Questions: Unit 04 Exam Question 3 of 7 (15 points) Question Attempt: 1 of 1 Testing math skills: In order to test the effectiveness of a program to improve mathematical skills, a simple random sample of 48 fifth graders was chosen to participate in the program. The students were given an exam at the beginning of the program and again at the end. The sample mean increase in the exam score was 13 points, with a sample standard deviation of 4.9 points. Part 1 of 2 (a) Construct a 99% confidence interval for the mean increase in score. Use tables or technology. Round your answers to the nearest tenth.

Solution Steps

To construct a 99% confidence interval for the mean increase in score, we will use the formula for the confidence interval of the mean with a known sample standard deviation. The formula is given by:

\[ \text{CI} = \bar{x} \pm z \left(\frac{s}{\sqrt{n}}\right) \]

where:

• \(\bar{x}\) is the sample mean,
• \(z\) is the z-score corresponding to the desired confidence level,
• \(s\) is the sample standard deviation,
• \(n\) is the sample size.

For a 99% confidence interval, we will use the z-score that corresponds to 99% confidence level. We will then calculate the margin of error and use it to find the confidence interval.

We are given the following data:

• Sample mean (\(\bar{x}\)) = 13
• Sample standard deviation (\(s\)) = 4.9
• Sample size (\(n\)) = 48
• Confidence level = 99%

For a 99% confidence interval, the z-score (\(z\)) is approximately 2.5758.

The margin of error is calculated using the formula: \[ \text{Margin of Error} = z \left(\frac{s}{\sqrt{n}}\right) \] Substituting the given values: \[ \text{Margin of Error} = 2.5758 \left(\frac{4.9}{\sqrt{48}}\right) \approx 1.8218 \]

The confidence interval is given by: \[ \text{CI} = \bar{x} \pm \text{Margin of Error} \] Substituting the values: \[ \text{CI} = 13 \pm 1.8218 \] This results in the interval: \[ (11.1782, 14.8218) \]

Final Answer