Question

TV advertising agencies face increasing challenges in reaching audience members because viewing TV programs viadigital streaming is gaining in popularity. The Harris poll reported on November 13, 2012, that 53% of 2343 Americanadults surveyed said they have watched digitally streamed TV programming on some type of devicea. Calculate and interpret a confidence interval at the 99% confidence level for the proportion of all adult Americans whowatched streamed programming up to that point in time." 9b. What sample size would be required for the width of a 99% C | to be at most .05 irrespective of the value p .

Answer 1

Answer with explanation:

Let p denotes the population proportion of all adult Americans who watched streamed programming up to that point in time.

Given : The Harris poll reported on November 13, 2012, that 53% of 2343 American adults surveyed said they have watched digitally streamed TV programming on some type of devices.

i.e. sample size :n= 2343

`\hat{p}=0.53`

We know that ,

Critical value for 99% confidence interval =
`z^*=2.576`

a) Confidence interval for population proportion:-

`\hat{p}\pm z^*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}`

`0.53\pm (2.576)\sqrt{(0.53(1-0.53))/(2343)}\\\\\approx0.53\pm0.0266\\\\=(0.53-0.0265,\ 0.53+ 0.0265)\\\\=(0.5035,\ 0.5565)`

i.e. Confidence interval : (0.5035, 0.5565)

Interpretation: A person can be 99% confident that the true proportion of all adult Americans who watched streamed programming up to that point in time lies between 0.5035 and 0.5565 .

b) Margin of error : E= half of width of CI=
`(0.05)/(2)=0.025`

The formula to find the sample size , if the prior estimate of population proportion is known:

`n=p(1-p)((z^*)/(E))^2`

Put all the value in the above formula :

`n=0.53(1-0.53)((2.576)/(0.025))^2\\\\=0.2491(103.04)^2\\\\=2644.75488256\approx2645`

Thus , the minimum sample size = 2645