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Vikentia [17]
3 years ago
10

PLEASE HURRY

Mathematics
1 answer:
Morgarella [4.7K]3 years ago
4 0

Answer:

1,3, and 5 are the ones that have to be checked

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A counselor at a community college claims that the mean GPA for students who have transferred to State University is more than 3
Lena [83]

Answer:

t=\frac{3.25-3.1}{\frac{0.3}{\sqrt{36}}}=3  

If we compare the p value and the significance level for example \alpha=1-0.99=0.01 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis.

And the best conclusion would be:

a) Yes. The counselor's claim is valid

Because we reject the null hypothesis in favor to the alternative hypothesis.

Step-by-step explanation:

Data given and notation  

\bar X=3.25 represent the sample mean

s=0.3 represent the sample standard deviation

n=36 sample size  

\mu_o =3.1 represent the value that we want to test  

\alpha represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is more than 3.1, the system of hypothesis would be:  

Null hypothesis:\mu \leq 3.1  

Alternative hypothesis:\mu > 3.1  

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

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

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

t=\frac{3.25-3.1}{\frac{0.3}{\sqrt{36}}}=3  

P-value  

The first step is calculate the degrees of freedom, on this case:  

df=n-1=36-1=35  

Since is a one right tailed test the p value would be:  

p_v =P(t_{(35)}>3)=0.00247  

Conclusion  

If we compare the p value and the significance level for example \alpha=1-0.99=0.01 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis.

And the best conclusion would be:

a) Yes. The counselor's claim is valid

Because we reject the null hypothesis in favor to the alternative hypothesis.

5 0
3 years ago
Quadrilateral ABCD is rotated 145° about point T. The result is quadrilateral A'B'C'D'. Which congruency statement is correct?
lilavasa [31]
<span>The congruency statement should be that which shows the congruency of the corresponding angle and the sides. In the rotation of the polygon, the angle A is congruent with angle A' still. The same is true for all the angles and sides. The congruency statement is therefore, ABCD is congruent to A'B'C'D' The answer to this item is letter A. </span>
3 0
3 years ago
Read 2 more answers
A scuba diver starts at 85 3/4 meters below the surface of the water and descends until he reaches 103 1/5
RSB [31]

Answer:

From his starting point of 85 3/4, he descended 17 9/20 meters

Step-by-step explanation:

We want to know how much farther he went down from his starting point.

103 1/5 - 85 3/4

S

tart off by making the denominators of both the numbers the same.

103 1/5 - 85 3/4

103 4/20 - 85 15/20

Make them into improper fractions

2064/20 - 1715/20 = 349/20

Make 349/20 into a mixed fraction

17 9/20

5 0
3 years ago
Racheal brought chocolate and vanilla to school for her birthday.80 students decided to take a cupcake and 56 of them took vanil
anyanavicka [17]

Answer:

70%

Step-by-step explanation:

56 vanilla cupcakes

------------------------------- = 0.7, or 0.7(100%), or 70%

     80 students

70% of students picked vanilla.

3 0
3 years ago
X y
Vanyuwa [196]

Answer:

a = 19

Step-by-step explanation:

If the rate of change is -8, subtract 8 from your y value of 27. And if 19 is correct, you should be able to subtract 8, and get 11 as the next number in the function, which you do, so 19 is the value of a.

5 0
3 years ago
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