Question

A random sample of size

nnequals=100100

yielded

ModifyingAbove p with caretpequals=0.760.76.

a. Is the sample size large enough to use the large sample approximation to construct a confidence interval for​ p? Explain.

b. Construct a

9090​%

confidence interval for

pp.

c. Interpret the

9090​%

confidence interval.

d. Explain what is meant by the phrase

​"9090​%

confidence​ interval."

Answer 1

a) Here, we have, n = 100, p = 0.76

Thus, np = (100 * 0.76) = 76 > 10, n(1-p) = (100 * 0.24)

= 24 > 10.

Hence, we can use the large sample approximation to

construct a confidence interval for p. (Ans).

b) 90% confidence interval for p is given by:

[, ], where, = 0.76, n = 100

= [0.76 - 0.07, 0.76 + 0.07] = [0.69, 0.83]. (Ans).

c) The interpretation of the 90% C.I. is that the probability

that the population proportion p will lie between 0.69 and

0.83 is 0.9. (Ans).

d) The phrase "90% confidence interval" implies that we are

90% confident that the interval will contain the population

parameter. (Ans).