Question

Marcelina has over 500 500500 songs on her mobile phone, and she wants to estimate the average length of the songs (in minutes). She takes an SRS of 28 2828 songs on her phone and calculates a sample mean of 3.4 3.43, point, 4 minutes and a standard deviation of 0.72 0.720, point, 72 minutes. The song lengths in the sample were roughly symmetric with no clear outliers. Based on this sample, which of the following is a 99 % 99%99, percent confidence interval for the mean length (in minutes) of the songs on her phone?

Answer 1

The 99% confidence interval for the mean length of the songs on Marcelina's phone is approximately (3.08 minutes, 3.78 minutes).

Explanation:

To calculate the 99% confidence interval for the mean length of the songs on Marcelina's phone, we can use the formula:

Confidence Interval = Sample Mean ± (Z * Standard Error)

Where:

Sample Mean is the mean length of the sample songs (3.43 minutes).

Z is the critical value associated with the desired confidence level (99% confidence level corresponds to a Z-value of approximately 2.576).

Standard Error is the standard deviation of the sample divided by the square root of the sample size.

Given that the sample size is 28 songs and the standard deviation is 0.72 minutes, we can calculate the standard error as:

Standard Error = Standard Deviation / √(Sample Size)

Standard Error = 0.72 / √28 ≈ 0.136

Now we can substitute the values into the formula:

Confidence Interval = 3.43 ± (2.576 * 0.136)

Calculating the confidence interval:

Confidence Interval = 3.43 ± 0.350

Confidence Interval = (3.08, 3.78)

Therefore, based on this sample, the 99% confidence interval for the mean length of the songs on Marcelina's phone is approximately (3.08 minutes, 3.78 minutes).

Hope this helps!