Question

in a survey conducted by the gallup organization, 1100 adult americans were asked how many hours they worked in the previous week. based on the results, a 95% confidence interval for the mean number of hours worked had a lower bound of 42.7 and an upper bound of 44.5. provide two recommendations for decreasing the margin of error of the interval
select the two recommendations that would decrease the margin of error of the interval
□ a. increase the sample size
□ b. increase the confidence level
□ c. use lower degrees of freedom
□ d. decrease the confidence level
□ e. decrease the standard deviation of hours worked
□ f. decrease the sample size

Explanation:

The margin of error formula for a confidence interval (assuming a normal distribution or large - sample \(z\) - interval) is \(E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\) (for \(z\) - intervals) or \(E = t_{\alpha/2}\frac{s}{\sqrt{n}}\) (for \(t\) - intervals).

• For option A:
• In the formula \(E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\) (or \(E = t_{\alpha/2}\frac{s}{\sqrt{n}}\)), as \(n\) (the sample size) increases, the value of \(\frac{1}{\sqrt{n}}\) decreases. For example, if \(n_1 = 100\), \(\frac{1}{\sqrt{n_1}}=\frac{1}{10}\); if \(n_2=400\), \(\frac{1}{\sqrt{n_2}}=\frac{1}{20}\). So, increasing \(n\) decreases the margin of error.
• For option B:
• Increasing the confidence level increases the value of \(z_{\alpha/2}\) (or \(t_{\alpha/2}\)). For a 95% confidence interval, \(z_{\alpha/2}=1.96\), and for a 99% confidence interval, \(z_{\alpha/2} = 2.576\). Since \(E\) is directly proportional to \(z_{\alpha/2}\) (or \(t_{\alpha/2}\)), increasing the confidence level increases the margin of error.
• For option C:
• Degrees of freedom (\(df=n - 1\)) is relevant for \(t\) - distributions. Using lower degrees of freedom (when \(n\) is small) actually increases the value of \(t_{\alpha/2}\) (from the \(t\) - distribution table). For example, for a 95% confidence interval, \(df = 10\), \(t_{\alpha/2}=2.228\); \(df=20\), \(t_{\alpha/2}=2.086\). So, using lower degrees of freedom (in the context of \(t\) - intervals) increases the margin of error.
• For option D:
• Decreasing the confidence level decreases the value of \(z_{\alpha/2}\) (or \(t_{\alpha/2}\)). For example, if we go from a 95% confidence interval (\(z_{\alpha/2}=1.96\)) to a 90% confidence interval (\(z_{\alpha/2}=1.645\)). Since \(E\) is directly proportional to \(z_{\alpha/2}\) (or \(t_{\alpha/2}\)), decreasing the confidence level decreases the margin of error.
• For option E:
• While \(\sigma\) (or \(s\)) is a measure of spread, in the context of a survey about hours worked, the standard deviation of hours worked is a characteristic of the population (or sample) of workers. It is not something that can be easily “decreased” by the researcher in the context of constructing the confidence interval (it is a descriptive measure of the data, not a parameter controlled by the researcher for the purpose of interval construction).
• For option F:
• Decreasing the sample size \(n\) increases the value of \(\frac{1}{\sqrt{n}}\). For example, if \(n_1 = 400\), \(\frac{1}{\sqrt{n_1}}=\frac{1}{20}\); if \(n_2 = 100\), \(\frac{1}{\sqrt{n_2}}=\frac{1}{10}\). So, decreasing \(n\) increases the margin of error.

So, the two recommendations that decrease the margin of error are increasing the sample size and decreasing the confidence level.

Answer:

A. Increase the sample size, D. Decrease the confidence level