The winning team's score in 13 high school basketball games was recorded. If the sample mean is 54.3 points and the sample standard deviation is 11.0 points, find the 98% confidence interval of the true mean.
37.9 70.7
To find the 98% confidence interval of the true mean, we will use the formula:
Confidence interval = sample mean ± (critical value) * (sample standard deviation / √n)
First, we need to determine the critical value for a 98% confidence interval.
Since the confidence interval is two-tailed, we need to divide the significance level (α) by 2.
The remaining probability (1 - α/2) represents the area under the standard normal distribution curve.
Using a standard normal distribution table or calculator, we find that the critical value for a 98% confidence interval is approximately 2.33.
Now, let's plug the values into the formula:
Confidence interval = 54.3 ± (2.33) * (11.0 / √13)
To calculate the confidence interval, we need to determine the standard error first. The standard error (SE) is given by the formula:
SE = sample standard deviation / √n
SE = 11.0 / √13
SE ≈ 3.05
Now, we can plug in this value into the confidence interval formula:
Confidence interval = 54.3 ± (2.33) * (3.05)
Confidence interval ≈ 54.3 ± 7.12
Lower bound of the confidence interval = 54.3 - 7.12 ≈ 47.18
Upper bound of the confidence interval = 54.3 + 7.12 ≈ 61.42
Therefore, the 98% confidence interval of the true mean is approximately 47.18 to 61.42 points.
To find the 98% confidence interval of the true mean, we can use the formula:
Confidence interval = sample mean ± (critical value * standard error)
First, let's calculate the standard error, which measures the variability of the sample mean:
Standard error = sample standard deviation / square root of the sample size
In this case, we don't know the sample size (n), so we need some additional information to proceed.
Could you provide the sample size (number of games) for which the winning team's scores were recorded?