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Tema [17]
3 years ago
7

Chuck has a gross pay of $815.70. By how much will Chuck’s gross pay be reduced if he has the following items withheld? federal

tax of $56 Social Security tax that is 6.2% of his gross pay Medicare tax that is 1.45% of his gross pay state tax that is 19% of his federal tax
Mathematics
2 answers:
drek231 [11]3 years ago
8 0

Answer:

THE ANSWER IS B.

$129.04

Step-by-step explanation:

tangare [24]3 years ago
3 0

Chuck has a gross pay of $815.70. His gross pay will be reduced by:

  • Federal tax of $56;
  • Social Security tax that is 6.2% of his gross pay;
  • Medicare tax that is 1.45% of his gross pay;
  • State tax that is 19% of his federal tax.

Let's count:

1. gross pay of $815.70 - 100%,

Social Security tax of $x - 6.2%.

Then

\dfrac{815.70}{x}=\dfrac{100}{6.2} ,\\ \\815.70\cdot 6.2=x\cdot 100,\\ \\x=\dfrac{815.70\cdot 6.2}{100} =50.5734.

2. gross pay of $815.70 - 100%,

Medicare tax of $y - 1.45%.

Then

\dfrac{815.70}{y}=\dfrac{100}{1.45} ,\\ \\815.70\cdot 1.45=y\cdot 100,\\ \\y=\dfrac{815.70\cdot 1.45}{100} =11.82765.

3. Federal tax of $56 - 100%,

State tax $z - 19%.

Then

\dfrac{56}{z}=\dfrac{100}{19} ,\\ \\56\cdot 19=z\cdot 100,\\ \\z=\dfrac{56\cdot 19}{100} =10.64.

4. Chuck’s gross pay will be reduced by

\$56+\$50.5734+\$11.82765+\$10.64=\$129.04105\approx \$129.04.

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(a) 95% confidence interval for the percentage of all car accidents that involve teenage drivers is [0.177 , 0.243].

(b) We are 95% confident that the percentage of all car accidents that involve teenage drivers will lie between 17.7% and 24.3%.

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Step-by-step explanation:

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           p = population percentage of all car accidents

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P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level

                                                     of significance are -1.96 & 1.96}  

P(-1.96 < \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} }} < 1.96) = 0.95

P( -1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }} < {\hat p-p} < 1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }} ) = 0.95

P( \hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }} < p < \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }} ) = 0.95

<u>95% confidence interval for p</u> = [\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }} , \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }}]

  = [ 0.21-1.96 \times {\sqrt{\frac{0.21(1-0.21)}{576} }} , 0.21+1.96 \times {\sqrt{\frac{0.21(1-0.21)}{576} }} ]

  = [0.177 , 0.243]

Therefore, 95% confidence interval for the percentage of all car accidents that involve teenage drivers is [0.177 , 0.243].

(b) We are 95% confident that the percentage of all car accidents that involve teenage drivers will lie between 17.7% and 24.3%.

(c) We are also provided that before you were hired in the company, the percentage of teenagers who where involved in car accidents was 18%.

The manager wants to see if the percentage of teenagers has changed since you join the company.

<u><em>Let p = percentage of teenagers who where involved in car accidents</em></u>

So, Null Hypothesis, H_0 : p = 18%    {means that the percentage of teenagers has not changed since you join the company}

Alternate Hypothesis, H_A : p \neq 18%    {means that the percentage of teenagers has changed since you join the company}

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where, \hat p = sample proportion teenage drivers = \frac{120}{576} = 0.21

           n = sample of accidents = 576

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Therefore, we conclude that the the percentage of teenagers has not changed since you join the company.

(d) Now at 0.1 significance level, the z table gives critical value of -1.6449 and 1.6449 for two-tailed test. Since our test statistics does not lie within the range of critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u>we reject our null hypothesis</u>.

Therefore, we conclude that the the percentage of teenagers has changed since you join the company.

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