PLEASE ANSWER BOTH QUESTIONS. Use Graph to help answer them. NO LINKS.

The number of calories K is needed each day by a moderately active women who weight w pounds, is h inches tall, and is a year ol

drek231 [11]

K = 917 + 6(w + h - a)
k = 1925
w = 119
a = 25

1925 = 917 + 6(119 + h - 25) =
1925 = 917 + 6(h + 94)
1925 = 917 + 6h + 564
1925 = 1481 + 6h
1925 - 1481 = 6h
444 = 6h
444/6 = h
74 = h........she is 74 inches tall

4 0

3 years ago

It has been reported that the probability that an individual will develop schizophrenia over their lifetime is 0.004. In a rando

Alinara [238K]

Answer:

Null hypothesis: $p=0.004$

Alternative hypothesis: $p \neq 0.004$

$z=\frac{0.00567 -0.004}{\sqrt{\frac{0.004(1-0.004)}{3000}}}=1.449$

$p_v =2*P(z>1.149)=0.2506$

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion if individuals that developed schizophrenia its not significantly different from 0.004.

Step-by-step explanation:

1) Data given and notation

n=3000 represent the random sample taken

X=17 represent the individuals developed schizophrenia in the sample

$\hat p=\frac{17}{3000}=0.00567$ estimated proportion of individuals developed schizophrenia in the sample

$p_o=0.004$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that true proportion of people who will develop schizophrenia is different from 0.004:

Null hypothesis: $p=0.004$

Alternative hypothesis: $p \neq 0.004$

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

3) Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.00567 -0.004}{\sqrt{\frac{0.004(1-0.004)}{3000}}}=1.449$

4) P value

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a bilateral test the p value would be:

$p_v =2*P(z>1.149)=0.2506$

5) Decision

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion if individuals that developed schizophrenia its not significantly different from 0.004.

4 0

4 years ago

Find x in the diagram below. T 1 35.4° 20.2° 252

Colt1911 [192]

Answer:

that is the answer in the photo above

3 0

3 years ago

Evaluate the expression <br> 15/5-1+50÷5=

ahrayia [7]

Answer:12

Step-by-step explanation:

7 0

3 years ago

Carlos paid an electrician x dollar per hour for a 7 hour job plus $125 for parts. The total charge was $520

cupoosta [38]

Answer:

7x + 125=520

Hope this helps! :D

6 0

4 years ago