The winning team's score in 11 high school basketball games were recorded. If the sample mean is 9.5 points and sample standard deviation is 0.25 points, what is the 98% confidence interval for the true population mean?
Boy, that is a really low mean for scoring in a basketball game! Are you sure you don't have a typo? Or is this the difference between the scores for the two teams?
98% = mean ± 2.33 SEm
SEm = SD/√n
To calculate the confidence interval for the true population mean, we can use the formula:
Confidence Interval = Sample Mean ± (Critical Value * Standard Error)
First, let's calculate the standard error, which represents the variability of your sample mean:
Standard Error = Sample Standard Deviation / Square Root of Sample Size
The sample mean is 9.5 points, and the sample standard deviation is 0.25 points. However, we don't know the sample size from the information given, so we'll assume it's a large enough value for the Central Limit Theorem to hold. For large sample sizes, we can use the standard deviation of the sample mean instead of the population standard deviation.
Next, we need to find the critical value associated with a 98% confidence level. This critical value corresponds to the z-score from the standard normal distribution.
Since we want a 98% confidence interval, there's 1% of the distribution in each tail. Therefore, we need to find the Z-score that leaves 1% in the upper tail. This can be found using a z-table or a statistical software.
In this case, the Z-score associated with a 98% confidence level is approximately 2.326.
Now, we have all the necessary values to calculate the confidence interval:
Confidence Interval = 9.5 ± (2.326 * Standard Error)
Keep in mind that the value of the standard error depends on the sample size if you have it.
Calculate the standard error using the formula mentioned earlier and substitute it into the confidence interval equation to find the range.