Question

In a random sample of 678 males, 58 of them have hypertension.

(a) Construct a 99% confidence interval for the proportion of male who has hypertension

(b) If the desired interval was 90% instead what would be the impact of the length of the interval?

(c) Does the interval from part (a) above support the statement from the AMA that fewer than 10% of males have hypertension?

Answer 1

a)

n=678

x=58

99 % confidence interval for proportion

Formula

Zc= 2.58 ( using z-table )

99 % confidence interval for proportion is

b)

Z-value corresponding to 0.99 C.I is 2.58 and z-value of 0..90 C.I is 1.96.z-value of confidence level 0.90 is small as compere to 0.99 confidence level.so length of interval is small as compere to confidence level 0.99

c)

claim : The AMA is fewer than 10% of males have hypertension .

The C.I is the value 10% lies tn the C.I so we failed to Reject null hypothesis .so AMA is fewer than 10% of males have hypertension