Question

Among college-age students (18-24 years old), 9.2% have hypertension. During a blood-donor program conducted during finals week, a blood-pressure reading is taken first, revealing that out of 2000 donors, 203 have hypertension.
Part One:
The sample proportion of successes (p-hat) is (to four decimal places)   

Assuming that hypertension in finals week is the same as at other times, what is the probability of getting a sample result as large as ours (p-value)?

Answer 1

The sample proportion of successes (p-hat) is

203 p= = 0.1015

Assuming that hypertension in finals week is the same as at other times,

Let o denotes the sample proportion for a random sample of size n =2000. Here, Let ø - Normal(0.09, .09.10.09)) or P ~ Normal(0.09. 0.00652) 2000 Now, 0.1015 – 0.09 0.0065 DP - 0.09 P(> 0.1015) = P(p > 0.1015)=1- P(p < 0.1015) =1- P”. 0.0065 + Plô > 0.1015) = 1 - P(Z < 0.1015 – 0.09 0.0065 → PCÔ > 0.1015) = 1 - P(Z < 1.469947) + PP > 0.1015) =1- 0.929212 = 0.070788 = 0.0708

The probability of getting a sample result as large as ours = 0.0708