Questions: A simple random sample of size n is drawn. The sample mean, x̄, is found to be 19.1, and the sample standard deviation, s, is found to be 4.7. Click the icon to view the table of areas under the t-distribution. Lower bound: Upper bound: 21.27 (Use ascending order. Round to two decimal places as needed.) Compare the results to those obtained in part (a). How does increasing the level of confidence affect the size of the margin of error, E? A. The margin of error decreases. B. The margin of error does not change. C. The margin of error increases. (d) If the sample size is 16, what conditions must be satisfied to compute the confidence interval? A. The sample must come from a population that is normally distributed and the sample size must be large. B. The sample data must come from a population that is normally distributed with no outliers. C. The sample size must be large and the sample should not have any outliers.

Solution Steps

To calculate the margin of error \( E \) at a 95% confidence level, we use the formula:

\[ E = \frac{Z \times \sigma}{\sqrt{n}} \]

where \( Z = 1.96 \) (Z-Score for 95% confidence), \( \sigma = 4.7 \) (sample standard deviation), and \( n = 16 \) (sample size).

Substituting the values, we have:

\[ E = \frac{1.96 \times 4.7}{\sqrt{16}} = 2.303 \]

The lower bound of the confidence interval is calculated as:

\[ \text{Lower Bound} = \bar{x} - E \]

where \( \bar{x} = 19.1 \) (sample mean). Thus,

\[ \text{Lower Bound} = 19.1 - 2.303 = 16.797 \]

For the 99% confidence level, we again use the margin of error formula:

\[ E = \frac{Z \times \sigma}{\sqrt{n}} \]

where \( Z = 2.5758 \) (Z-Score for 99% confidence). Substituting the values, we have:

\[ E = \frac{2.5758 \times 4.7}{\sqrt{16}} = 3.0266 \]

The lower bound for the 99% confidence interval is calculated as:

\[ \text{Lower Bound} = \bar{x} - E \]

Thus,

\[ \text{Lower Bound} = 19.1 - 3.0266 = 16.0734 \]

Comparing the margins of error at the two confidence levels, we find:

• Margin of Error at 95% confidence level: \( 2.303 \)
• Margin of Error at 99% confidence level: \( 3.0266 \)

Since \( 3.0266 > 2.303 \), we conclude that:

\[ \text{C. The margin of error increases.} \]

Final Answer