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pav-90 [236]
3 years ago
14

Quarters are currently minted with weights normally distributed and having a standard deviation of 0.065. New equipment is being

tested in an attempt to improve quality by reducing variation. A simple random sample of 25 quarters is obtained from those manufactured with the new​ equipment, and this sample has a standard deviation of 0.047 . Use a 0.05 significance level to test the claim that quarters manufactured with the new equipment have weights with a standard deviation less than 0.065. Does the new equipment appear to be effective in reducing the variation of​ weights?
Mathematics
1 answer:
Ilya [14]3 years ago
5 0

Answer:

Yes, the new equipment appear to be effective in reducing the variation of​ weights.

Step-by-step explanation:

We are given that Quarters are currently minted with weights normally distributed and having a standard deviation of 0.065.

A simple random sample of 25 quarters is obtained from those manufactured with the new​ equipment, and this sample has a standard deviation of 0.047.

Let \sigma = <u><em>standard deviation of weights of new equipment.</em></u>

SO, Null Hypothesis, H_0 : \sigma \geq 0.065      {means that the new equipment have weights with a standard deviation more than or equal to 0.065}

Alternate Hypothesis, H_A : \sigma < 0.065      {means that the new equipment have weights with a standard deviation less than 0.065}

The test statistics that would be used here <u>One-sample chi-square</u> test statistics;

                           T.S. =  \frac{(n-1)s^{2} }{\sigma^{2} }  ~ \chi^{2}__n_-_1

where, s = sample standard deviation = 0.047

           n = sample of quarters = 25

So, <u><em>the test statistics</em></u>  =  \frac{(25-1)\times 0.047^{2} }{0.065^{2} }  ~  \chi^{2}__2_4   

                                     =  12.55

The value of chi-square test statistics is 12.55.

Now, at 0.05 significance level the chi-square table gives critical value of 13.85 at 24 degree of freedom for left-tailed test.

Since our test statistic is less than the critical value of chi-square as 12.55 < 13.85, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u><em>we reject our null hypothesis</em></u>.

Therefore, we conclude that the new equipment have weights with a standard deviation less than 0.065.

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Answer:

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Step-by-step explanation:

* At first lets revise the standard and general forms of the

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- c is the y-intercept ⇒ means x = 0

- h = -b/2a

- k = f(h)

* Lets solve the problem

- We will find the equation of the parabola

∵ The vertex is (-1 , -108)

∴ h = -1 and k = -108

∵ y-intercept = -105

- Equate the two forms

∵ ax² + bx + c = a(x - h)² + k ⇒ solve the (   )²

∴ ax² + bx + c = a(x² - 2hx + h²) + k ⇒ open the bracket

∴ ax² + bx + c = ax² - 2ahx + ah² + k ⇒ by comparing the two sides

∴ c = ah² + k

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∴ -105 = a(-1)² + -108

∴ -105 = a - 108 ⇒ add 108 to the both sides

∴ 3 = a

- Lets write the equation in the standard form

∴ y = 3(x - -1)² + -108

∴ y = 3(x + 1)² - 108

* To find the x-intercepts means the parabola intersects the x-axis,

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∴ 3(x + 1)² - 108 = 0 ⇒ Add 108 to the both sides

∴ 3(x + 1)² = 108 ⇒ divide the both sides by 3

∴ (x + 1)² = 36 ⇒ take square root for both sides

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∵ x + 1 = 6 ⇒ subtract 1 from both sides

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∴

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         \rule{50}{1}\large\blue\textsf{\textbf{\underline{Question:-}}}\rule{50}{1}

       <em>Write any 4 laws, or properties, of exponents.</em>

<em />

<em>        </em>\rule{50}{1}\large\blue\textsf{\textbf{\underline{Answer and how to solve:-}}}\rule{50}{1}

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\Large\text{It states that:-}

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<h3>Good luck with your studies.</h3>

#TogetherWeGoFar

\rule{50}{1}\smile\smile\smile\smile\smile\smile\rule{50}{1}

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