Among collegeage students 1824 years old 92 have hypertensio

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 200 donors, 29 have hypertension.

(1) The sample proportion of successes (p-hat) is exactly 0.145

2.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)??????

This is the problem i am stuck on

.3. which comes from a test-statistic of z=2.59

Solution

here, no need of taking 9.2% in account. it just a statistical general data.

Now, p = 29/200 =0.145

q= 1-p = 0.8555

P(x=29)= 200 C 29 * p^29 * q^171 = 200C29 * 0.145^29 * 0.855^171 = 0.0798

Now taking general case,

p=0.092

q=0.908

n=200

P(x=29) = 200C29 * p^29 * q^171 = 200C29 * 0.092^29 * 0.908^171 = 0.00435

This problem is not related with normal distribution. So need to worry about z value.. Just forget that value is given. This problem is related with binominal distribution