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3 years ago

Plz help me thank youuu

Mathematics

1 answer:

kobusy [5.1K]3 years ago

5 0

You have to show the question more I can’t see the graph

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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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