Question

Grades on a standardized test are known to have a mean of 990 for students in the United States. The test is administered to 473 randomly selected students in Florida; in this sample, the mean is 1002.87 and the standard deviation (s) is 106.92. The 95% confidence interval for the average test score for Florida students is ____ , ______ . (Round your responses to two decimal places.)

Answer 1

To calculate the 95% confidence interval for Florida students' average test scores, we use the sample mean, standard deviation, and size, combined with the z-distribution critical value of 1.96. The interval is found by subtracting and adding the margin of error to the sample mean, resulting in the interval of 993.24 to 1012.50.

Step-by-step explanation:

To calculate the 95% confidence interval for the average test score for Florida students, we use the sample mean, the sample standard deviation, and the size of the sample (n). The confidence interval is calculated as

Confidence Interval = Sample mean ± (Critical value * (Sample standard deviation / sqrt(n)))

Since the sample size is large (n > 30), we can use the z-distribution to find the critical value. For a 95% confidence level, the critical value (z*) is approximately 1.96. First, we calculate the standard error (SE), which is the standard deviation divided by the square root of the sample size:

SE = s / sqrt(n)

SE = 106.92 / sqrt(473)

SE = 106.92 / 21.753 = 4.914

Next, we calculate the margin of error (ME) using the critical value and the standard error:

ME = z* * SE

ME = 1.96 * 4.914 = 9.63144

Finally, the confidence interval is calculated by adding and subtracting the margin of error from the sample mean:

Lower bound = 1002.87 - 9.63144 = 993.24

Upper bound = 1002.87 + 9.63144 = 1012.50

Therefore, the 95% confidence interval for the average test score for Florida students is 993.24, 1012.50.