$x^{-2}=(-6)^{-2}=\dfrac{1}{(-6)^2}=\dfrac{1}{6}\qquad\text{used}\ a^{-n}=\dfrac{1}{a^n}\\=====================\\x^{-1}=(-6)^{-1}=\dfrac{1}{(-6)^1}=-\dfrac{1}{6}\\=====================\\x^0=(-6)^0=1\qquad a^0=1\ \text{for all real numbers except 0}\\=====================\\x^1=(-6)^1=-6\qquad a^1=a\ \text{for all real numbers}\\=====================\\x^2=(-6)^2=36$

5 0

3 years ago

You select a random sample of n=14 families in your neighborhood and find the following family sizes (number of people in the fa

satela [25.4K]

Answer:

a. Best estimate for the mean family size for the population of all families in a country is 16.3571428571

b. 95% confidence interval for the estimate is 16.3571428571±0.7106240840. That is between 15.6465187731 and 17.0677669411

c. The sample is unlikely to be very representative of all families in a country.

Step-by-step explanation:

a. mean family size for the population of all families in a country can be estimated as sample mean, and calculated as:

$\frac{15+ 16+ 19+ 18+ 17+ 15+ 16+ 16+ 15+ 16+ 15+ 16+ 17+ 18 }{14}$ ≈ 16.3571428571

b. 95% confidence interval for the estimate can be calculated using the equation CI=M± $\frac{t*s}{\sqrt{N} }$ where

M is the mean estimate (16.3571428571)

t is the corresponding statistic in the 95% confidence level and with 14 degrees of freedom (2.160)

s is the standard deviation of the sample (1.2309777100)

N is the sample size (14)

s can be calculated as square root of mean squared differences from the sample mean.

Thus 95% CI=16.3571428571± $\frac{2.160*1.2309777100}{\sqrt{14} }$ ≈ 16.3571428571±0.7106240840

The sample is unlikely to be very representative of all families in a country because as the sample data increases, estimate will improve.

7 0

3 years ago

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