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A population is normally distributed with a standard deviation of 2.8. A random sample is obtained from this population and the observations are: 8, 9, 10, 13, 14, 16, 17, 20, 21. Construct the 95% confidence interval for the mean of this population. Construct the 99% confidence interval for the mean of this population.

User Ivantedja
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Answer:

The 95% confidence interval for the mean of this population is between 12.39 and 16.05.

The 99% confidence interval for the mean of this population is between 11.82 and 16.62.

Explanation:

The first step is finding the mean of the sample:

There are 9 observations. So


\mu_(x) = (8+9+10+13+14+16+17+20+21)/(9) = 14.22

95% confidence interval:

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


\alpha = (1-0.95)/(2) = 0.025

Now, we have to find z in the Ztable as such z has a pvalue of
1-\alpha.

So it is z with a pvalue of
1-0.025 = 0.975, so
z = 1.96

Now, find 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 = 1.96*(2.8)/(√(9)) = 1.83

The lower end of the interval is the sample mean subtracted by M. So it is 14.22 - 1.83 = 12.39

The upper end of the interval is the sample mean added to M. So it is 14.22 + 1.83 = 16.05

The 95% confidence interval for the mean of this population is between 12.39 and 16.05.

99% confidence interval:

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 Ztable as such z has a pvalue of
1-\alpha.

So it is z with a pvalue of
1-0.005 = 0.995, so
z = 2.575

Now, find 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 = 2.575*(2.8)/(√(9)) = 2.40

The lower end of the interval is the sample mean subtracted by M. So it is 14.22 - 2.40 = 11.82

The upper end of the interval is the sample mean added to M. So it is 14.22 + 2.40 = 16.62

The 99% confidence interval for the mean of this population is between 11.82 and 16.62.

User Yanilda
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