Suppose (as is roughly true) that 88% of college students were employed last sum

Q: Suppose (as is roughly true) that 88% of college students were employed last sum

a) p^ = x/n = 324/400 = 0.81 [ANSWER] ***************** b) Here, n = 400 p = 0.88 u = mean = p = 0.88 s = standard deviation = sqrt(p(1-p)/n) = 0.016248077 Hence, p^ ~ N(0.88, 0.16248077) [ANSWER] ********************** c) We then add/subtract two standard deviations from the mean, so it is between 0.88 - 2*0.016248 and 0.88 + 2*0.016248 0.847504 and 0.912496 or in percent, 84.75% and 91.25% [ANSWER]

ID: 3129948 • Letter: S

Question

Suppose (as is roughly true) that 88% of college students were employed last summer. A samply survey interviews SRS of 400 college students at OSU. In the SRS, 324 college students were found to be employed last summer. What is the sample proportion p of college students at OSU who worked last summer? What is the approximate distribution of the proportion p of college students at OSU who worked last summer? Using the 68 - 95 - 99.7, if you had drawn an SRS of 400 college students in the US, between what two percents would you expect p to fall about 95% of the time?

Explanation / Answer

p^ = x/n = 324/400 = 0.81 [ANSWER]

*****************

Here,
n =    400
p =    0.88

u = mean = p =    0.88

s = standard deviation = sqrt(p(1-p)/n) =    0.016248077

Hence, p^ ~ N(0.88, 0.16248077) [ANSWER]

**********************

We then add/subtract two standard deviations from the mean, so it is between

0.88 - 2*0.016248 and 0.88 + 2*0.016248

0.847504 and 0.912496 or in percent,

84.75% and 91.25% [ANSWER]

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Suppose (as is roughly true) that 88% of college students were employed last sum

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