Question

A newspaper reported on the results of an opinion poll in which adults were asked what one thing they are most likely to do when they are home sick with a cold or the flu. In the​ survey, ​% said that they are most likely to sleep and ​% said that they would watch television. Although the sample size was not​ reported, typically opinion polls include approximately​ 1,000 randomly selected respondents.
a. Assuming a sample size of​ 1,000 for this​ poll, construct a 95 ​% confidence interval for the true percentage of all adults who would choose to sleep when they are at home sick.
b. If the true percentage of adults who would choose to sleep when they are at home sick is 74​%, would you be​ surprised?

Answer 1

From the given question. The missing values include:

In the survey 62% attest that they are most likely to sleep; &

20% attest that they would watch television.

Answer:

Step-by-step explanation:

Now, the percentage of adults that choose to sleep is:

p = 62%

p = 0.62

q = 1 - 0.62 = 0.38

(a)

Assume Sample size n =  1000

The 95% C.I for the true percentage can be computed as:

$= \Bigg (p - Z_{\alpha/2} \times \sqrt{\dfrac{p \times q}{n}} \ , \ p + Z_{\alpha/2} \times \sqrt{\dfrac{p \times q}{n}} \Bigg )$

$= (0.62 - 1.96 \times \sqrt{\dfrac{0.62 \times 0.38}{1000}} \ , \ 0.62 + 1.96 \times \sqrt{\dfrac{0.62 \times 0.38}{1000}})$

$= (0.62 -0.030085\ , \ 0.62 +0.030085)$

= (0.5899, 0.6501)

= (58.99%, 65.01%)

b.) Yes, I will be suprised.