The Residential Energy Consumption Survey found in 2001 that 47% of American household had internet access. A market survey organization repeated this study in a certain town with 25,000 households, using a simple random sample of 500 households: 239 of the same households had internet access. a) The percentage of households in the town with internet access is estimated at ; this is likely to be off by ___ or so. b) If possible, find a 95%-confidence interval for the percentage of all 25,000 households with internet access. If this is not possible, explain why or why not.

47.8%

2.23%

43.34% ; 52.26%

Step-by-step explanation:

A.) percentage household with internet access:

Numner with internet access / samples size

239 / 500

= 0.478 ; 0.478 * 100% = 47.8%

Standard deviation :

Sqrt(Proportion with internet access) * proportion without internet access)

Sqrt[(0.478) * (1 - 0.478)]

= 0.4995

Standard Error of estimate = 0.4995 * √sample size

= 0.4995 * √500

= 11.16

To percentage ;

(11.16/ 500) * 100

= 2.23%

Confidence interval :

Sample percentage ± 2(percentage standard error)

= 47.8 + 2(2.23) = 52.26%

= 47.8 - 2(2.23) = 43.34%

i got 10.28 but that rounds up to 10.3