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otez555 [7]
3 years ago
10

Consider the following hypothesis test. H0: μ ≥ 55 Ha: μ < 55 A sample of 36 is used. Identify the p-value and state your con

clusion for each of the following sample results. Use α = 0.01.
(a) x = 54 and s = 5.3 Find the value of the test statistic. (Round your answer to three decimal places.)
Find the p-value. (Round your answer to four decimal places.)
State your conclusion.
(b) x = 53 and s = 4.6
Find the value of the test statistic. (Round your answer to three decimal places.)
Find the p-value. (Round your answer to four decimal places.)
p-value =
(c) x = 56 and s = 5.0
Find the value of the test statistic.
Find the p-value. (Round your answer to four decimal places.)
p-value =
State your conclusion.
Mathematics
1 answer:
Ket [755]3 years ago
6 0

Answer:

Step-by-step explanation:

Given that:

H_o: \mu \ge 55 \ \\ \\ H_1 : \mu < 55

(a) For x = 54 and s = 5.3

The test statistics can be computed as:

Z = \dfrac{\overline x - \mu}{\dfrac{s}{\sqrt{n}} }

Z = \dfrac{54-55}{\dfrac{5.3}{\sqrt{36}} }

Z = \dfrac{-1}{\dfrac{5.3}{6} }

Z = -1.132

degree of freedom = n - 1

= 36 - 1

= 35

Using the Excel Formula:

P-Value = T.DIST(-1.132,35,1) = 0.1326

Decision: p-value is greater than significance level; do not reject \mathbf{H_o}

Conclusion: \  There  \ is \  insufficient \  evidence  \  to \  conclude \  that \ \mu < 55

b

For x = 53 and s = 4.6

The test statistics can be computed as:

Z = \dfrac{\overline x - \mu}{\dfrac{s}{\sqrt{n}} }

Z = \dfrac{53-55}{\dfrac{4.6}{\sqrt{36}} }

Z = \dfrac{-2}{\dfrac{4.6}{6} }

Z = -2.6087

degree of freedom = n - 1

= 36 - 1

= 35

Using the Excel Formula:

P-Value = T.DIST(-2.6087,35,1) =0.0066

Decision: p-value is < significance level; we reject the null hypothesis.

Conclusion: \  There  \ is \  sufficient \  evidence  \  to \  conclude \  that \mu < 55

c)

For x = 56 and s = 5.0

The test statistics can be computed as:

Z = \dfrac{\overline x - \mu}{\dfrac{s}{\sqrt{n}} }

Z = \dfrac{56-55}{\dfrac{5.0}{\sqrt{36}} }

Z = \dfrac{56-55}{\dfrac{5.0}{\sqrt{36}} }

Z = 1.2

degree of freedom = n - 1

= 36 - 1

= 35

Using the Excel Formula:

P-Value = T.DIST(1.2,35,1) = 0.88009

Decision: p-value is greater than significance level; do not reject \mathbf{H_o}

Conclusion: \  There  \ is \  insufficient \  evidence  \  to \  conclude \  that \ \mu < 55

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2 years ago
Pollution of the rivers in the United States has been a problem for many years. Consider the following events: a: the river is p
vampirchik [111]

Answer:

The answer is below

Step-by-step explanation:

P(A') = 1 - P(A) = 1 - 0.3 = 0.7

P(B'/A) = 1 - P(B/A) = 1 - 0.75 - 0.25

P(B/A') = 0.2, P(B'/A') = 1 - 0.2 = 0/8

P(C|A∩B) = 0.20,

P(C|A'∩B) = 0.15,

P(C|A∩B') = 0.80, and

P(C|A'∩B') = 0.90.

A) From conditional probability;

P(B/A) = P(B ∩ A) / P(A)

P(B∩A) = P(B/A) × P(A) = 0.75 × 0.3 = 0.225

P(C/A∩B) = P(A∩B∩C)/P(A∩B)

P(A∩B∩C) =  P(C/A∩B) × P(A∩B)

P(A∩B∩C) = 0.2 ×0.225 = 0.045

B) P(A∩B∩C) = P(A) × P(B/A) × P(C/A∩B)

P(B'∩C) = P(A∩(B'∩C)) + P((A'∩B')∩C)

P(B'∩C) = P(A) × P(B'/A) × P(C/A∩B') + P(A') × P(B'/A') × P(C/A'∩B')

P(B'∩C) = 0.3 × 0.25 × 0.8 + (0.7 × 0.8 × 0.9) = 0.06 + 0.504 = 0.564

C) P(C) =  P(A∩B∩C) +  P(A'∩B∩C)  +  P(A∩B'∩C) +  P(A'∩B'∩C)

P(A'∩B∩C) = P(A') × P(B/A') × P(C/A'∩B) = 0.7 × 0.2 × 0.15 = 0.021

P(A∩B'∩C) = P(A) × P(B'/A) × P(C/A∩B') =  0.3 × 0.25 × 0.8 = 0.06

P(A'∩B'∩C) = P(A') × P(B'/A') × P(C/A'∩B') = 0.7 × 0.8 × 0.9 = 0.504

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D) P(A/B'∩C) = P(A∩B'∩C)/P(B'∩C)

P(A/B'∩C) = 0.06 / 0.564 = 0.106

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