Question 757032: A recent survey of 800 randomly selected households in the United States found that approximately 50.5% of the households have Internet access.
a. Check whether the three conditions for computing a confidence interval are met in this case.
b. Regardless of your answer to part a, compute the 95% confidence interval for the proportion of households in the United States that have Internet access. What is the margin of error?
c. What is it that you are 95% sure is in the confidence interval?
d. Interpret this confidence interval.
e. Explain the meaning of 95% confidence.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A recent survey of 800 randomly selected households in the United States found that approximately 50.5% of the households have Internet access.
a. Check whether the three conditions for computing a confidence interval are met in this case.
b. Regardless of your answer to part a, compute the 95% confidence interval for the proportion of households in the United States that have Internet access. What is the margin of error?
p-hat = 0.505
ME = 1.96*sqrt[0.505*0.495/800] = 0.0346
95% CI: 0.505-0.0346 < p < 0.505+0.0346
95% CI: 0.4704 < p < 0.5396
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c. What is it that you are 95% sure is in the confidence interval?
Ans: The population proportion of households that have Internet access.
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d. Interpret this confidence interval.
We have 95% confidence that the population proportion
is between 47.04% and 53.96%
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e. Explain the meaning of 95% confidence.
Texts use various statements to describe the meaning of such
an interval. Please consult your text.
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Cheers,
Stan H.
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