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39 votes
39 votes
(c) Construct a 99% confidence interval for u if the sample
size, n, is 35.

User Adam Shamsudeen
by
3.0k points

1 Answer

14 votes
14 votes

Answer:

The confidence interval is
(\overline{x} - 1.99(\sigma)/(√(35)), \overline{x} + 1.99(\sigma)/(√(35))), in which
\overline{x} is the sample mean and
\sigma is the standard deviation for the population.

Explanation:

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


\alpha = (1 - 0.99)/(2) = 0.005

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 pvalue of
1 - 0.005 = 0.995, so Z = 2.575.

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.

In this question:


M = 1.99(\sigma)/(√(35))

The lower end of the interval is the sample mean subtracted by M, while upper end of the interval is the sample mean added to M. Thus, the confidence interval is
(\overline{x} - 1.99(\sigma)/(√(35)), \overline{x} + 1.99(\sigma)/(√(35))), in which
\overline{x} is the sample mean and
\sigma is the standard deviation for the population.

User Ran Davidovitz
by
3.1k points