Mathematics, 12.03.2020 01:02 shywelch123

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.

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Answer from: ayoismeisalex

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.

Step-by-step explanation:

The first step is finding the mean of the sample:

There are 9 observations. So

$\mu_{x} = \frac{8+9+10+13+14+16+17+20+21}{9} = 14.22$

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:

$\alpha = \frac{1-0.95}{2} = 0.025$

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

$M = z*\frac{\sigma}{\sqrt{n}}$

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

$M = 1.96*\frac{2.8}{\sqrt{9}} = 1.83$

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:

$\alpha = \frac{1-0.99}{2} = 0.005$

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

$M = z*\frac{\sigma}{\sqrt{n}}$

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

$M = 2.575*\frac{2.8}{\sqrt{9}} = 2.40$

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.

Answer from: Quest

answer: 414,377,008

step-by-step explanation:

just multiply it

Answer from: Quest

answer 1 : option "c"

answer 2 : option "d"

answer 3 : option "b"

answer 4 : option "d"

answer 5 : option "b"

answer 6 : option "a"

answer 7 : option "c"

answer 8 : option "d"

answer 9 : option "c"

answer 10 : option "d"

answer 11 : option "c"

answer 12 option "b"

answer 13 option "c"

answer 14 option "c"

answer 15 option "d"

answer 16 option "a"

answer 17 option "b"

answer 18 option "c"

answer 19 option "a"

answer 20 option "a"

answer 21 option "c"

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