students are reluctant to report cheating by other students. a simple survey put
ID: 3124766 • Letter: S
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students are reluctant to report cheating by other students. a simple survey puts this question to an SRS of 400 undergraduates: "you witness two students cheating on a quiz. do you go to the professor?" when the population proportion p is 12% it can be shown that the sample proportion p hat has approximately N(0.12, 0.016). what is the probability that the sample proportion p hat is different from the population proportion p by 5%? Continuous Random Variables (EXAMPLE: Sample Proportion) Students are reluctant to report cheating by other students. A sample survey puts this question to an SRS of 400 undergraduates: "You witness two students cheating on a quiz. Do you go to the professor?" When the population proportion p is 12%, it can be shown that the sample proportion P has approximately N (0.12, 0.016). What is the probability that the sample proportion p is different from the population proportion p by 5%? lp-pl:.05Explanation / Answer
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 0.12 - 0.05 = 0.07
x2 = upper bound = 0.12 + 0.05 = 0.17
u = mean = 0.12
s = standard deviation = 0.016
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = -3.125
z2 = upper z score = (x2 - u) / s = 3.125
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.000889025
P(z < z2) = 0.999110975
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.998221949
Thus, those outside this interval is the complement = 0.001778051 [ANSWER]
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