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Answer :
Final answer:
A 90 percent confidence interval for the proportion of households that receive the printed newspaper regular basis is calculated using the sample proportion of 0.6,
with a margin of error obtained from a standard error of 0.0365 and a z-score of 1.645 for 90% confidence.
The interval is (0.5399, 0.6601), meaning there is 90% confidence that the true proportion lies between 53.99% and 66.01%.
Explanation:
To construct a 90 percent confidence interval for the proportion of households that receive the newspaper in a printed form, we can use the sample proportion and the standard error of the proportion to find the margin of error and then add and subtract this margin of error to our sample proportion.
The sample proportion (π) is calculated as the number of households receiving the newspaper in printed form divided by the total number of households surveyed. Given that 108 out of 180 households receive the printed newspaper, the sample proportion is 108/180 = 0.6.
The standard error (SE) for the proportion is calculated using the following formula: SE = √[π(1 - π) / n], where n is the sample size. Therefore, SE = √[0.6(1 - 0.6) / 180] = √(0.24 / 180) = √0.001333 = 0.0365.
To obtain the confidence interval, we need the z-score that corresponds to a 90% confidence level. Since a two-tailed standard normal distribution leaves 5% in each tail for a 90% confidence level, we look up the z-score that corresponds to leaving 5% (or 0.05) in one tail, which is z = 1.645.
The margin of error (E) is then calculated as E = z * SE = 1.645 * 0.0365 = 0.0601.
We then construct the confidence interval (CI): CI = π ± E = 0.6 ± 0.0601, resulting in a CI of (0.5399, 0.6601).
This confidence interval indicates that we can be 90% confident that the true proportion of households in Pete's area that receive a printed newspaper on a regular basis is between 53.99% and 66.01%.
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