Someone please help me with this question

A sociologist wants to determine if the life expectancy of people in africa is less than the life expectancy of people in asia.

Colt1911 [192]

Answer:

The sample statistics are attached.

First, we need to determine the hypothesis.

The null hypothesis should be: $H_{o} : u_{africa} \geq u_{asia}$ .

The alternative hypothesis should be: $H_{a}: u_{africa}$

The next step is about calculating the t statistics based on the samples. This t test will give the probability value or p-value, which determines if there's enough to reject the null hypothesis. The formula we need to use to find the t-value is: $t=\frac{x_{1}-x_{2} }{\sqrt{\frac{s_{1} ^{2} }{n_{1} } +\frac{s_{2} ^{2} }{n_{2} } } }$

Replacing all the values into the formula, we have:

$t=\frac{52.4-58.1}{\sqrt{\frac{(7.4)^{2} }{45}+\frac{(8.8)^{2} }{35} } } \\t=\frac{-5.7}{\sqrt{1.2+2.2} } =\frac{-5.7}{1.8} =-3.2$

Therefore, the t-value is -3.2.

Now, we using a level of significance of 0.01, and a degree of freedom (df) of 34, we use the t-table to find the p-value for this results. (df = N -1; in this case, we take the smaller sample, which is 35, giving us 34 of df).

Therefore, according to the table attached, the p-value is less than 0.01, which is less than our level of significance. This result means that the null hypothesis is rejected. In other words, there's enough evidence to say that the life expectancy of people in Africa is less than the life of expectancy of people in Asia.

3 0

3 years ago

Công ty A vay ngân hàng số tiền 25.000.000.000 đồng, lãi suất 12%/năm, trả nợ dần trong 5 năm. Hợp đồng qui định công ty A phải

svetoff [14.1K]

Bfbdbcjgjdnsldk backside

7 0

2 years ago

2.6.5 A plant physiologist grew birch seedlings in the green-house and measured the ATP content of their roots. (See Example 1.1

Ilya [14]

Answer:

$(a)\ \bar x = 1.19$

$(b)\ \sigma_x = 0.18$

Step-by-step explanation:

Given

$n = 4$

$x: 1.45\ 1.19\ 1.05\ 1.07$

Solving (a): The mean

This is calculated as:

$\bar x = \frac{\sum x}{n}$

So, we have:

$\bar x = \frac{1.45 + 1.19 + 1.05 + 1.07}{4}$

$\bar x = \frac{4.76}{4}$

$\bar x = 1.19$

Solving (b): The standard deviation

This is calculated as:

$\sigma_x = \sqrt{\frac{\sum(x - \bar x)^2}{n - 1}}$

So, we have:

$\sigma_x = \sqrt{\frac{(1.45 -1.19)^2 + (1.19 -1.19)^2 + (1.05 -1.19)^2 + (1.07 -1.19)^2}{4 - 1}}$

$\sigma_x = \sqrt{\frac{(0.1016}{3}}$

$\sigma_x = \sqrt{0.033867}$

$\sigma_x = 0.18$

4 0

2 years ago

70 pointsss for the correct answer

nignag [31]

Answer:

f(x) = 12x³ - 19x² + 17x - 17

Step-by-step explanation:

We can the numerator by multiplying the denominator by the quotient, and then adding the remainder.

We're given a denominator of 4x - 5

a quotient of 3x² - x + 3

and a remainder of -2

so the original numerator is:

(4x - 5)(3x² - x + 3) - 2

= 12x³ - 4x² + 12x - 15x² + 5x - 15 - 2

= 12x³ - 19x² + 17x - 17

Let's test that by dividing it by 4x -5 with long division:

$\ \ \ \ \ \ \ \ \ \ 3x^2 - x + 3\\4x - 5\overline{)12x^3 - 19x^2 + 17x - 17}\\.\ \ \ \ \ \ \ \ \ \underline{2x^3 - 15x^2}\\.\ \ \ \ \ \ \ \ \ \ \ \ \ \ -4x^2 + 17x\\.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \underline{-4x^2 + 5x} \\.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 12x - 17\\.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \underline{12x - 15}\\.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2R$

That matches, so we know our answer's right, and:

f(x) = 12x³ - 19x² + 17x - 17

3 0

3 years ago

Please tell me the answers to 4 and 5 no links

Cerrena [4.2K]

Answer:

4) choice a 30%

5) choice b 62.5%

Step-by-step explanation:

4) 3/10 = 0.30 or 30%

5) 5/8 = 0.625 or 62.5%

5 0

3 years ago