A sample of 50 elements from a population with a standard deviation of 20 is selected. The sample mean is 150. The 90% confidence interval for is: a.165.0 to 185.0. b.146.4 to 1…

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asked Jan 6, 2022 202k views

1 Answer

Ilana · Answer 1 · 2022-01-12T04:34:29+0000

Answer:

c.145.3 to 154.7.

Explanation:

We have to find our
$\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = (1 - 0.9)/(2) = 0.05$

Now, we have to find z in the Z-table as such z has a p-value of
$1 - \alpha$ .

That is z with a p-value of
1 - 0.05 = 0.95 , so Z = 1.645.

Now, find the margin of error M as such

$M = z(\sigma)/(√(n))$

In which
$\sigma$ is the standard deviation of the population and n is the size of the sample.

M = 1.645(20)/(√(50)) = 4.7

The lower end of the interval is the sample mean subtracted by M. So it is 150 - 4.7 = 145.3.

The upper end of the interval is the sample mean added to M. So it is 150 + 4.7 = 154.7.

Thus the correct answer is given by option c.