Questions: An admissions director wants to estimate the mean age of all students enrolled at a college. The estimate must be within 1.7 years of the population mean. Assume the population of ages is normally distributed. (a) Determine the minimum sample size required to construct a 90% confidence interval for the population mean. Assume the population standard deviation is 1.8 years. (b) The sample mean is 21 years of age. Using the minimum sample size with a 90% level of confidence, does it seem likely that the population mean could be within 10% of the sample mean? within 11% of the sample mean? Explain. Click here to view page 1 of the Standard Normal Table. Click here to view page 2 of the Standard Normal Table. (a) The minimum sample size required to construct a 90% confidence interval is students. (Round up to the nearest whole number.)

Solution Steps

To estimate the mean age of all students enrolled at a college within a margin of error of \(1.7\) years at a \(90\%\) confidence level, we first calculate the minimum sample size required. The formula for the minimum sample size \(n\) is given by:

\[ n = \left( \frac{z \cdot \sigma}{E} \right)^2 \]

where:

• \(z\) is the z-score corresponding to the confidence level (\(z \approx 1.645\) for \(90\%\)),
• \(\sigma = 1.8\) years (population standard deviation),
• \(E = 1.7\) years (margin of error).

Substituting the values:

\[ n = \left( \frac{1.645 \cdot 1.8}{1.7} \right)^2 \approx 4 \]

Thus, the minimum sample size required is:

\[ \boxed{n = 4} \]

Using the minimum sample size of \(n = 4\) and the sample mean \(\bar{x} = 21\) years, we calculate the \(90\%\) confidence interval for the population mean. The confidence interval is given by:

\[ \bar{x} \pm z \cdot \frac{\sigma}{\sqrt{n}} \]

Substituting the values:

\[ 21 \pm 1.645 \cdot \frac{1.8}{\sqrt{4}} = 21 \pm 1.645 \cdot 0.9 \]

Calculating the margin of error:

\[ 1.645 \cdot 0.9 \approx 1.4805 \]

Thus, the confidence interval is:

\[ (21 - 1.4805, 21 + 1.4805) = (19.5195, 22.4805) \]

Rounding to two decimal places, we have:

\[ \text{Confidence Interval: } (19.52, 22.48) \]

Next, we check if the population mean could be within \(10\%\) and \(11\%\) of the sample mean.

Calculating \(10\%\) and \(11\%\) of the sample mean:

\[ 10\% \text{ of } 21 = 2.1 \quad \text{and} \quad 11\% \text{ of } 21 = 2.31 \]

Thus, the ranges are:

• For \(10\%\): \( (21 - 2.1, 21 + 2.1) = (18.9, 23.1) \)
• For \(11\%\): \( (21 - 2.31, 21 + 2.31) = (18.69, 23.31) \)

Now, we compare these ranges with the confidence interval \( (19.52, 22.48) \):

• For \(10\%\): The confidence interval \( (19.52, 22.48) \) does not fully cover \( (18.9, 23.1) \), hence it is not likely that the population mean could be within \(10\%\) of the sample mean.
• For \(11\%\): The confidence interval \( (19.52, 22.48) \) does not fully cover \( (18.69, 23.31) \), hence it is not likely that the population mean could be within \(11\%\) of the sample mean.

Final Answer