Question

HELP PLEASE.....EASY QUESTION!!!!

Answer 1

Answer:

90% of confidence interval for the true population of club members who use compost

(0.44424 , 0.45576)

Step-by-step explanation:

Step(i):-

Given that the size of the sample 'n' =200

Given that the sample proportion

            p = 45% = 0.45

Level of significance = 90% or 10%

Critical value Z₀.₁₀ = 1.645

Step(ii):-

90% of confidence interval for the true population of club members who use compost

$(p^{-} - Z_{0.10} \sqrt{\frac{p(1-p^{-} )}{n} } , p^{-} + Z_{0.10} \sqrt{\frac{p(1-p^{-} )}{n} })$

$((0.45 - 1.645\sqrt{\frac{0.45(1-0.45)}{200} } , 0.45 + 1.645\sqrt{\frac{0.45(1-0.45)}{200} })$

(0.45 - 0.00576 , 0.45 +0.00576)

(0.44424 , 0.45576)

Final answer:-

90% of confidence interval for the true population of club members who use compost

(0.44424 , 0.45576)