Joseph built a model of a pyramid where

A nutrional label states that there are 36 grams of carbs in each serving. This accounts for 12% of a daily value. how many gram

Nina [5.8K]

Answer:

300 gm

Hope I helped :)

Step-by-step explanation:

3 0

3 years ago

a sample of 546 boys aged 6–11 was weighed, and it was determined that 89 of them were overweight. A sample of 508 girls aged 6–

AleksAgata [21]

Answer:

p-value: 0.6527

Step-by-step explanation:

Hello!

You have two samples to study, from each sample the weight of each child was measured and counted the total of overweight kids (x: "success") in each group:

Sample 1 (Boys aged 6-11)

n₁= 546

x₁= 89

^p₁= x₁/n₁ = 89/546 ≅0.16

Sample 2 (girls aged 6-11)

n=508

x= 74

^p= x/n = 74/508 ≅ 0.15

If the hypothesis statement is "The proportion of boys that are overweight differs from the proportion of girls that are overweight", the test hypothesis is:

H₀: ρ₁ = ρ₂

H₁: ρ₁ ≠ ρ₂

This type of hypothesis leads to a two-tailed rejection region, then the p-value will also be two-tailed. To calculate the p-value you have to first calculate the value of the statistic under the null hypothesis, in this case, is a test for the difference between two proportions:

Z= (^ρ₁ - ^ρ₂) - (ρ₁ - ρ₂) ≈ N(0;1)

√(ρ` * (1 - ρ`) * (1/n₁ + 1/n₂))

ρ`= x₁ + x₂ = 89+74 = 0.154 ≅ 0.15

n₁ + n₂ 546 + 508

Z⁰ᵇ = (0.16-0.15) - (0)

√(0.15 * (1 - 0.15) * (1/546 + 1/508))

Z⁰ᵇ = 0.45

I've mentioned before that in this test you have a two-tailed p-value. The value calculated (0.45) corresponds to the right or positive tail and the left tail is symmetrical to it concerning the distribution mean, in this case, is 0, so it is -0.45. To obtain the p-value you need to calculate the probability of both values and add them:

P(Z>0.45) + P(Z<-0.45) = (1- P(Z<0.45)) + P(Z<-0.45) = (1-0.67364) + 0.32636 = 0.65272 ≅ 0.6527

p-value: 0.6527

Since there is no signification level in the problem, I'll use the most common to reach a decision. α: 0.05

Since the p-value is greater than α, you do not reject the null Hypothesis, in other words, there is no significative difference between the proportion of overweight boys and the proportion of overweight girls.

I hope it helps!

3 0

3 years ago

Which equation can be used to find how much of a 50 g sample of nitrogen-16 is left after 7 minutes?

miv72 [106K]

Answer:

Step-by-step explanation:

50(0.5) square 50

8 0

3 years ago

What is the formula for compound interest?

Valentin [98]

This is the formula.

8 0

3 years ago

Blood pressure: High blood pressure has been identified as a risk factor for heart attacks and strokes. The proportion of U.S. a

Y_Kistochka [10]

Answer:

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

For this case we know this:

$n=37 , p=0.2$

We can find the standard error like this:

$SE = \sqrt{\frac{\hat p (1-\hat p)}{n}}= \sqrt{\frac{0.2*0.8}{37}}= 0.0658$

So then our random variable can be described as:

$p \sim N(0.2, 0.0658)$

Let's suppose that the question on this case is find the probability that the population proportion would be higher than 0.4:

$P(p>0.4)$

We can use the z score given by:

$z = \frac{p -\mu_p}{SE_p}$

And using this we got this:

$P(p>0.4) = 1-P(z< \frac{0.4-0.2}{0.0658}) = 1-P(z$

And we can find this probability using the Ti 84 on this way:

2nd> VARS> DISTR > normalcdf

And the code that we need to use for this case would be:

1-normalcdf(-1000, 3.04; 0;1)

Or equivalently we can use:

1-normalcdf(-1000, 0.4; 0.2;0.0658)

Step-by-step explanation:

We need to check if we can use the normal approximation:

$np = 37 *0.2 = 7.4 \geq 5$

$n(1-p) = 37*0.8 = 29.6\geq 5$

We assume independence on each event and a random sampling method so we can conclude that we can use the normal approximation and then ,the population proportion have the following distribution :

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

For this case we know this:

$n=37 , p=0.2$

We can find the standard error like this:

$SE = \sqrt{\frac{\hat p (1-\hat p)}{n}}= \sqrt{\frac{0.2*0.8}{37}}= 0.0658$

So then our random variable can be described as:

$p \sim N(0.2, 0.0658)$

Let's suppose that the question on this case is find the probability that the population proportion would be higher than 0.4:

$P(p>0.4)$

We can use the z score given by:

$z = \frac{p -\mu_p}{SE_p}$

And using this we got this:

$P(p>0.4) = 1-P(z< \frac{0.4-0.2}{0.0658}) = 1-P(z$

And we can find this probability using the Ti 84 on this way:

2nd> VARS> DISTR > normalcdf

And the code that we need to use for this case would be:

1-normalcdf(-1000, 3.04; 0;1)

Or equivalently we can use:

1-normalcdf(-1000, 0.4; 0.2;0.0658)

7 0

3 years ago