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3 years ago

13

Isaac won 122 pieces of gum playing hoops at the county fair. At school he gave four to every student in his math class. He only

has 2 remaining. How many students are in his class?

2 answers:

Lera25 [3.4K]3 years ago

5 0

122 divid by 4 = 30.5

Svetach [21]3 years ago

3 0

There are 30 students in his math class.

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A manufacturer produces crankshafts for an automobile engine. the crankshafts wear after 100,000 miles (0.0001 inch) is of inter

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Part A:

Significant level:

<span>α = 0.05

Null and alternative hypothesis:

</span><span>h0 : μ = 3 vs h1: μ ≠ 3

Test statistics:

$z= \frac{\bar{x}-\mu}{\sigma/\sqrt{n}} \\ \\ = \frac{2.78-3}{0.9/\sqrt{15}} \\ \\ = \frac{-0.22}{0.2324} =-0.9467$

P-value:

P(-0.9467) = 0.1719

Since the test is a two-tailed test, p-value = 2(0.1719) = 0.3438

Conclusion:

Since the p-value is greater than the significant level, we fail to reject the null hypothesis and conclude that there is no sufficient evidence that the true mean is different from 3.

Part B:

The power of the test is given by:

$\beta=\phi\left(Z_{0.025}+ \frac{3-3.25}{0.9/\sqrt{15}}\right) -\phi\left(-Z_{0.025}+ \frac{3-3.25}{0.9/\sqrt{15}}\right) \\ \\ =\phi\left(1.96+ \frac{-0.25}{0.2324} \right)-\phi\left(-1.96+ \frac{-0.25}{0.2324} \right)=\phi(0.8842)-\phi(-3.0358) \\ \\ =0.8117-0.0012=0.8105$

Therefore, the power of the test if </span><span>μ = 3.25 is 0.8105.

Part C:

</span>The <span>sample size that would be required to detect a true mean of 3.75 if we wanted the power to be at least 0.9 is obtained as follows:

$1-0.9=\phi\left(Z_{0.025}+ \frac{3-3.75}{0.9/\sqrt{n}}\right) -\phi\left(-Z_{0.025}+ \frac{3-3.75}{0.9/\sqrt{n}}\right) \\ \\ \Rightarrow0.1=\phi\left(1.96+ \frac{-0.75}{0.9/\sqrt{n}}\right)-\phi\left(-1.96+ \frac{-0.75}{0.9/\sqrt{n}} \right) \\ \\ =\phi\left(1.96+(-3.2415)\right)-\phi\left(1.96+(-3.2415)\right) \\ \\ \Rightarrow\frac{-0.75}{0.9/\sqrt{n}}=-3.2415 \\ \\ \Rightarrow\frac{0.9}{\sqrt{n}}=\frac{-0.75}{-3.2415}=0.2314 \\ \\ \Rightarrow\sqrt{n}=\frac{0.9}{0.2314}=3.8898$

$\Rightarrow n=(3.8898)^2=15.13$

Therefore, the </span>s<span>ample size that would be required to detect a true mean of 3.75 if we wanted the power to be at least 0.9 is 16.</span>

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3 years ago