Suppose that an accounting firm does a study to determine the time needed to complete one person's tax forms. It randomly surveys 50 people. The sample mean is 24.6 hours. There is a known standard deviation of 8 hours. The population distribution is assumed to be normal. Round answers to 3 decimal places.
a. Find the following:
i. $\bar{x}=$
ii. $\sigma=$
iii. $n=$
b. Construct a $90 \%$ confidence interval for the population mean time to complete tax forms.
i. State the confidence interval.
$\mathrm{Cl}:$
ii. Calculate the error bound.
EBM:
c. Suppose that the firm decided that it needed to be at least $99 \%$ confident of the population mean length of time to within one hour. How would the number of people the firm surveys change?
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Solving this gives us the new sample size as approximately \(\boxed{425}\) people.
Step 1 :Given that the sample mean (\(\bar{x}\)) is 24.6, the standard deviation (\(\sigma\)) is 8, and the sample size (\(n\)) is 50.
Step 2 :To construct a 90% confidence interval for the population mean time to complete tax forms, we use the formula \(\bar{x} \pm z*\frac{\sigma}{\sqrt{n}}\), where \(z\) is the z-score corresponding to the desired level of confidence. For a 90% confidence level, the z-score is approximately 1.645.
Step 3 :Substituting the given values into the formula, we get \(24.6 \pm 1.645*\frac{8}{\sqrt{50}}\).
Step 4 :Solving this gives us the 90% confidence interval for the population mean time to complete tax forms as \(\boxed{(22.739, 26.461)}\).
Step 5 :The error bound is the difference between the upper limit of the confidence interval and the sample mean, which is \(\boxed{1.861}\).
Step 6 :If the firm wants to be 99% confident of the population mean length of time to within one hour, we use the formula for the sample size needed for a given margin of error, which is \(n = \left(\frac{z*\sigma}{E}\right)^2\), where \(E\) is the desired margin of error. For a 99% confidence level, the z-score is approximately 2.576, and the desired margin of error is 1.
Step 7 :Substituting these values into the formula, we get \(n = \left(\frac{2.576*8}{1}\right)^2\).
Step 8 :Solving this gives us the new sample size as approximately \(\boxed{425}\) people.