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3 years ago

6

A rule for nth term of -2 , -5/4 ,-1/2,1/4

1 answer:

miv72 [106K]3 years ago

6 0

Answer:

Step-by-step explanation:

This is an arithmetic sequence

a1 = -2

Difference = 2nd term - first term

$=\frac{-5}{4}-[-2]\\\\=\frac{-5}{4}+2\\\\=\frac{-5}{4}+\frac{2*4}{1*4}\\\\=\frac{-5+8}{4}=\frac{3}{4}$

$a_{n}=a+(n-1)d\\\\=-2+(n-1)*(\frac{3}{4})\\\\=-2+\frac{3}{4}n-\frac{3}{4}\\\\=\frac{-2*4}{1*4}-\frac{3}{4}+\frac{3}{4}n\\\\=\frac{-8-3}{4}+\frac{3}{4}n\\\\=\frac{-11}{4}+\frac{3}{4}n$

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A simple random sample of items resulted in a sample mean of . The population standard deviation is . a. Compute the confidence

Varvara68 [4.7K]

Answer:

(a): The 95% confidence interval is (46.4, 53.6)

(b): The 95% confidence interval is (47.9, 52.1)

(c): Larger sample gives a smaller margin of error.

Step-by-step explanation:

Given

$n = 30$ -- sample size

$\bar x = 50$ -- sample mean

$\sigma = 10$ --- sample standard deviation

Solving (a): The confidence interval of the population mean

Calculate the standard error

$\sigma_x = \frac{\sigma}{\sqrt n}$

$\sigma_x = \frac{10}{\sqrt {30}}$

$\sigma_x = \frac{10}{5.478}$

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$z = 1.960$

Calculate margin of error (E)

$E = z * \sigma_x$

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$E = 3.577$

The confidence bound is:

$Lower = \bar x - E$

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$Lower = 46.4$ --- approximated

$Upper = \bar x + E$

$Upper = 50 + 3.577$

$Upper = 53.577$

$Upper = 53.6$ --- approximated

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Solving (b): The confidence interval of the population mean if mean = 90

First, calculate the standard error of the mean

$\sigma_x = \frac{\sigma}{\sqrt n}$

$\sigma_x = \frac{10}{\sqrt {90}}$

$\sigma_x = \frac{10}{9.49}$

$\sigma_x = 1.054$

The 95% confidence interval for the z value is:

$z = 1.960$

Calculate margin of error (E)

$E = z * \sigma_x$

$E = 1.960 * 1.054$

$E = 2.06584$

The confidence bound is:

$Lower = \bar x - E$

$Lower = 50 - 2.06584$

$Lower = 47.93416$

$Lower = 47.9$ --- approximated

$Upper = \bar x + E$

$Upper = 50 + 2.06584$

$Upper = 52.06584$

$Upper = 52.1$ --- approximated

<em>So, the 95% confidence interval is (47.9, 52.1)</em>

Solving (c): Effect of larger sample size on margin of error

In (a), we have:

$n = 30$ $E = 3.577$

In (b), we have:

$n = 90$ $E = 2.06584$

<em>Notice that the margin of error decreases when the sample size increases.</em>

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