Find the 90% two-sided confidence interval for the mean score.

[ianrice]

The distribution of grades is know to follow a normal distribution, but the mean and variance are unknown. A random sample of 12 students produced the following scores:

59, 84, 68, 93, 49, 77, 82, 75, 81, 58, 70, 80

Find the 90% two-sided confidence interval for the mean score.

 

[Stephen Tashi]

The distribution of grades is know to follow a normal distribution

Does this grading system allow negative grades? If not, the the grades can't actually be normally distributed.

Of course, perhaps this is a homework problem where such technicalities are ignored. If so, you should post it somewhere in the "Homework & Coursework Questions" section of the forum and show your attempt at solution.

 

[statdad]

"Does this grading system allow negative grades? If not, the the grades can't actually be normally distributed."

Nothing is ever really normally distributed: if the standard deviation is small enough then to a very good approximation the normal distribution can be quite useful.

 

[DrDu]

You might consider a Student's t-distribution

 

[blue_raver22]

Based on the given information, we can use the t-distribution to find the 90% two-sided confidence interval for the mean score. First, we calculate the sample mean and standard deviation using the given scores:

Sample mean, x̄ = (59+84+68+93+49+77+82+75+81+58+70+80)/12 = 73.5
Sample standard deviation, s = √[(1/(12-1)) * ((59-73.5)^2 + (84-73.5)^2 + (68-73.5)^2 + (93-73.5)^2 + (49-73.5)^2 + (77-73.5)^2 + (82-73.5)^2 + (75-73.5)^2 + (81-73.5)^2 + (58-73.5)^2 + (70-73.5)^2 + (80-73.5)^2)] = 13.9

Next, we need to determine the critical value for a 90% confidence interval with 12-1 = 11 degrees of freedom. Using a t-table, we find the critical value to be 1.796.

Finally, we can calculate the confidence interval using the formula:

x̄ ± tα/2 * (s/√n)

Where x̄ is the sample mean, s is the sample standard deviation, n is the sample size, and tα/2 is the critical value.

Plugging in the values, we get:

73.5 ± 1.796 * (13.9/√12) = (67.4, 79.6)

Therefore, we can be 90% confident that the true mean score for all students falls within the interval of 67.4 to 79.6. This means that if we were to take multiple random samples of 12 students and calculate the confidence interval each time, 90% of those intervals would contain the true mean score.