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andrew-mc [135]

3 years ago

13

A house painting company charges $376 plus $12 per hour. Another company $280 plus $15 per hour. A. how long is a job for which

both companies will charge the same amount? B.What will the cost be?

1 answer:

Anna35 [415]3 years ago

5 0

For company a ( $376 + $12 an hr) it would take 2 hours because 376+12= 388+12= 400 (376 + (12 x 2) )

for company b, ( $280 + $15 ) it would take 9 hours because 280 + 15 = 295 + 15 =
310 + 15 = 325 + 15 = 340 + 15 = 355 + 15 = 370 + 15 = 385 + 15 = 400 (280 + (15 x 8))

376 x (12 x 2)
376 x 24 = 400

280 x (15 x 8)
280 x 120 = 400

1) a = 2 hrs
b = 8 hrs

2) $400

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

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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.

<em> Null Hypothesis, </em> $H_0$ <em> : </em> $\mu_A = \mu_B$ <em> or </em> $\mu_D$ <em> = 0 </em>

<em> Alternate Hypothesis, </em> $H_1$ <em> : </em> $\mu_A \neq \mu_B$ <em> or </em> $\mu_D \neq$ <em> 0</em>

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

<em>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 .</em>

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

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