Question

asked Jan 10, 2021 228k views

1 Answer

Yanilda · Answer 1 · 2021-01-16T21:19:07+0000

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.

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