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3 years ago

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Use the t-distribution and the sample results to complete the test of the hypotheses. Use a significance level. Assume the resul

ts come from a random sample, and if the sample size is small, assume the underlying distribution is relatively normal.Test H0: μ= 4 vs Ha: μ≠4 using the sample results x= 4.8, s= 2.3, with n= 15.Give the test statistic and the p-value.

1 answer:

Ksenya-84 [330]3 years ago

3 0

Answer:

Test statistic = 1.3471

P-value = 0.1993

Accept the null hypothesis.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 4

Sample mean, $\bar{x}$ = 4.8

Sample size, n = 15

Alpha, α = 0.05

Sample standard deviation, s = 2.3

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 4\\H_A: \mu \neq 4$

We use two-tailed t test to perform this hypothesis.

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{4.8 - 4}{\frac{2.3}{\sqrt{15}} } = 1.3471$

Now, we calculate the p-value.

P-value = 0.1993

Since the p-value is greater than the significance level, we fail to reject the null hypothesis and accept it.

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Step-by-step explanation:

Straightforward use of the inverse tangent function of a calculator will tell you θ ≈ -0.08678 radians. This is an angle in the 4th quadrant, where your restriction on θ places it. (To comply with the restriction, you would need to consider the angle value to be 2π-0.08678 radians. The trig values for this angle are the same as the trig values for -0.08678 radians.)

Likewise, straightforward use of the calculator to find the other function values gives ...

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<em>Note on inverse tangent</em>

Depending on the mode setting of your calculator, the arctan or tan⁻¹ function may give you a value in degrees, not radians. That doesn't matter for this problem. sin(arctan(-0.087)) is the same whether the angle is degrees or radians, as long as you don't change the mode in the middle of the computation.

We have shown radians in the above answer because the restriction on the angle is written in terms of radians.

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<em>Alternate solution</em>

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$\cos{\theta}=\dfrac{1}{\sqrt{1+\tan^2{\theta}}}\\\\\sin{\theta}=\dfrac{\tan{\theta}}{\sqrt{1+\tan^2{\theta}}}$

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This more complicated computation gives the same result as above.

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Step-by-step explanation:

To prove that $w_1,\dots w_n$ form a basis for W, we must check that this set is a set of linearly independent vector and it generates the whole space W. We are given that T is an isomorphism. That is, T is injective and surjective. A linear transformation is injective if and only if it maps the zero of the domain vector space to the codomain's zero and that is the only vector that is mapped to 0. Also, a linear transformation is surjective if for every vector w in W there exists v in V such that T(v) =w

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$\lambda_1 = \cdots = \lambda_n$ .

Recall that $w_i = T(v_i)$ for i=1,...,n. Consider $T^{-1}$ to be the inverse transformation of T. Consider the equation

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If we apply $T^{-1}$ to this equation, then, we get

$T^{-1}(\lambda_1w_1+\dots \lambda_n w_n) =T^{-1}(0) = 0$

Since T is linear, its inverse is also linear, hence

$T^{-1}(\lambda_1w_1+\dots \lambda_n w_n) = \lambda_1T^{-1}(w_1)+\dots + \lambda_nT^{-1}(w_n)=0$

which is equivalent to the equation

$\lambda_1v_1+\dots + \lambda_nv_n =0$

Since $v_1,\dots,v_n$ are linearly independt, this implies that $\lambda_1=\dots \lambda_n =0$ , so the set $\{w_1, \dots, w_n\}$ is linearly independent.

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$w = a_1w_1+\dots+a_nw_n$

Since T is surjective, there exists a vector v in V such that T(v) = w. Since $v_1,\dots, v_n$ is a basis of v, there exist $a_1,\dots a_n$ , such that

$a_1v_1+\dots a_nv_n=v$

Then, applying T on both sides, we have that

$T(a_1v_1+\dots a_nv_n)=a_1T(v_1)+\dots a_n T(v_n) = a_1w_1+\dots a_n w_n= T(v) =w$

which proves that $w_1,\dots w_n$ generate the whole space W. Hence, the set $\{w_1, \dots, w_n\}$ is a basis of W.

Consider the linear transformation $T:\mathbb{R}^2\to \mathbb{R}^2$ , given by T(x,y) = T(x,0). This transformations fails to be injective, since T(1,2) = T(1,3) = (1,0). Consider the base of $\mathbb{R}^2$ given by (1,0), (0,1). We have that T(1,0) = (1,0), T(0,1) = (0,0). This set is not linearly independent, and hence cannot be a base of $\mathbb{R}^2$

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3 years ago