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AveGali [126]
1 year ago
14

A survey conducted by the American Automobile Association showed that a family of four spends an average of $215.60 per day whil

e on vacation. Suppose a sample of 64 families of four vacationing at Niagara Falls resulted in a sample mean of $252.45 per day and a sample standard deviation of $77.50.
a. Develop a 95% confidence interval estimate of the mean amount spent per day by a family of four visiting Niagara Falls (to 2 decimals). ( , )

b. Based on the confidence interval from part (a), does it appear that the population mean amount spent per day by families visiting Niagara Falls differs from the mean reported by the American Automobile Association?
Mathematics
1 answer:
Naddika [18.5K]1 year ago
4 0

Answer:

a) (233.09,271.80)

b) Yes  

Step-by-step explanation:

We are given the following information:

Population mean = $215.60

Sample mean = $252.45

Sample standard deviation = $77.50.

n = 64

Formula:

\bar{x} \pm t_{critical}\frac{s}{\sqrt{n}}

Putting the values, we get,

t_{critical}\text{ at degree of freedom 63 and}~\alpha_{0.05} = 1.998

252.45 \pm 1.998(\frac{77.50}{\sqrt{64}} ) = 252.45 \pm 19.355 = (233.09,271.80)

b) The mean amount spent per day by families visiting Niagara Falls differs from the mean reported by the American Automobile Association because the confidence interval lies above  $215.60

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Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

(3*x+7)^2+5-((7*x-9)^2*x+5)=0

Step by step solution :

STEP1:

Equation at the end of step 1

(((3x+7)2)+5)-(x•(7x-9)2+5) = 0

STEP2:

Checking for a perfect cube

2.1 -49x3+135x2-39x+49 is not a perfect cube

Trying to factor by pulling out :

2.2 Factoring: -49x3+135x2-39x+49

Thoughtfully split the expression at hand into groups, each group having two terms :

Group 1: -39x+49

Group 2: -49x3+135x2

Pull out from each group separately :

Group 1: (-39x+49) • (1) = (39x-49) • (-1)

Group 2: (49x-135) • (-x2)

Factoring by pulling out fails :

The groups have no common factor and can not be added up to form a multiplication.

Polynomial Roots Calculator :

Find roots (zeroes) of : F(x) = -49x3+135x2-39x+49

Polynomial Roots Calculator is a set of methods aimed at finding values of x for which F(x)=0

Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers x which can be expressed as the quotient of two integers

The Rational Root Theorem states that if a polynomial zeroes for a rational number P/Q then P is a factor of the Trailing Constant and Q is a factor of the Leading Coefficient

In this case, the Leading Coefficient is -49 and the Trailing Constant is 49.

The factor(s) are:

of the Leading Coefficient : 1,7 ,49

of the Trailing Constant : 1 ,7 ,49

Let us test ....

Equation at the end of step

2

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-49x3 + 135x2 - 39x + 49 = 0

STEP

3

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Cubic Equations:

3.1 Solve -49x3+135x2-39x+49 = 0

Future releases of Tiger-Algebra will solve equations of the third degree directly.

Meanwhile we will use the Bisection method to approximate one real solution.

Approximating a root using the Bisection Method :

We now use the Bisection Method to approximate one of the solutions. The Bisection Method is an iterative procedure to approximate a root (Root is another name for a solution of an equation).

The function is F(x) = -49x3 + 135x2 - 39x + 49

At x= 3.00 F(x) is equal to -176.00

At x= 2.00 F(x) is equal to 119.00

Intuitively we feel, and justly so, that since F(x) is negative on one side of the interval, and positive on the other side then, somewhere inside this interval, F(x) is zero

Procedure :

(1) Find a point "Left" where F(Left) < 0

(2) Find a point 'Right' where F(Right) > 0

(3) Compute 'Middle' the middle point of the interval [Left,Right]

(4) Calculate Value = F(Middle)

(5) If Value is close enough to zero goto Step (7)

Else :

If Value < 0 then : Left <- Middle

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(6) Loop back to Step (3)

(7) Done!! The approximation found is Middle

Next Middle will get us close enough to zero:

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The desired approximation of the solution is:

x ≓ 2.596896529

Note, ≓ is the approximation symbol

One solution was found :

x ≓ 2.596896529

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