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PIT_PIT [208]
1 year ago
13

A teacher is interested in performing a hypothesis test to compare the mean math score of the girls and the mean math score of t

he boys. She randomly selects 10 girls from the class and then randomly selects 10 boys. She arranges the girls' names alphabetically and uses this list to assign each girl a number between 1 and 10. She does the same thing for the boys.
a. Paired t-test. Since there are 10 boys and 10 girls, we can link the two samples.
b. Paired t-test. Since the boys and girls are in the same class, and are hence dependent samples, they are can be linked.
c. 1-sample t-test. The teacher should compare the sample mean for the girls against the population mean for the boys.
d. Two-sample t-test. There is no natural pairing between the two populations.
Mathematics
1 answer:
tamaranim1 [39]1 year ago
8 0

Answer:

d. Two-sample t-test. There is no natural pairing between the two populations.

Step-by-step explanation:

A two-sample t - test is a test performed on the data of two random samples, each independently obtained from a different given population. The purpose of the test is to determine whether the difference between these two populations is statistically significant.

Independent samples are samples that do not affect one another. The mean math scores of the samples of boys and girls do not affect each other. They are independent samples, hence the correct test procedure is two - sample t - test

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Which properties are present in a table that represents an exponential function in the form y-b* when b > 1?
Oksana_A [137]

Answer:

<u>Properties that are present are </u>

Property I

Property IV

Step-by-step explanation:

The function given is  y=b^x  where b > 1

Let's take a function, for example,  y=2^x

Let's check the conditions:

I. As the x-values increase, the y-values increase.

Let's put some values:

y = 2 ^ 1

y = 2

and

y = 2 ^ 2

y = 4

So this is TRUE.

II. The point (1,0) exists in the table.

Let's put 1 into x and see if it gives us 0

y = 2 ^ 1

y = 2

So this is FALSE.

III. As the x-value increase, the y-value decrease.

We have already seen that as x increase, y also increase in part I.

So this is FALSE.

IV. as the x value decrease the y values decrease approaching a singular value.

THe exponential function of this form NEVER goes to 0 and is NEVER negative. So as x decreases, y also decrease and approached a value (that is 0) but never becomes 0.

This is TRUE.

Option I and Option IV are true.

7 0
2 years ago
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Recipe ingredients remain in a constant ratio no matter how many servings are prepared. Which table shows a possible
Tpy6a [65]

Answer:

Table N 4

Step-by-step explanation:

we know that

A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form y/x=k or y=kx

<em>Verify the table 4</em>

For x=1, y=2

so

y/x=2/1=2

For x=2, y=4

so

y/x=4/2=2

For x=3, y=6

so

y/x=6/3=2

therefore

The constant of proportionality k is equal to 2 and the equation is equal to

y=2x

The table 4 represent a direct variation, therefore is a possible ratio table for ingredients X and Y

6 0
1 year ago
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A team of 10 players is to be selected from a class of 6 girls and 7 boys. Match each scenario to its probability. You have to d
tankabanditka [31]
The selection of r objects out of n is done in

C(n, r)= \frac{n!}{r!(n-r)!} many ways.

The total number of selections 10 that we can make from 6+7=13 students is 

C(13,10)= \frac{13!}{3!(10)!}= \frac{13*12*11*10!}{3*2*1*10!}= \frac{13*12*11}{3*2}=  286
thus, the sample space of the experiment is 286

A. 
<span>"The probability that a randomly chosen team includes all 6 girls in the class."

total number of group of 10 which include all girls is C(7, 4), because the girls are fixed, and the remaining 4 is to be completed from the 7 boys, which can be done in C(7, 4) many ways.


</span>C(7, 4)= \frac{7!}{4!3!}= \frac{7*6*5*4!}{4!*3*2*1}= \frac{7*6*5}{3*2}=35
<span>
P(all 6 girls chosen)=35/286=0.12

B.
"</span>The probability that a randomly chosen team has 3 girls and 7 boys.<span>"

with the same logic as in A, the number of groups were all 7 boys are in, is 

</span>C(6, 3)= \frac{6!}{3!3!}= \frac{6*5*4*3!}{3!3!}= \frac{6*5*4}{3*2*1}=20
<span>
so the probability is 20/286=0.07

C.
"</span>The probability that a randomly chosen team has either 4 or 6 boys.<span>"

case 1: the team has 4 boys and 6 girls

this was already calculated in part A, it is </span>0.12.
<span>
case 2, the team has 6 boys and 4 girls.

there C(7, 6)*C(6, 4) ,many ways of doing this, because any selection of the boys which can be done in C(7, 6) ways, can be combined with any selection of the girls. 

</span>C(7, 6)*C(6, 4)= \frac{7!}{6!1}* \frac{6!}{4!2!} =7*15= 105
<span>
the probability is 105/286=0.367

since  case 1 and case 2 are disjoint, that is either one or the other happen, then we add the probabilities:

0.12+0.367=0.487 (approximately = 0.49)

D.
"</span><span>The probability that a randomly chosen team has 5 girls and 5 boys.</span><span>"

selecting 5 boys and 5 girls can be done in 

</span>C(7, 5)*C(6,5)= \frac{7!}{5!2} * \frac{6!}{5!1}=21*6=126

many ways,

so the probability is 126/286=0.44
6 0
2 years ago
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What is the value of the 2s in 42,256
Anna71 [15]
2,000 and 200 are the values of the 2s in 42256

4 0
2 years ago
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The GPA of accounting students in a university is known to be normally distributed. A random sample of 20 accounting students re
Vladimir79 [104]

Answer:

The 95% of confidence intervals

(2.84 ,2.99)

Step-by-step explanation:

A random sample of 20 accounting students results in a mean of 2.92 and a standard deviation of 0.16

given small sample size n =20

sample mean x⁻ =2.92

sample standard deviation 'S' =0.16

level of significance ∝ =  0.95

The 95% of confidence intervals

x^{-}  ± t_{\alpha } \frac{S}{\sqrt{n} }

the degrees of freedom γ=n-1 =20-1=19

t-table 2.093

(x^{-}  - t_{\alpha } \frac{S}{\sqrt{n} },x^{-}  + t_{\alpha } \frac{S}{\sqrt{n} })

(2.92 - 2.093(\frac{0.16}{\sqrt{20} } ,2.92+2.093(\frac{0.16}{\sqrt{20} } )

(2.92-0.0748,2.92+0.0748)

(2.84 ,2.99)

Therefore the 95% of confidence intervals

(2.84 ,2.99)

4 0
2 years ago
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