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Olenka [21]
3 years ago
14

A college library has five copies of a certain text on reserve. Three copies (1, 2, and 3) are first printings, and the remainin

g two (4, and 5) are second printings. A student examines these books in random order, stopping only when a second printing has been selected. One possible outcome is 4, and another is 214. (a) List the outcomes in S. (b) Let A denote the event that exactly one book must be examined. What outcomes are in A
Mathematics
1 answer:
dezoksy [38]3 years ago
6 0

Answer:

S = { (4), (5), (1,4), (1,5), (2,4), (2,5), (3,4), (3,5), (1,2,3,4), (1,2,3,5), (2,1,3,4) (2,1,3,5), (3,1,2,4), (3,1,2,5) }

A ={(4), (5)}

Step-by-step explanation:

Given that:

Among the three copies, (1,2,3) are the first printings, and (4,5) are the second printings.

A student who examines these books in random order stops when a second printing has been selected.

Thus, we can compute the sample space associated with these experiments as:

S = { (4), (5), (1,4), (1,5), (2,4), (2,5), (3,4), (3,5), (1,2,3,4), (1,2,3,5), (2,1,3,4) (2,1,3,5), (3,1,2,4), (3,1,2,5) }

Suppose A represents the event that we must examine exactly one book.

Then the outcomes of A are:

A ={(4), (5)}

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Answer:

Rule

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

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Null hypothesis: H0 = 0.60

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Since z at 0.10 significance level is between -1.645 and +1.645 and the z score for the test (z = 1.35) falls with the region bounded by Z at 0.1 significance level. And also the two-tailed hypothesis P-value is 0.177 which is greater than 0.1. Then we can conclude that we don't have enough evidence to FAIL or reject the null hypothesis, and we can say that at 0.10 significance level the null hypothesis is valid.

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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