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MrRissso [65]
2 years ago
13

Fifty specimens of a new computer chip were tested for speed in a certain application, along with 50 specimens of chips with the

old design. The average speed, in MHz, for the new chips was 495.6, and the standard deviation was 19.4. The average speed for the old chips was 481.2, and the standard deviation was 14.3.Can you conclude that the mean speed of the new chips is greater than that of the old chips?
Mathematics
1 answer:
Murljashka [212]2 years ago
4 0

Answer:

Under 99% confidence level we can say that mean speed of the new chips is greater than that of the old chips

Step-by-step explanation:

H_{0}: Mean speed of the new chip is the same as the old chip

H_{a}: Mean speed of the new chip is greater than the old chip

to calculate the z-statistic we can use the formula:

z=\frac{M_{n}-M_{o}}{\sqrt{\frac{s_{n}^2}{N_{n}} +\frac{s_{o}^2}{N_{o}} } }

if we put the numbers then

z=\frac{495.6-481.2}{\sqrt{\frac{19.4^2}{65}} +\frac{14.3^2}{65} } } =4.8171

The probability of this z-statistic is < 0.001 Therefore Under 99% confidence level we can say that mean speed of the new chips is greater than that of the old chips

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d=\sqrt{(11--1)^2+(4--1)^2}&#10;\\=\sqrt{(11+1)^2+(4+1)^2}=\sqrt{12^2+5^2}=\sqrt{144+25}=\sqrt{169}=13

Our side lengths, from least to greatest, are 5, 12 and 13.

To be similar but not congruent, the side lengths must have the same ratio between corresponding sides but not be the same length.  10, 24 and 26 are all 2x the original side lengths, so this works.
5 0
2 years ago
What additional information could you use to show that ΔSTU ≅ ΔVTU using SAS? Check all that apply.
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2 years ago
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8090 [49]

Answer:

Step-by-step explanation:

<u><em>Step(i):-</em></u>

<em>Given first random sample size n₁ = 500</em>

Given  Roper survey reported that 65 out of 500 women ages 18-29 said that they had the most say when purchasing a computer.

<em>First sample proportion </em>

<em>                              </em>p^{-} _{1} = \frac{65}{500} = 0.13

<em>Given second sample size n₂ = 700</em>

<em>Given a sample of 700 men (unrelated to the women) ages 18-29 found that 133 men said that they had the most say when purchasing a computer.</em>

<em>second sample proportion </em>

<em>                              </em>p^{-} _{2} = \frac{133}{700} = 0.19

<em>Level of significance = α = 0.05</em>

<em>critical value = 1.96</em>

<u><em>Step(ii)</em></u><em>:-</em>

<em>Null hypothesis : H₀: There  is no significance difference between these proportions</em>

<em>Alternative Hypothesis :H₁: There  is significance difference between these proportions</em>

<em>Test statistic </em>

<em></em>Z = \frac{p_{1} ^{-}-p^{-} _{2}  }{\sqrt{PQ(\frac{1}{n_{1} } +\frac{1}{n_{2} } )} }<em></em>

<em>where </em>

<em>         </em>P = \frac{n_{1} p^{-} _{1}+n_{2} p^{-} _{2}  }{n_{1}+ n_{2}  } = \frac{500 X 0.13+700 X0.19  }{500 + 700 } = 0.165<em></em>

<em>        Q = 1 - P = 1 - 0.165 = 0.835</em>

<em></em>Z = \frac{0.13-0.19  }{\sqrt{0.165 X0.835(\frac{1}{500 } +\frac{1}{700 } )} }<em></em>

<em>Z =  -2.76</em>

<em>|Z| = |-2.76| = 2.76 > 1.96 at 0.05 level of significance</em>

<em>Null hypothesis is rejected at 0.05 level of significance</em>

<em>Alternative hypothesis is accepted at 0.05 level of significance</em>

<u><em>Conclusion:</em></u><em>-</em>

<em>There is there is a difference between these proportions at α = 0.05</em>

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2 years ago
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