UpStudy Free Solution:

To find the margin of error and construct a 98% confidence interval using the standard normal distribution (since the population standard deviation is known), we will go through the following steps:

Identify the necessary parameters:

Sample mean (\(\bar { x} \)) = 33.5 minutes

Population standard deviation (\(\sigma \)) = 8.7 minutes

Sample size (\(n\)) = 8

Confidence level = 98%

Determine the critical value (\(z^ { *} \)) for a 98% confidence interval:

For a 98% confidence level, the critical value \(z^ { *} \) from the standard normal distribution is approximately 2.33. This value can be found using standard normal distribution tables or a calculator.

Calculate the standard error of the mean (SEM):

\[\text { SEM} = \frac { \sigma } { \sqrt { n} } = \frac { 8.7} { \sqrt { 8} } \approx 3.08\]

Calculate the margin of error (ME):

\[\text { ME} = z^ \times \text { SEM} = 2.33 \times 3.08 \approx 7.18\]

Construct the 98% confidence interval:

\[\text { CI} = \left ( \bar { x} - \text { ME} , \bar { x} + \text { ME} \right ) = \left ( 33.5 - 7.18, 33.5 + 7.18 \right ) = \left ( 26.32, 40.68 \right ) \]

Margin of Error:

The margin of error (ME) is \(7.18\) minutes (rounded to two decimal places).

Comparison of Results:

Using the t-distribution (with sample standard deviation), the 98% confidence interval was (25.7, 41.3).

Using the standard normal distribution (with known population standard deviation), the 98% confidence interval is (26.32, 40.68).

The intervals are similar, but the interval using the standard normal distribution is slightly narrower. This is expected because the population standard deviation provides a more precise measure of variability compared to the sample standard deviation, leading to a smaller standard error and thus a narrower confidence interval.

Supplemental Knowledge

In statistics, a confidence interval provides a range of values that is likely to contain the population parameter with a certain level of confidence. When the population standard deviation is known, the standard normal distribution (z-distribution) is used instead of the t-distribution.

Given:

Sample mean (\(\bar { x} \)) = 33.5 minutes

Sample size (\(n\)) = 8

Population standard deviation (\(\sigma \)) = 8.7 minutes

Confidence level = 98%

The margin of error (ME) for a confidence interval using the z-distribution is calculated as:

\[ME = z^ \cdot \frac { \sigma } { \sqrt { n} } \]

where \(z^ { *} \) is the critical value from the z-distribution for the desired confidence level.

For a 98% confidence level, \(z^ { *} \) can be found using standard normal distribution tables or z-score calculators. The critical value \(z^ { *} \) for a 98% confidence interval is approximately 2.33.

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