Answer:
Note: The full question is attached as picture below
a) Hо : p = 0.71
Ha : p ≠ 0.71
<em>p </em>= x / n
<em>p </em>= 91/110
<em>p </em>= 0.83.
1 - Pо = 1 - 0.71 = 0.29.
b) Test statistic = z
= <em>p </em>- Pо / [√Pо * (1 - Pо ) / n]
= 0.83 - 0.71 / [√(0.71 * 0.29) / 110]
= 0.12 / 0.043265
= 2.77360453
Test statistic = 2.77
c) P-value
P(z > 2.77) = 2 * [1 - P(z < 2.77)] = 2 * 0.0028
P-value = 0.0056
∝ = 0.01
P-value < ∝
Reject the null hypothesis. There is sufficient evidence to support the researchers claim at the 1% significance level.
Answer:
C
Step-by-step explanation:
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I will give an instance to answer this. Suppose that we have 1,000 days. And it was given above that machine A has 10 % chances to malfunctioned and B has a 7%. So machine A has 100 days that malfunctioned while machine B has 70 days. And 7 days that both machines malfunctioned (10%x7%). Next is to add 100 days from machine A and 70 days from machine B and subtract with the 7 days where both of them malfunctioned. So the result is 163 days out of 1000 that we expect an incidence of failure. So the chance of machine B to malfunctioned is 70 days. 70/163 is 42.93%
