Confidence interval with finite populations As a rule of thumb when the sample s
ID: 3130511 • Letter: C
Question
Confidence interval with finite populations
As a rule of thumb when the sample size is more than 5% of the population size and the sampling process does not allow replacement, which is referred as a finite population situation, we should adjust the standard error of interest for constructing confidence intervals. In the following scenarios, as the samples are relatively large relative to the populations, i.e. n 0.05N, you will need to use the finite population correction factor to make necessary adjustments.
(a) In a survey of graduates' starting salaries at a business school, a simple random sample of 40 graduates is selected from the 2015 class of 200 students. The class's average starting salary () is unknown. However, based on past studies, the standard deviation of starting salaries () has been quite stable at $15,000. The sample's average starting salary is $57,000 per year. Construct a 92% confidence interval for the class's average starting salary (2 decimals). Make a verbal statement about the 92% confidence interval. Based on the 92% CI, make a statement on a hypothesized average salary of $72,000. Based on the 92% CI, make a statement on a hypothesized average salary of $58,000. Based on the 92% CI, make a statement on a hypothesized average salary of $45,000.
(b) A normally distributed population of 100 has mean and standard deviation which are both unknown. A simple random sample of 23 is drawn from the population without replacement. The sample average is 73.24 while the sample standard deviation is 18.68. Construct the 95% confidence interval for the population mean (2 decimals). Make a verbal statement about the 95% CI.
(c) A binomial population of 500 has unknown proportion . A simple random sample of 300 is drawn without replacement. The number of successes in the sample is 134. Construct the 98% confidence interval for the population proportion (5 decimals). Make a verbal statement about the 98% CI.
Explanation / Answer
(a) In a survey of graduates' starting salaries at a business school, a simple random sample of 40 graduates is selected from the 2015 class of 200 students. The class's average starting salary () is unknown. However, based on past studies, the standard deviation of starting salaries () has been quite stable at $15,000. The sample's average starting salary is $57,000 per year. Construct a 92% confidence interval for the class's average starting salary (2 decimals).
Here, N = 200, n = 40,
fpc = sqrt((N-n)/(N-1)) = sqrt((200-40)/(200-1)) = 0.896671679
Hence, the corrected standard deviation is
S = s*fpc = 15000*0.896671679 = 13450.07519
Note that
Margin of Error E = z(alpha/2) * S / sqrt(n)
Lower Bound = X - z(alpha/2) * S / sqrt(n)
Upper Bound = X + z(alpha/2) * S / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.04
X = sample mean = 57000
z(alpha/2) = critical z for the confidence interval = 1.750686071
S = corrected sample standard deviation = 13450.07519
n = sample size = 40
Thus,
Margin of Error E = 3723.085355
Lower bound = 53276.91464
Upper bound = 60723.08536
Thus, the confidence interval is
( 53276.91464 , 60723.08536 ) [ANSWER CONFIDENCE INTERVAL]
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Make a verbal statement about the 92% confidence interval.
Hence, we are 92% confident that the true mean average starting salary for the graduates of the business school is between $53276.91464 and $60723.08536. [CONCLUSION]
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Based on the 92% CI, make a statement on a hypothesized average salary of $72,000.
As $72000 is not inside the confidence interval, we are 92% confident that the true man average starting salary is not $72000.
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Based on the 92% CI, make a statement on a hypothesized average salary of $58,000.
As $58000 is inside the confidence interval,there is no significant evidence that that true mean is not $58000.
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Based on the 92% CI, make a statement on a hypothesized average salary of $45,000.
As $45000 is not inside the confidence interval, we are 92% confident that the true man average starting salary is not $45000.
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