Question

To estimate the mean of a population with unknown distribution shape and unknown standard deviation, we take a random sample of size 64. The sample mean is 22.3 and the sample standard deviation is 8.8.
If we wish to compute a 92% confidence interval for the population mean, what will be the t multiplier?

Answer 1

Answer:

$t_{critical} = \pm 1.7793$                        

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 64

Sample mean = 22.3

Sample standard deviation = 8.8

We want to estimate 92% confidence interval.

92% Confidence interval:  

$\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}$  

Putting the values, we get,

Degree of freedom = n - 1 = 63

$t_{critical}\text{ at degree of freedom 63 and}~\alpha_{0.01} = \pm 1.7793$  

$22.3 \pm 1.7793(\dfrac{8.8}{\sqrt{64}} ) = 22.3 \pm 1.95723 = (20.34,24.26)$