Ask question

Ask question

All categories

3 years ago

12

Colleen recorded the type of trees in the woods near her house. She counted 20 spruce trees and 50 maple trees. What is the expe

rimental probability that the next tree Colleen finds will be a spruce tree?
A) 2/7
B) 3/5
C)5/7
D)2/5

1 answer:

tankabanditka [31]3 years ago

3 0

Answer:

The answer is A $\frac{2}{7}$

Step-by-step explanation:

You might be interested in

Which of the following is a function?

Pie

Relationship 2 is a function because x isn’t being repeated.

5 0

4 years ago

Using appreciate property -2/3 × 3/5 + 5/2 - 3/5 × 1/6

o-na [289]

Answer:

The answer for your question is 2.

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Evaluate the expression <br> 5.4+ab, if a = 4 and b = 8

dsp73

Answer:

37.4

Step-by-step explanation:

a = 4 and b = 8

change the equation

5.4 + (4) (8)

use P.E.M.D.A.S. and multiply

4*8 = 32

add

5.4 + 32 = 37.4

7 0

3 years ago

Solve the quadratic equation algebraically 5-x<2(x-3)+5

vodka [1.7K]

Answer:

x>2

Step-by-step explanation:

5-x<2(x-3)+5

5-x<2x-6+5

5-x<2x-1

6-x<2x

6<3x

x>2

7 0

3 years ago

A genetic experiment involving peas yielded one sample of offspring consisting of 420 green peas and 174 yellow peas. Use a 0.01

slavikrds [6]

Answer:

a) $z=\frac{0.293 -0.23}{\sqrt{\frac{0.23(1-0.23)}{594}}}=3.649$

b) For this case we need to find a critical value that accumulates $\alpha/2$ of the area on each tail, we know that $\alpha=0.01$ , so then $\alpha/2 =0.005$ , using the normal standard table or excel we see that:

$z_{crit}= \pm 2.58$

Since the calculated value is higher than the critical value we have enough evidence to reject the null hypothesis at 1% of significance.

Step-by-step explanation:

Data given and notation

n=420+174=594 represent the random sample taken

X=174 represent the number of yellow peas

$\hat p=\frac{174}{594}=0.293$ estimated proportion of yellow peas

$p_o=0.23$ is the value that we want to test

$\alpha=0.01$ represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

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

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion of yellow peas is 0.23:

Null hypothesis: $p=0.23$

Alternative hypothesis: $p \neq 0.23$

When we conduct a proportion test we need to use the z statisitc, 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 $p_o$ .

Calculate the statistic

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

$z=\frac{0.293 -0.23}{\sqrt{\frac{0.23(1-0.23)}{594}}}=3.649$

Statistical decision

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:

$p_v =2*P(z>3.649)=0.00026$

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

b) Critical value

For this case we need to find a critical value that accumulates $\alpha/2$ of the area on each tail, we know that $\alpha=0.01$ , so then $\alpha/2 =0.005$ , using the normal standard table or excel we see that:

$z_{crit}= \pm 2.58$

Since the calculated value is higher than the critical value we have enough evidence to reject the null hypothesis at 1% of significance.

5 0

4 years ago