The population of a certain town increased by 2% from 35,000. What was the population after the increase?

Mathematics

1 answer:

olga nikolaevna [1]3 years ago

3 0

35,000 X .02 = 700, so 2% of 35,000 is 700. With that, the population would be 35,700.

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People tend to evaluate the quality of their lives relative to others around them (Frieswijk et al., 2004). In one study, resear

Alborosie

Answer:

Step-by-step explanation:

Hello!

Research.

n=9 frail elderly were interview and compared to a fictitious person who was worse off then the interviewee, a life-satisfaction score was determined for each person.

18, 23, 24, 22, 19, 27, 23, 26, 25

Assuming that the population average score is μ= 20, the researchers think that the elderly in the sample are more or less satisfied than others in the general population.

a. You have the information of one sample, assuming this sample has a normal distribution and each elderly interviewed is independent, then the t-test of choice is a one-sample t-test.

b. and c. If you say that the elderly are "more or less" satisfied than the others, this means that they are either as satisfied as to the general population or not satisfied as to the general population. Symbolically:

H₀: μ = 20

H₁: μ ≠ 20

This is a two-tailed test, meaning, you will have two critical regions.

α: 0.05

Left critical value: $t_{n-1;/\alpha 2} = t_{8; 0.025}= -2.306$

Right critical value: $t_{n-1;1-\alpha /2} = t_{8;0.975} = 2.306$

$t_{H_0}= \frac{X[bar]-Mu}{\frac{S}{\sqrt{n} } } ~t_{n-1}$

X[bar]= 23

S= 3

$t_{H_0}= \frac{23-20}{\frac{3}{\sqrt{9} } }= 3$

Considering that the calculated t-value is greater than the right critical value, the decision is to reject the null hypothesis, so using a significance level of 5% you can conclude that the average life-satisfaction score of the elderly is different than 20.

I hope it helps!