The high temperatures in degrees Fahrenheit) of a random sample of 13 middle-schoolers is given below. Assume high temperatures are normally distributed. Based on this data, find the 95% confidence interval of the mean high temperature of middle-schoolers. <> 99.6 99.0 97.0 98.1 98.5 98.2 96.4 97.5 98.4 98.7 99.2 97.3 96.8 I'm 95% confident the true mean temperature is between and degrees. 6) A student was asked to find a 99% confidence interval for widget width using data from a random sample of size n= 19. Which of the following is a correct interpretation of the interval 11.4> n = test takers

To find the 95% confidence interval of the mean high temperature of middle-schoolers, we need to calculate the sample mean, sample standard deviation, and the critical value from the t-distribution table.

Given data: 99.6, 99.0, 97.0, 98.1, 98.5, 98.2, 96.4, 97.5, 98.4, 98.7, 99.2, 97.3, 96.8

Step 1: Calculate the sample mean (x) and sample standard deviation (s).

x = (99.6 + 99.0 + 97.0 + 98.1 + 98.5 + 98.2 + 96.4 + 97.5 + 98.4 + 98.7 + 99.2 + 97.3 + 96.8) / 13 = 98.4

s = √((99.6 - 98.4)² + (99.0 - 98.4)² + ... + (96.8 - 98.4)²) / (13 - 1) = 0.972

Step 2: Find the critical value (t) from the t-distribution table for a 95% confidence level and 12 degrees of freedom (13 - 1).

t = 2.179

Step 3: Calculate the confidence interval using the formula CI = x ± (t * (s/√n)).

CI = 98.4 ± (2.179 * (0.972/√13)) = 98.4 ± 0.897

Step 4: Round the confidence interval to 1 decimal place.

CI = (97.5, 99.3)

Therefore, we can say with 95% confidence that the true mean high temperature of middle-schoolers is between 97.5 and 99.3 degrees Fahrenheit.

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