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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:

You can do this through synthetic division. It is a shorthand way of dividing polynomials.

First you need to make your divisor equal to zero so you can solve for what goes into the division box:

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Now you can proceed to synthetic division. Make an upside down division box with the divisor outside and the coefficients of the dividend listed, along with their sign. Leave a space below the divident

+2 |   +1     0     -1     0     0     -5

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Next you bring down the first coefficient:

+2 |   +1     0     -1     0     0     -5

    |<u>                                              </u>

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Then you multiply it by the divisor and write the product under the next coefficient:

+2 |   +1     0     -1     0     0     -5

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Next add the column and put the sum below it:

+2 |   +1     0     -1     0     0     -5

    |<u>          +2                                   </u>

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Then multiply again and repeat until you reach the last coefficient:

+2 |   +1     0     -1     0     0     -5

    |<u>          +2    +4   +6   +12   +24 </u>

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Now that you have your results, add  in the x and their powers. The powers will start with the highest power but 1 less than the dividend. Since the dividend's highest power is 5, then the quotient's highest power will be 4. Then write it in descending order :

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Now the last coefficient is your remainder. So your results will be:

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<h3>How to solve algebra</h3>

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brainly.com/question/4344214

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