Algebra 2 question need help thanks.

The College Board SAT college entrance exam consists of three parts: math, writing and critical reading (The World Almanac 2012)

Wittaler [7]

Answer:

Yes, there is a difference between the population mean for the math scores and the population mean for the writing scores.

Test Statistics = $\frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } }$ follows $t_n_- _1$ .

Step-by-step explanation:

We are provided with the sample data showing the math and writing scores for a sample of twelve students who took the SAT ;

Let A = Math Scores ,B = Writing Scores and D = difference between both

So, $\mu_A$ = Population mean for the math scores

$\mu_B$ = Population mean for the writing scores

Let $\mu_D$ = Difference between the population mean for the math scores and the population mean for the writing scores.

Null Hypothesis, $H_0$ : $\mu_A = \mu_B$ or $\mu_D$ = 0

Alternate Hypothesis, $H_1$ : $\mu_A \neq \mu_B$ or $\mu_D \neq$ 0

Hence, Test Statistics used here will be;

$\frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } }$ follows $t_n_- _1$ where, Dbar = Bbar - Abar

$s_D$ = $\sqrt{\frac{\sum D_i^{2}-n*(Dbar)^{2}}{n-1}}$

n = 12

Student Math scores (A) Writing scores (B) D = B - A

1 540 474 -66

2 432 380 -52

3 528 463 -65

4 574 612 38

5 448 420 -28

6 502 526 24

7 480 430 -50

8 499 459 -40

9 610 615 5

10 572 541 -31

11 390 335 -55

12 593 613 20

Now Dbar = Bbar - Abar = 489 - 514 = -25

Bbar = $\frac{\sum B_i}{n}$ = $\frac{474+380+463+612+420+526+430+459+615+541+335+613}{12}$ = 489

Abar = $\frac{\sum A_i}{n}$ = $\frac{540+432+528+574+448+502+480+499+610+572+390+593}{12}$ = 514

∑ $D_i^{2}$ = 22600 and $s_D$ = $\sqrt{\frac{\sum D_i^{2}-n*(Dbar)^{2}}{n-1}}$ = $\sqrt{\frac{22600 - 12*(-25)^{2} }{12-1} }$ = 37.05

So, Test statistics = $\frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } }$ follows $t_n_- _1$

= $\frac{-25 - 0}{\frac{37.05}{\sqrt{12} } }$ follows $t_1_1$ = -2.34

Now at 5% level of significance our t table is giving critical values of -2.201 and 2.201 for two tail test. Since our test statistics doesn't fall between these two values as it is less than -2.201 so we have sufficient evidence to reject null hypothesis as our test statistics fall in the rejection region .

Therefore, we conclude that there is a difference between the population mean for the math scores and the population mean for the writing scores.

8 0

3 years ago

If the point A is located at ( –5, – 3) and A’ is the image of A after being rotated about the origin by 270° (counter clockwise

Ede4ka [16]

Answer:

A'(- 3, 5 )

Step-by-step explanation:

Under a counterclockwise rotation about the origin of 270°

a point (x, y ) → (y, - x ), hence

A(- 5, - 3 ) → A'(- 3, 5 )

5 0

3 years ago

The Cunninghams are moving across the country. Mr. Cunningham leaves 3 hours before Mrs. Cunningham. If he averages 40mph and sh

svp [43]

Recall your d = rt, distance = rate * time

now, if say, by the time they meet, Mr Cunningham has travelled "d" miles, that means Mrs Cunningham must also had travelled "d" miles as well.

However, he left 3 hours earlier, so by the time he travelled "d" miles, and took say "t" hours, for her it took 3 hour less, because she started driving 3 hours later, so, she's been on the road 3 hours less than Mr Cunningham, so by the time they meet, Mrs Cunningham has travelled then "t - 3" hours.

$\bf \begin{array}{lccclll}
&\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\
&------&------&------\\
\textit{Mr Cunningham}&d&40&t\\
\textit{Mrs Cunningham}&d&80&t-3
\end{array}
\\\\\\
\begin{cases}
\boxed{d}=40t\\
d=80(t-3)\\
----------\\
\boxed{40t}=80(t-3)
\end{cases}
\\\\\\
\cfrac{40t}{80}=t-3\implies \cfrac{t}{2}=t-3\implies t=2t-6\implies 6=2t-t
\\\\\\
6=t$

5 0

3 years ago

Find the value of n . 6/n = 24/28 y = ___.

il63 [147K]

Answer:

$n=7$

Step-by-step explanation:

Cross multiply, isolate the variable, and divide by the coefficient to solve.

$\frac{6}{n}=\frac{24}{28} \\ \\ 24n=168 \\ \\ n=7$

Plug back in to check.

$\frac{6}{7}=0.857142857 \\ \\ \frac{24}{28} =0.857142857$

5 0

3 years ago

Divide (- 6 + i)/(1 - 2i) =

Kaylis [27]

$~~~\dfrac{(-6 +i)}{(1- 2i)}\\\\\\=\dfrac{(-6+i)(1+2i)}{(1-2i)(1+2i)}\\\\\\=\dfrac{-6-12i+i+2i^2}{1-(2i)^2}\\\\\\=\dfrac{-6-11i-2}{1+4}~~~~~~~~~~~~~;[i^2=-1]\\\\\\=\dfrac{-8-11i}{5}\\\\\\=-\dfrac 85-\dfrac{11}{5}i$

4 0

1 year ago

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Algebra 2 question need help thanks.​

Algebra 2 question need help thanks.