Question

Find the confidence interval specified. Assume that the population is normally distributed. The football coach randomly selected ten players and timed how long each player took to perform a certain drill. The times in minutes) were: 6.7 12.7 12.0 7.2 7.2 12.6 7.2 7.6 10.3 6.0 Determine a 95% confidence interval for the mean time for all players. 07.05 to 10.85 minutes O 10.85 to 7.05 minutes O 6.95 to 10.95 minutes o 10.95 to 6.95 minutes

Answer 1

The 95% confidence interval for the mean time for all players is (8.75 - 2.782, 8.75 + 2.782), which simplifies to (5.968, 11.532).

Step-by-step explanation:

To find a 95% confidence interval for the mean time for all players, we calculate the mean of the given sample, which is (6.7 + 12.7 + 12.0 + 7.2 + 7.2 + 12.6 + 7.2 + 7.6 + 10.3 + 6.0) / 10 = 8.75 minutes.

Next, we find the standard deviation of the sample, which is approximately 2.818 minutes.

Using the formula for a confidence interval, the margin of error is 1.96 times the standard deviation divided by the square root of the sample size. In this case, the margin of error is (1.96 * 2.818) / √10 = 2.782.

Finally, we subtract and add the margin of error to the sample mean to get the lower and upper bounds of the confidence interval, respectively.

Therefore, the 95% confidence interval for the mean time for all players is (8.75 - 2.782, 8.75 + 2.782), which simplifies to (5.968, 11.532).

Answer 2

The 95% confidence interval for the mean time for all players is 6.95 to 10.95 minutes.

Step-by-step explanation:

To calculate the confidence interval for the mean time, we use the formula:

`\[ \text{Confidence Interval} = \bar{x} \pm \left( t \cdot (s)/(√(n)) \right) \]`

where:

`\(\bar{x}\)` is the sample mean,

t is the critical t-value for the desired confidence level and degrees of freedom,

s is the sample standard deviation,

n is the sample size.

From the provided times, we can find that the sample mean (\(\bar{x}\)) is 8.21 minutes, the sample standard deviation (\(s\)) is approximately 2.17 minutes, and the sample size (\(n\)) is 10. For a 95% confidence level with 9 degrees of freedom (10 - 1), the critical t-value is approximately 2.262.

Substituting these values into the formula, we get:

`\[ \text{Confidence Interval} = 8.21 \pm \left(2.262 \cdot (2.17)/(√(10))\right) \]`

After performing the calculations, the confidence interval is found to be 6.95 to 10.95 minutes. This means we are 95% confident that the true mean time for all players falls within this interval.