A random sample of size 15 taken from a normally distributed population revealed a sample mean of 75 and a sample variance of 25. The upper limit of a 95% confidence interval for the population mean would equal?

The upper limit of a 95% confidence interval for the population mean would equal 83.805.

Step-by-step explanation:

The standard deviation is the square root of the variance. Since the variance is 25, the sample's standard deviation is 5.

We have the sample standard deviation, not the population, so we use the t-distribution to solve this question.

T interval:

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 15 - 1 = 14

Now, we have to find a value of T, which is found looking at the t table, with 14 degrees of freedom(y-axis) and a confidence level of 0.95(). So we have T = 1.761

The margin of error is:

M = T*s = 1.761*5 = 8.805.

The upper end of the interval is the sample mean added to M. So it is 75 + 8.805 = 83.805.

The upper limit of a 95% confidence interval for the population mean would equal 83.805.

step-by-step explanation:

y = | -2(-2) |     →     y = | 4 |     →     y = 4       ⇒   (-2, 4)

y = | -2(-1) |     →     y = | 2 |     →       y =   2       ⇒   (-1, 2)    

y = | -2(0) |     →     y = | 0 |     →       y = 0         ⇒   (0, 0)

y = | -2(1) |   c →     y = | -2 |     →     y =   2         ⇒   (1, 2)    

y = | -2(2) |     →     y = | - 4 |     →     y =   4         ⇒   (2, 4)