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GenaCL600 [577]
2 years ago
14

Here's the question sorry​

Mathematics
1 answer:
Lunna [17]2 years ago
6 0

L =  total number of slices for Luke

K = total number of slices for Kira

A = total number of slices for Ali

\stackrel{\textit{two thirds of all slices}}{\cfrac{2}{3}L}=4\implies 2L=12\implies L=\cfrac{12}{2}\implies L=6 \\\\\\ \stackrel{\textit{two thirds of all slices}}{\cfrac{2}{3}K}=6\implies 2K=18\implies K=\cfrac{18}{2}\implies K=9 \\\\\\ \stackrel{\textit{two thirds of all slices}}{\cfrac{2}{3}A}=8\implies 2A=24\implies A=\cfrac{24}{2}\implies A=12

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Answer:

Rule

If;

P-value > significance level --- accept Null hypothesis

P-value < significance level --- reject Null hypothesis

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Question; A sample of 1100 computer chips revealed that 62% of the chips fail in the first 1000 hours of their use. The company's promotional literature states that 60% of the chips fail in the first 1000 hours of their use. The quality control manager wants to test the claim that the actual percentage that fail is different from the stated percentage. Determine the decision rule for rejecting the null hypothesis, H0, at the 0.10 level.

Step-by-step explanation:

Given;

n=1100 represent the random sample taken

Null hypothesis: H0 = 0.60

Alternative hypothesis: Ha <> 0.62

Test statistic z score can be calculated with the formula below;

z = (p^−po)/√{po(1−po)/n}

Where,

z= Test statistics

n = Sample size = 1100

po = Null hypothesized value = 0.60

p^ = Observed proportion = 0.62

Substituting the values we have

z = (0.62-0.60)/√{0.60(1-0.60)/1100}

z = 1.354

z = 1.35

To determine the p value (test statistic) at 0.01 significance level, using a two tailed hypothesis.

P value = P(Z<-1.35) + P(Z>1.35) = 0.0885 + 0.0885= 0.177

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