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Bezzdna [24]
3 years ago
13

A certain alloy contains 14.25% nickel. How much nickel is there in a piece weighing 375 pounds? round to the nearest hundredth

Mathematics
1 answer:
grandymaker [24]3 years ago
3 0
Namely, what is 14.25% of 375lbs

well, if we take 375 to be 100%, how much is 14.25%?

well  \bf \begin{array}{ccllll}
amount&\%\\
\textendash\textendash\textendash\textendash\textendash\textendash&\textendash\textendash\textendash\textendash\textendash\textendash\\
375&100\\
x&14.25
\end{array}\implies \cfrac{375}{x}=\cfrac{100}{14.25}

solve for "x"
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Answer:

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(B) The test statistic is 1.617.

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Step-by-step explanation:

We are given that a random sample of eight cars from each manufacturer is selected, and eight drivers are selected to drive each automobile for a specified distance.

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Driver         Manufacturer A               Manufacturer B

   1                      32                                       28

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  8                      25                                       27

Let \mu_1 = mean MPG for the fuel efficiency of Manufacturer A brand

\mu_2 = mean MPG for the fuel efficiency of Manufacturer B brand

SO, Null Hypothesis, H_0 : \mu_1-\mu_2=0  or  \mu_1= \mu_2    {means that there is a not any significant difference in the mean MPG (miles per gallon) when testing for the fuel efficiency of these two brands of automobiles}

Alternate Hypothesis, H_A : \mu_1-\mu_2\neq 0  or  \mu_1\neq  \mu_2   {means that there is a significant difference in the mean MPG (miles per gallon) for the fuel efficiency of these two brands of automobiles}

The test statistics that will be used here is <u>Two-sample t test statistics</u> as we don't know about the population standard deviations;

                      T.S.  = \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}  } }  ~ t__n__1+_n__2-2

where, \bar X_1 = sample mean MPG for manufacturer A = \frac{\sum X_A}{n_A} = 27.625

\bar X_2 = sample mean MPG for manufacturer B =\frac{\sum X_B}{n_B} = 25.625

s_1 = sample standard deviation for manufacturer A = \sqrt{\frac{\sum (X_A-\bar X_A)^{2} }{n_A-1} } = 2.72

s_2 = sample standard deviation manufacturer B = \sqrt{\frac{\sum (X_B-\bar X_B)^{2} }{n_B-1} } = 2.20

n_1 = sample of cars selected from manufacturer A = 8

n_2 = sample of cars selected from manufacturer B = 8

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(A) The mean for the differences is = 27.625 - 25.625 = 2

(B) <u><em>The test statistics</em></u>  =  \frac{(27.625-25.625)-(0)}{2.474 \times \sqrt{\frac{1}{8}+\frac{1}{8}  } }  ~  t_1_4

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(C) Now at 10% significance level, the t table gives critical values between -1.761 and 1.761 at 14 degree of freedom for two-tailed test. Since our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u>we fail to reject our null hypothesis</u>.

Therefore, we conclude that there is a not any significant difference in the mean MPG (miles per gallon) when testing for the fuel efficiency of these two brands of automobiles.

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