Question

It is found that on average, 9% of the population recycles its garbage. Under this assumption, if a random sample of 120 households in Iowa City is taken, would you be surprised to find that fewer than 3% of households in the sample recycle their garbage? Use the fact that p-hat has an approximately normal distribution. Group of answer choices I would not be surprised because there are lots of lazy students in Iowa City. I would not be surprised because 3% is not that far away from 9%. I would be surprised because Iowa City is a very clean community which means lots of people recycle their garbage. I would be surprised because if the 9% average is true, the chance of a sample proportion being less than 3% is very small (only 1%). Who cares? It's only garbage!

Answer 1

Complete Question

It is found that on average, 9% of the population recycles its garbage. Under this assumption, if a random sample of 120 households in Iowa City is taken, would you be surprised to find that fewer than 3% of households in the sample recycle their garbage? Use the fact that p-hat has an approximately normal distribution.

Group of answer choices

A I would not be surprised because there are lots of lazy students in Iowa City.

B I would not be surprised because 3% is not that far away from 9%.

C I would be surprised because Iowa City is a very clean community which means lots of people recycle their garbage.

D I would be surprised because if the 9% average is true, the chance of a sample proportion being less than 3% is very small (only 1%). Who cares? It's only garbage!

Answer:

The correct option is D

Explanation:

From the question we are told that

The sample size is n = 120

The mean of the proportion is p = 0.09

Generally the standard deviation is mathematically represented as

`\sigma  =  \sqrt{(p(1-p))/(n) }`

=>
`\sigma  =  \sqrt{(0.09(1-0.09))/(120 ) }`

=>
`\sigma  = 0.02612`

Generally the chance that fewer than 3% of households in the sample recycle their garbage is mathematically represented as

`P(X < 0.03) =  P( (X - p)/(\sigma)  <  (0.03 -0.09)/(0.02612) )`

`P(X < 0.03) =  P( Z  < -2.297 )`

From the z-table

`P( Z  < -2.297 ) = 0.01`

=> [tex]P(X < 0.03) = 0.01/tex]

=> [tex]P(X < 0.03) = 1\% /tex]