An opinion poll asks an SRS of 1500 adults, "Do you happen to jog?" Suppose that the population proportion who jog (a parameter) is p = 0.15. To estimate p, we use the proportion p-hat in the sample who answer "Yes." Justify the use of a normal approximation and find the following probabilities. a) P(p > 0.16)
b) P(0.14

0.14007 ; 0.71986

Explanation:

Given that :

Justification :

Mean = np ≥ 10

(1 - np) ≥ 10

n = sample size ; p = 0.15

q 1 - 0.15 = 0.85

np = 1500 * 0.15 = 225 > 10

nq = 1500 * 0.85 = 1275 > 10

Hence, the Justification

P(p" > 0.16)

P(Z > (p" - p) /sqrt(p(q) /n))

P(Z > (0.16- 0.15) /sqrt(0.15(0.85) /1500))

P(Z > 0.01 / 0.0092195)

P(Z > 1.08) = 0.14007

P(0.14 < p < 0.16)

P(Z (0.14- 0.15) /sqrt(0.15(0.85) /1500)) < Z < (0.16 - 0.15) /sqrt(0.15(0.85) /1500))

P(Z <-0.01 / 0.0092195) - 0.01 / 0.0092195)

P(Z < 1.08) - P(Z < - 1.08)

Using the Z probability m calculator :

0.85993 - 0.14007

= 0.71986

= 0.72208