Question

A recent Gallup poll asked 1006 Americans, "During the past year, about how many books, either hardcover or paperback, did you read either all or part of the way through?" Results of the survey indicated that x = 13.4 books and s = 16.6 books. Construct a 99% confidence interval for the mean number of books Americans read either all or part of during the preceding year. Interpret that interval.

Answer 1

Answer:

The required interval is $[12.0519,14.7481]$

Step-by-step explanation:

Consider the provided information.

It is given that:

n = 1006, mean = 13.4 and s = 16.6  

We need to construct a 99% confidence interval.

That means α = 0.01 and α/2 = 0.005

$Z_{\alpha/2}=2.5758$

First find the standard error as shown:

$SE=\frac{16.6}{\sqrt{1006}} =0.5234$

Now to calculate confidence interval calculate: $Mean \pm Z_{\alpha/2} \times SE$

$=13.4 \pm 2.5758 \times 0.5234$

$=13.4 \pm 1.34817372$

$[12.0519,14.7481]$

Hence, the required interval is $[12.0519,14.7481]$