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natali 33 [55]
2 years ago
10

One of your company's salesmen estimates he will drive 20,000 miles this year and each of the next two years. As the service man

ager for the company's fleet of six-cylinder compact automobiles, you are asked to determine the total costs for each of the three years. Year 1 Total Cost: $ Year 2 Total Cost: $ Year 3 Total Cost: $

Mathematics
1 answer:
grigory [225]2 years ago
4 0

Therefore, the total costs for year 1 will be $2,240, for year 2 they will be $2,040, and for year 3 they will be $2,120.

<h3><u /></h3><h3><u>Costs</u></h3>

Given that one of your company's salesmen estimates he will drive 20,000 miles this year and each of the next two years, and as the service manager for the company's fleet of six-cylinder compact automobiles, you are asked to determine the total costs for each of the three years, to determine said cost, the following calculation must be made:

  • Year 1:
  • 20000 x 0.112 = X
  • 2,240 = X

  • Year 2:
  • 20000 x 0.102 = X
  • 2,040 = X

  • Year 3:
  • 20000 x 0.106 = X
  • 2,120 = X

Therefore, the total costs for year 1 will be $2,240, for year 2 they will be $2,040, and for year 3 they will be $2,120.

Learn more about costs in brainly.com/question/4557688

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To test Upper H 0​: muequals50 versus Upper H 1​: muless than50​, a random sample of size nequals23 is obtained from a populatio
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Answer:

Step-by-step explanation:

Hello!

1)

<em>To test H0: u= 50 versus H1= u < 50, a random sample size of n = 23 is obtained from a population that is known to be normally distributed. Complete parts A through D. </em>

<em> A) If  ¯ x = 47.9  and s=11.9, compute the test statistic .</em>

For thistest the corresponsing statistis is a one sample t-test

t= \frac{X[bar]-Mu}{\frac{S}{\sqrt{n} } }~~t_{n-1}

t_{H_0}= \frac{47.9-50}{\frac{11.9}{\sqrt{23} } } = -0.846= -0.85

B) If the researcher decides to test this hypothesis at the a=0.1 level of significance, determine the critical value(s).

This test is one-tailed to the left, meaning that you'll reject the null hypothesis to small values of the statistic. The ejection region is defined by one critical value:

t_{n-1;\alpha }= t_{22;0.1}= -1.321

Check the second attachment. The first row shows α= Level of significance; the First column shows ν= sample size.

The t-table shows the values of the statistic for the right tail. P(tₙ≥α)

But keep in mind that this distribution is centered in zero, meaning that the right and left tails are numerically equal, only the sign changes. Since in this example the rejection region is one-tailed to the left, the critical value is negative.

C) What does the distribution graph appear like?

Attachment.

D) Will the researcher reject the null hypothesis?

As said, the rejection region is one-tailed to the right, so the decision rule is:

If t_{H_0} ≤ -1.321, reject the null hypothesis.

If t_{H_0} > -1.321, do not reject the null hypothesis.

t_{H_0}= -0.85, the decision is to not reject the null hypothesis.

2)

To test H0​: μ=100 versus H1​:≠​100, a simple random sample size of nequals=24 is obtained from a population that is known to be normally distributed. Answer parts​ (a)-(d).

a) If x =104.2 and s=9.6, compute the test statistic.

For this example you have to use a one sample t-test too. The formula of the statistic is the same:

t_{H_0}= \frac{X[bar]-Mu}{\frac{S}{\sqrt{n} } } = \frac{104.2-100}{\frac{9.6}{\sqrt{24} } = } = 2.143

b) If the researcher decides to test this hypothesis at the α=0.01 level of​ significance, determine the critical values.

This hypothesis pair leads to a two-tailed rejection region, meaning, you'll reject the null hypothesis at either small or big values of the statistic. Then the rejection region is divided into two and determined by two critical values (the left one will be negative and the right one will be positive but the module of both values will be equal).

t_{n-1;\alpha/2 }= t_{23; 0.005}= -2.807

t_{n-1;1-\alpha /2}= t_{23;0.995}= 2.807

c) Draw a​ t-distribution that depicts the critical​ region(s). Which of the following graphs shows the critical​ region(s) in the​t-distribution?

Attachment.

​(d) Will the researcher reject the null​ hypothesis?

The decision rule for the two-tailed hypotheses pair is:

If t_{H_0} ≤ -2.807 or if t_{H_0} ≥ 2.807, reject the null hypothesis.

If -2.807 < t_{H_0} < 2.807, do not reject the null hypothesis.

t_{H_0}= 2.143 is greater than the right critical value, the decision is to reject the null hypothesis.

Correct option:

B. The researcher will reject the null hypothesis since the test statistic is not between the critical values.

3)

Full text in attachment. The sample size is different by 2 but it should serve as a good example.

H₀: μ = 20

H₁: μ < 20

a) n= 18, X[bar]= 18.3, S= 4, Compute statistic.

t_{H_0}= \frac{X[bar]-Mu}{\frac{S}{\sqrt{n} } }= \frac{18.3-20}{\frac{4}{\sqrt{18} } } = -1.80

b) The rejection region in this example is one-tailed to the left, meaning that you'll reject the null hypothesis to small values of t.

Out of the three graphics, the correct one is A.

c)

To resolve this you have to look for the values in the t-table that are the closest to the calculated t_{H_0}

Symbolically:

t_{n-1;\alpha_1 } \leq t_{H_0}\leq t_{n-1;\alpha _2}

t_{H_0}= -1.80

t_{17; 0.025 }= -2.110

t_{17;0.05}= -1.740

Roughly defined you can say that the p-value is the probability of obtaining the value of t_{H_0}, symbolically: P(t₁₇≤-1.80)

Under the distribution the calculated statistic is between the values of -2.110 and -1.740, then the p-value will be between their cumulated probabilities:

A. 0.025 < p-value < 0.05

d. The researcher decides to test the hypothesis using a significance level of α: 0.05

Using the p-value approach the decision rule is the following:

If p-value ≤ α, reject the null hypothesis.

If p-value > α, do not reject the null hypothesis.

We already established in item c) that the p-value is less than 0.05, so the decision is to reject the null hypothesis.

Correct option:

B. The researcher will reject the null hypothesis since the p-value is less than α.

I hope this helps!

6 0
3 years ago
Mustafa’s soccer team is planning a school dance as a fundraiser. The DJ charges $200 and decorations cost $100. The team decide
statuscvo [17]

The formula can be used to determine how many students are required to earn $1,500 in profit.t:

5.n - 100 - 200 = 1500.

<h3>What is linear equation?</h3>

An equation is deemed to be linear if a variable's maximum power consistently equals 1. Another name for it is a one-degree equation. The form of the linear equation Ax + B = 0 is standard for one-variable linear equations. In this instance, as well as A are variables while B is a constant.

Ax + By = C is the standard form of a two-variable linear equation. This equation contains the variables x and y, the coefficients Both a B, as well as the constant C.

<h3>According to the given data:</h3>

Given that the no. of students is n.

DJ charge - $200 Decorations cost - $100

Charge each student should pay - $5.00

Therefore the equation that find the no. of students to make a profit of $1,500.5.n - 100 - 200 = 1500

To learn more about linear equation visit:

brainly.com/question/2030026

#SPJ9

7 0
1 year ago
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