Question

Suppose that the mean time that visitors stay at a museum is 94.2 minutes with a standard deviation of 15.5 minutes. The standard error of the mean,ox, is 3.1. A random sample of 25 of the times chosen. What interval captures 68% of the means for random samples of 25 scores?

Answer 1

Answer:

$94.2 -0.994*3.1 = 91.1186$

$94.2 +0.994*3.1 = 97.2814$

And the 68% confidence interval is given by (91.1186, 97.2814)

Step-by-step explanation:

For this case we know that mean time that visitors stay at a museum is given by:

$\bar X = 94.2$

The standard deviation is given by:

$s= 15.5$

And the standard error is given by:

$SE = \frac{s}{\sqrt{n}} =3.1$

And we want to interval captures 68% of the means for random samples of 25 scores and for this case the critical value can be founded like this using the normal standard distribution or excel:

$z_{\alpha/2}= \pm 0.994$

We can find the interval like this:

$\bar X \pm ME$

And replacing we got:

$94.2 -0.994*3.1 = 91.1186$

$94.2 +0.994*3.1 = 97.2814$

And the 68% confidence interval is given by (91.1186, 97.2814)