525 subjects owned a pet 1991 did not. Among pet owners 298 women;989 of non pet

Question

525 subjects owned a pet 1991 did not. Among pet owners 298 women;989 of non pet owner were women. p^1: 0.568, p^2: 0.497
I need help with second part
"Give a 95% confidence interval for the difference in the two proportions. Dom not use rounded values. Round final answer to three decimal places.

***And If You Can Explain Formula Or Which Function To Use On Calc*****

the Heath ABc5eady, 525 sublects owned a pet and 1991 subjects dd not. Among the pet owners, there were 2ge womena 989 of the non-pet owners were women, Rand the proportion of pet owners who Mere women. Do the same for the non-pet owners (Be sure to let Population 1 correspond to the group with the higher proportion so that the dnference will be positive. Round your answers to three decimal Give a 95% conndence interval for the deference in the two proportions (Do not use rounded values. Round your final answers to three decimal places

Explanation / Answer

Sigma(diff) = sqrt(p1cap*(1-p1cap)/n1 + p2cap*(1-p2cap)/n2)

Sigma(diff) = sqrt(0.5676*(1-0.5676)/525 + 0.4967*(1-0.4967)/1991) = 0.0244

p1cap - p2cap = 0.07088

For 95%, z-value = 1.96

lower bound = 0.07088 - 1.96*0.0244 = 0.0231
upper bound = 0.07088 + 1.96*0.0244 = 0.1187

Related Questions

Navigate

• Browse (All)
• Browse 5
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.