G A random sample of size 16 taken from a normally distributed population revealed a sample mean of 50 and a sample variance of 36. The upper limit of a 95% confidence interval …

Question

asked Jun 6, 2021 130k views

1 Answer

Guylhem · Answer 1 · 2021-06-12T20:54:04+0000

Answer:

The upper limit is

k = 52.94

Explanation:

From the question we told that

The sample size is
n = 16

The sample mean is
$\= x = 50$

The sample variance is
$\sigma ^2 = 36$

For a 95% confidence interval the confidence level is 95%

Given that the confidence level is 95% then the level of significance is mathematically evaluated as

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha = 0.05$

Next we obtain the critical value of
$(\alpha )/(2)$ from the normal distribution table(reference- math dot armstrong dot edu), the value is

$Z_{( \alpha )/(2) } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{(\alpha )/(2) } * (\sigma)/(√(n) )$

Here
$\sigma$ is the standard deviation which is mathematically evaluated as

$\sigma = √(\sigma^2)$

substituting values

$\sigma = √(36)$

=>
$\sigma = 6$

So

E = 1.96 * (6)/(√(16) )

E = 2.94

The 95% confidence interval is mathematically represented as

$\= x - E < \mu < \= x + E$

substituting values

$50 -2.94 < \mu <50 +2.94$

$47.06 < \mu <52.94$

The upper limit is

k = 52.94