is smaller than the absolute value of the z-score for Williams (0.288). Thus, this student's rating for Clinton is closer to the mean admiration rating of Clinton compared to the mean admiration rating of Williams.
Similarly, both the z-score for Clinton and Williams have larger absolute values than the z-score for Lopez (0.147). This suggests that the student's admiration rating for Lopez is closer to the mean admiration rating of Lopez compared to the other two celebrities.
Therefore, based on the z-scores, this student most admires Jennifer Lopez compared to Hillary Clinton and Venus Williams.
Which value (1 or 7) is the high value?
A, B, C. Z = (score-mean)/SD
D. Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions/probabilities related to the Z scores.