Question

A random sample of 1400 Internet users was selected from the records of a large Internet provider and asked whether they would use the Internet or the library to obtain information about health issues. Of these, 872 said they would use the Internet

1. The standard error ˆp SE of the proportion pˆ that would use the Internet rather than the library is:_______

a. 0.013.
b. 0.25.
c. 0.485.
d. 0.623.

2. If the Internet provider wanted an estimate of the proportion p that would use the Internet rather than the library, with a margin of error of at most 0.02 in a 99% confidence interval, how large a sample size would be required? (Assume that we don’t have any prior information about p).

a. 33
b. 3909
c. 2401
d. 4161

Answer 1

1
`\sigma_(\= x ) = 0.0130`

2
`n = 3908.5`

Explanation:

From the question we are told that

The sample size is
`n_p = 1400`

The number of those that said the would use internet is
`k = 872`

The margin of error is
`E = 0.02`

Generally the sample proportion is mathematically evaluated as

`\r p = (k)/(n_p)`

substituting values

`\r p = ( 872)/(1400)`

substituting values

`\r p = 0.623`

Generally the standard error of
`\r p` is mathematically evaluated as

`\sigma_(\= x ) = \sqrt{(\r p (1- \r p))/(n) }`

substituting values

`\sigma_(\= x ) = \sqrt{(0.623 (1- 0.623))/(1400) }`

`\sigma_(\= x ) = 0.0130`

For a 95% confidence interval the confidence level is 95%

Given that the confidence interval is 95% the we can evaluated the level of confidence as

`\alpha = 100 - 99`

`\alpha = 1\%`

`\alpha = 0.01`

Next we obtain the critical value of
`(\alpha )/(2)` from normal distribution table (reference math dot armstrong dot edu) , the value is

`Z_{(\alpha )/(2) } = 2.58`

Give that the population size is very large the sample size is mathematically represented as

`n = [ \frac{Z_{(\alpha )/(2) ^2 * \r p ( 1 - \r p )}}{E^2} ]`

substituting values

`n = [ (2.58 ^2 * 0.623 ( 1 -0.623 ))/(0.02^2) ]`

`n = 3908.5`