Question 1132222: The city of Providence plans to conduct a study on population expenditures related to the use of the city�s public transit such as bus and subway services. Data will be obtained by the mean of a random sample of the families living in Providence city. To determine the sample size, it was hypothesized that on average, families spend annually 2000 QAR with a standard deviation of 1200 QAR.
1. Calculate the sample size allowing Providence city to have an estimation of average family�s expenditures (sample mean) on public transit that falls within an interval of 200 QAR above or below average population expenditures with a 95% level of confidence? Repeat the same analysis with a 90% level of confidence instead of 95%.
2. Assuming the precision of the estimation was relaxed from 200 QAR to 400 QAR, what is the required sample size at a 95% level of confidence?
3. Considering the result obtained in (1) and (2), what are the impacts of levels of confidence and precision on sample size?
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! The formula for sample size(n) is
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n = ((critical value(CV) * standard deviation)/margin of error(ME))^2
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CV is the z-value associated with the given confidence level
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Note we assume the mean family income is Normally distributed
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CV for 95% confidence level is the z-score associated with critical probability(p*) of 0.975 which is 1.96
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CV for 90% confidence level is z-score associated with p* of 0.95 which is 1.645
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1.a 95% confidence level, ME is 200
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n = (1.96 * 1200 / 200)^2 = 138.2976 is approximately 138
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1.b 90% confidence level, ME is 200
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n = (1.645 * 1200 /200)^2 = 97.4169 is approximately 97
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2. 95% confidence level, ME is 400
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n = (1.96 * 1200 / 400)^2 = 34.5744 is approximately 35
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3. A greater degree of confidence with the same margin of error requires more samples while relaxing the precision of the margin of error requires fewer samples.
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