Please help last question:((

Mathematics

1 answer:

jasenka [17]3 years ago

5 0

Answer:

7. C(N(h))=33hx+460h

8. about $13.94 is the cost

Step-by-step explanation:

7. C(x)=(33x+460)(N(h))=(N(40)h)--C(x)=(33x+460)(40h)--C(x)=1320hx+18400h--simplified to C(N(h))=33hx+460h

8. C(N(10))=33(10)x+460(10)--C(N(10))=330x+4600--4600/330=about13.94

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The diameters of bearings used in an aircraft landing gear assembly have a standard deviation of ???? = 0.0020 cm. A random samp

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Answer:

(a) We conclude after testing that mean diameter is 8.2500 cm.

(b) P-value of test = 2 x 0.0005% = 1 x $10^{-5}$ .

(c) 95% confidence interval on the mean diameter = [8.2525 , 8.2545]

Step-by-step explanation:

We are given with the population standard deviation, $\sigma$ = 0.0020 cm

Sample Mean, Xbar = 8.2535 cm and Sample size, n = 15

(a) Let Null Hypothesis, $H_0$ : Mean Diameter, $\mu$ = 8.2500 cm

Alternate Hypothesis, $H_1$ : Mean Diameter, $\mu$ $\neq$ 8.2500 cm{Given two sided}

So, Test Statistics for testing this hypothesis is given by;

$\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ follows Standard Normal distribution

After putting each value, Test Statistics = $\frac{8.2535-8.2500}{\frac{0.0020}{\sqrt{15} } }$ = 6.778

Now we are given with the level of significance of 5% and at this level of significance our z score has a value of 1.96 as it is two sided alternative.

<em>Since our test statistics does not lie in the rejection region{value smaller than 1.96} as 6.778>1.96 so we have sufficient evidence to accept null hypothesis and conclude that the mean diameter is 8.2500 cm.</em>

(b) P-value is the exact % where test statistics lie.

For calculating P-value , our test statistics has a value of 6.778

So, P(Z > 6.778) = Since in the Z table the highest value for test statistics is given as 4.4172 and our test statistics has value higher than this so we conclude that P - value is smaller than 2 x 0.0005% { Here 2 is multiplied with the % value of 4.4172 because of two sided alternative hypothesis}

Hence P-value of test = 2 x 0.0005% = 1 x $10^{-5}$ .

Probability(-1.96 < N(0,1) < 1.96) = 0.95 { This indicates that at 5% level of significance our Z score will lie between area of -1.96 to 1.96.

P(-1.96 < $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96\frac{\sigma}{\sqrt{n} }$ < $Xbar - \mu$ < $1.96\frac{\sigma}{\sqrt{n} }$ ) = 0.95