Question

A 2011 NBC News survey found that 80% of a sample of 4500 American teens said they owned an MP3 player such as an iPad. Assume that the sample was an SRS. The large-sample 90% confidence interval for the proportion of all American teens who owned an MP3 player is Options: a) 0.785 to 0.815 b) 0.792 to 0.808 c) 0.798 to 0.802 d) 0.805 to 0.795

Answer 1

The 90% confidence interval for the true proportion of American teens who own an MP3 player, from a sample where 80% out of 4500 teens said they owned an MP3 player, is 0.792 to 0.808. Option b is correct.

Step-by-step explanation:

The question requires calculating the confidence interval for the true proportion of American teens who own an MP3 player, given a sample proportion and sample size.

To find the correct answer, you need to use the formula for a confidence interval for a proportion:

CI = ± z * √[p(1-p)/n]

Given that 80% of the 4500 American teens surveyed said they owned an MP3 player,

we have:

where CI is the confidence interval,

z is the z-score corresponding to the confidence level (which is 1.645 for 90% confidence),

p is the sample proportion (0.80), and

n is the sample size (4500).

Plugging these values into the formula, we get:

CI=0.80±1.645×√[0.80(1-0.80)/4500]

Calculating this gives us a confidence interval of approximately 0.792 to 0.808.

Upon computing, we get the interval as 0.792 to 0.808, which corresponds to option (b).