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Advocard [28]
3 years ago
10

Help me I wasn't here

Mathematics
1 answer:
Anna007 [38]3 years ago
6 0

Answer:

6/8 or 3/4 or 75%

Step-by-step explanation:

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<h2><u>Answer with explanation</u>:</h2>

Let \mu be the population mean.

As per given , we have

H_0:\mu \leq4\\\\ H_a: \mu >4

Since the alternative hypothesis is right-tailed , so the test is a right-tailed test.

Also, population standard deviation is given \sigma=1 , so we perform one-tailed z-test.

Test statistic : z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}

, where \mu = Population mean

\sigma = Population standard deviation

n= sample size

\overline{x} = Sample mean

For n= 18 , \overline{x}=4.50 , \sigma=1 , \mu =4, we have

z=\dfrac{4.5-4}{\dfrac{1}{\sqrt{18}}}\approx2.12

P-value (for right tailed test): P(z>2.12) = 1-P(z≤ 2.12)  [∵ P(Z>z)=1-P(Z≤z)]\

=1- 0.0340=0.9660

Decision : Since P-value(0.9660) > Significance level (0.01), it means we are failed to reject the null hypothesis.

[We reject null hypothesis if p-value is larger than the significance level . ]

Conclusion : We do not have sufficient evidence to show that the goal is not being met at α = .01 .

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3 years ago
House
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I can answer this for u
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