Final answer:
To find the 95% confidence interval for the population mean of the major test scores, calculate the sample mean and the standard deviation, then apply the t-distribution and degrees of freedom to get the t-score and use the formula CI = μ ± (t * s / √n) to obtain the interval, rounding to the nearest whole number.
Step-by-step explanation:
To calculate the 95% confidence interval for the population mean of the major test scores, we first find the sample mean (μ) and the sample standard deviation (s) using the provided test scores. With the data given: 486, 499, 483, 491, 478, 415, 473, 398, 450, 467, 421, 492, 436, 496, the sample mean and standard deviation are calculated as follows:
- Mean: (μ) = (Sum of all scores) / (Number of scores)
- Standard Deviation: (s) = sqrt[(Σ(x - μ)2) / (n-1)]
After computing these values, we then use the t-distribution and degrees of freedom (df = n - 1) to determine the t-score relevant for a 95% confidence interval. Lastly, we apply the formula for the 95% confidence interval:
CI = μ ± (t * s / √n)
This calculation will give us the lower and upper bounds of the confidence interval, which we round to the nearest whole number as requested.
As an example, if the sample mean is 67 and the standard deviation is 5 for a sample size of 14, the degrees of freedom would be 13. Using the t-score for 13 degrees of freedom at a 95% confidence level, let's say it's approximately 2.160. The confidence interval is calculated as:
CI = 67 ± (2.160 * 5 / √14)
Therefore, after computation, the confidence interval might be approximated and rounded as:
Lower bound: 64 (rounded)
Upper bound: 70 (rounded)
The actual numbers will vary based on the calculated mean and standard deviation for the provided scores.