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2 years ago

HELP

Mathematics

1 answer:

Ulleksa [173]2 years ago

3 0

Answer:

Ok, to find MAD, you have to add up all of the decimals in the set, and then divide by how much decimals there are in the set.

Adding them all up...

The final result of the adding is 56.7

Now, we have to divide by how much decimals.

We now have. 56.7 / 6

Do the division.

56.7 / 6 ===> 9.45

No need to round, the final answer is 9.4

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What is the value of t? 45.19

Mariana [72]

Alternating Angles?
If so, its the same, so still 45.19
If its not Alternating Angles, then I’m lost.
Or is T referring to the line?

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3 years ago

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Families USA, a monthly magazine that discusses issues related to health and health costs, surveyed 20 of its subscribers. It fo

Harman [31]

Answer:

(a) The 90 percent confidence interval for the population mean yearly premium is ($10,974.53, $10983.47).

(b) The sample size required is 107.

Step-by-step explanation:

(a)

The (1 - α)% confidence interval for population mean is:

$CI=\bar x\pm t_{\alpha/2, (n-1)}\times \frac{s}{\sqrt{n}}$

Given:

$\bar x=\$10,979\\s=\$1000\\n=20$

Compute the critical value of t for 90% confidence level as follows:

$t_{\alpha/2, (n-1)}=t_{0.10/2, (20-1)}=t_{0.05, 19}=1.729$

*Use a t-table.

Compute the 90% confidence interval for population mean as follows:

$CI=\bar x\pm t_{\alpha/2, (n-1)}\times \frac{s}{\sqrt{n}}$

$=10979\pm 1.729\times \frac{1000}{\sqrt{20}}\\=10979\pm4.47\\ =(10974.53, 10983.47)$

Thus, the 90 percent confidence interval for the population mean yearly premium is ($10,974.53, $10983.47).

(b)

The margin of error is provided as:

MOE = $250

The confidence level is, 99%.

The critical value of z for 99% confidence level is:

$z_{\alpha/2}=z_{0.01/2}=z_{0.005}=2.58$

Compute the sample size as follows:

$MOE= z_{\alpha/2}\times \frac{s}{\sqrt{n}}$

$n=[\frac{z_{\alpha/2}\times s}{MOE} ]^{2}$

$=[\frac{2.58\times 1000}{250}]^{2}$

$=106.5024\\\approx107$

Thus, the sample size required is 107.

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3 years ago