a survey of 2323 adults in a certain large country aged 18 and older conducted b

Question

a survey of 2323 adults in a certain large country aged 18 and older conducted by a reputable polling organization found 425 have donated blood in the past two years. complete parts (a) - (c) below.

(a) obtain a point estimate for the population proportion of adults in the country aged 18 and older who have donated blood in the past two years. p= [ ] ( round to three decimal places as needed)

(b) verify that the requirements for constructing a confidence interval about p are satisfied.

-The sample [can be assumed to be, is stated to be, is stated to not be, cannot be assumed to be] a simple random sample, the value of [np(1-p), n, p, np, p(1-p)] is [ ], which is [greater than or equal to, less than] 10, and the [sample proportion, population proportion, sample size, population size] [cannot be assumed to be, is stated to not be, can be assumed to be, is stated to be] less than or equal to 5% of the [sample size, population size, sample proportion, population proportion]

construct and interpret a 90% confidence interval for the population proportion of adults in the country who have donated blood in the past two years. select the correct coice below and fill in any anser boxes within your choice.( round to 3 decimal places as needed)

A. we are [ ]% confident the proportion of adults in the country aged 18 and older who have donated blood in the past two years is between [ ] & [ ].

B. there is a [ ]% chance the proportion of adults in the country aged 18 and older who have donated blood in the past two years is between [ ] and [ ].

Explanation / Answer

let x=donated blood in the past two years=425

n=adults in a certain large country aged 18=2323

(a) point estimate for the population proportion of adults in the country aged 18 and older who have donated blood in the past two years

p=x/n=425/2323=0.183

(b)np>10

sample proportion cannot be assumed to be less than or equal to 5% of the sample size

(c) SE(p)=sqrt(p(1-p)/n)=sqrt(0.183*(1-0.183)/2323)=0.008

(1-alpha)*100% confidence interval p=p± z(alpha/2)*SE(p)

90% confidence interval for p=0.0183±z(0.1/2)*0.008=0.0183±1.645*0.008=0.0183±0.013=(0.170,0.196)

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