Please help!

never [62]

Three loaves since that’s what’s available to the family

5 0

3 years ago

In 2008, the average household debt service ratio for homeowners was 13.2. The household debt service ratio is the ratio of debt

ankoles [38]

Answer:

$t=\frac{13.88-13.2}{\frac{3.14}{\sqrt{44}}}=1.436$

$df=n-1=44-1=43$

$p_v =P(t_{(43)}>1.436)=0.079$

We see that the p value i higher than the significance level so then we FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly higher than 13.2 *the value of 2008 ).

Step-by-step explanation:

Information given

$\bar X=13.88$ represent the sample mean

$s=3.14$ represent the sample standard deviation

$n=44$ sample size

$\mu_o =13.2$ represent the value that we want to test

$\alpha=0.05$ represent the significance level

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test

Hypothesis to test

We want to conduct a hypothesis in order to check if the true mean has increased from 2008 , and the system of hypothesi are:

Null hypothesis: $\mu \leq 13.2$

Alternative hypothesis: $\mu > 13.2$

The statistic for this case is:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Calculating the statistic

Replacing the info given we got:

$t=\frac{13.88-13.2}{\frac{3.14}{\sqrt{44}}}=1.436$

P-value

The degrees of freedom are:

$df=n-1=44-1=43$

Since is a right tailed test the p value is:

$p_v =P(t_{(43)}>1.436)=0.079$

Decision

We see that the p value i higher than the significance level so then we FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly higher than 13.2 *the value of 2008 ).

8 0

4 years ago

A study was performed to investigate whether teens and adults had different habits when it comes to consuming meat-free meals. I

Elden [556K]

Answer:

H₀ should be rejected at CI 95% .

Then the proportion of adults with 95 % CI is bigger than the proportion of teen

Step-by-step explanation:

From adult sample:

n₂ = 2323

x₂ = 1601

p₂ = 1601 / 2323 p₂ = 0,689 or p₂ = 68,9%

From teen sample:

n₁ = 875

x₁ = 555

p₁ = 555/ 875 p₁ = 0,634 or p₁ = 63,4

Values of p₁ and p₂ suggest that the proportion of adults consuming at least one meat free meal per week is bigger than teen proportion.

To either prove or reject the above statement we have to develop a difference of proportion test according to:

Hypothesis test:

Null Hypothesis H₀ p₁ = p₂

Alternative Hypothesis Hₐ p₂ > p₁

So is a one-tail test to the right

We can establish a confidence interval of 95 % then α = 5 %

or α = 0,05

As the samples are big enough we will develop a z test

Then z(c) for α = 0,05 from z table is z(c) = 1,64

To calculate z(s)

z(s) = ( p₂ - p₁ ) / √p*q* ( 1/n₁ + 1/n₂ )

where p = ( x₁ + x₂ ) / n₁ + n₂ p = 555 + 1601 / 875 + 2323

p = 2156/3198 p = 0,674

and q = 1 - 0,674 q = 0,326

z(s) = ( 0,689 - 0,634 ) / √0,674*0,326 ( 1/875 + 1 / 2323

z(s) = 0,055/ √ 0,2197 ( 0,00114 + 0,00043)

z(s) = 0,055/ √0,2197* 0,00157

z(s) = 0,055/ √ 3,45*10⁻⁴

z(s) = 0,055 / 1,85*10⁻²

z(s) = 5,5/1,85

z(s) = 2,97

Comparing z(s) and z(c) z(s) > z(c)

Then z(s) is in the rejection region we reject H₀.

We can claim that the proportion of adult eating at least one meat-free meal is bigger than the proportion of teen

6 0

3 years ago

Which point is on the graph of f(x)=2x= 2.5%?

katovenus [111]

Answer: The parabola has its concavity downwards, so we need a function in the model:

With a negative value of 'a'

The vertex is (0,0), so we have that:

The x-coordinate of the vertex is given by the equation:

So we have a function in the model:

With a < 0

The only option with this format is B:

Step-by-step explanation:

4 0

3 years ago

Can anyone help me in pre-cal <br>tan^2 x = tan x<br>sin^2 x = sin x

vagabundo [1.1K]

They will equal each other at the angles when they are equal to 1. This is because 1^2=1. Tan is 1 at pi/4 and 5pi/4. Sin is 1 at pi/2

3 0

3 years ago