Suppose the average lifetime of a certain type of car battery is known to be 60 months. Consider conducting a two-sided test on it based on a sample of size 25 from a normal distribution with a population standard deviation of 4 months.a) If the true average lifetime is 62 months and ďˇ =0.01, what is the probability of a type II error
Answers
Answer:
0.53187998
Step-by-step explanation:
We are given;
Population mean; μ = 60
Sample size; n = 25
Population standard deviation; σ = 4
Significance level; α = 0.01
Let's define the hypothesis;
Null hypothesis;H0: μ = 60
Alternative hypothesis;Ha: μ ≠ 60
Z-score for a significance level of 0.01 from tables is z = 2.575
Since it's a 2 sided test, then;
Z-score at lower limit = -2.575
Z-score at upper limit = 2.575
We know that z-score formula is;
z = (x¯ - μ)/(σ/√n)
At lower limit;
-2.575 = (x¯ - 60)/(4/√25)
-2.575 × 0.8 = x¯ - 60
-2.06 = x¯ - 60
x¯ = 60 - 2.06
x¯ = 57.94
At upper-limit;
2.575 = (x¯ - 60)/(4/√25)
2.575 × 0.8 = x¯ - 60
2.06 = x¯ - 60
x¯ = 60 + 2.06
x¯ = 62.06
the probability of a type II error is expressed as:
β = P(57.94 < x¯ < 62.06)
If the true average lifetime is 62 months, then;
β = P((57.94 - 62)/(4/√25)) < z < (62.06 - 62)/(4/√25)
β = P(-5.08 < z < 0.08)
β = P(Z < 0.08) - P(Z < -5.08)
From z-distribution table attached, we have; P(Z < 0.08) = 0.53188
Also, from z-distribution calculator online, we have;
P(Z < -5.08) = 0.0000000189
Thus;
β = 0.53188 - 0.0000000189
β ≈ 0.53187998
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Answers
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you add the x and then you subtract the 1
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What is an Equation?An equation is a mathematical statement containing two expressions on either sides which are connected with an equal to sign.
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~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Answer:
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Answer:
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Step-by-step explanation:
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The question is incomplete. Here is the complete question.
Uninhibited growth can be modeled by exponential functions other than [tex]A(t)=A_{0}e^{kt}[/tex]. for example, if an initial population P₀ requires n units of time to triple, then the function [tex]P(t)=P_{0}(3)^{\frac{t}{n} }[/tex] models the size of the population at time t. An insect population grows exponentially. Complete the parts a through d below.
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b) What will the population be in 47 days?
c) When wil the population reach 750?
d) Express the model from part (a) in the form [tex]A(t)=A_{0}e^{kt}[/tex].
Answer: a) [tex]P(t)=50(3)^{\frac{t}{30} }[/tex]
b) P(t) = 280 insects
c) t = 74 days
d) [tex]A(t)=50e^{0.037t}[/tex]
Step-by-step explanation:
a) n is time necessary to triple the population of insects, i.e., n = 30 and P₀ = 50. So, Exponential equation for growth is
[tex]P(t)=50(3)^{\frac{t}{30} }[/tex]
b) In t = 47 days:
[tex]P(t)=50(3)^{\frac{t}{30} }[/tex]
[tex]P(47)=50(3)^{\frac{47}{30} }[/tex]
[tex]P(47)=50(3)^{1.567}[/tex]
P(47) = 280
In 47 days, population of insects will be 280
c) P(t) = 750
[tex]750=50(3)^{\frac{t}{30} }[/tex]
[tex]\frac{750}{50}=(3)^{\frac{t}{30} }[/tex]
[tex](3)^{\frac{t}{n} }=15[/tex]
Using the property Power Rule of logarithm:
[tex]log(3)^{\frac{t}{30} }=log15[/tex]
[tex]\frac{t}{30}log(3)=log15[/tex]
[tex]t=\frac{log15}{log3} .30[/tex]
t = 74
To reach a population of 750 insects, it will take 74 days
d) To express the population growth into the described form, determine the constant k, using the following:
A(t) = 3A₀ and t = 30
[tex]A(t)=A_{0}e^{kt}[/tex]
[tex]3A_{0}=A_{0}e^{30k}[/tex]
[tex]3=e^{30k}[/tex]
Use Power Rule again:
[tex]ln3=ln(e^{30k})[/tex]
[tex]ln3=30k[/tex]
[tex]k=\frac{ln3}{30}[/tex]
k = 0.037
Equation for exponential growth will be:
[tex]A(t)=50e^{0.037t}[/tex]
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Answers
y=-3 cos(2x) is the equation has a maximum at (4π , 3) and a minimum at (4π , -3)
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The sine function's parameters, a, b, c, and d, must be adjusted using the sliders.
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Answer: y=-3sin(2x)
Step-by-step explanation:
Answered correctly on edge
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Answer:
D is the correct answer
Answer:
B
I Double CHecked For WHoever Is In My Comments and I'm Correct
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Step-by-step explanation:
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Answer:
x = 6 +3√ 10 , 6 − 3 √ 10
decimal form
X= 15.48683298...