Mathematics, 08.07.2020 02:01 dward5823

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 a =0.01, what is the probability of a type II error?b) What is the required sample size to satisfy and the type II error probability of b(62) = 0.1?

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Answer from: GbTaco

a. the probability of a type II error is 0.5319

b. the required sample size to satisfy and the type II error probability is 59.4441

Step-by-step explanation:

From the information given; we have:

sample size n = 25

Population standard deviation $\sigma$ = 4

true average lifetime = Sample Mean $\bar X$ = 62

We can state our null hypothesis and alternative hypothesis as follows:

Null hypothesis:

$\mathbf{H_o : \mu = 60}$

Alternative hypothesis

$\mathbf{H_1 : \mu \neq 60}$

Where ;

∝ = 0.01

From the standard normal tables at critical value ∝ = 0.01 ; the level of significance is -2.575 lower limit and 2.575 upper limit

The z statistics for the lower limit is:

$lower \ limit = \dfrac{\bar X - \mu }{\dfrac{\sigma}{\sqrt {n}}}$

$-2.575= \dfrac{\bar x - 60 }{\dfrac{4}{\sqrt 25}}}$

$-2.575= \dfrac{\bar x - 60 }{0.8}}}$

$-2.575*0.8= {\bar x - 60 }{}}}$

$-2.06= {\bar x - 60 }{}}}$

$\bar x = 60-2.06$

$\bar x = 57.94$

The z statistics for the upper limit is:

$lower \ limit = \dfrac{\bar X - \mu }{\dfrac{\sigma}{\sqrt {n}}}$

$2.575= \dfrac{\bar x - 60 }{\dfrac{4}{\sqrt 25}}}$

$2.575= \dfrac{\bar x - 60 }{0.8}}}$

$2.575*0.8= {\bar x - 60 }{}}}$

$2.06= {\bar x - 60 }{}}}$

$\bar x = 60-(-2.06)$

$\bar x = 60+2.06$

$\bar x = 62.06$

Thus; the probability of a type II error is determined as follows:

β = P ( $57.94 < \bar x < 62.06$ )

$= P ( \dfrac{57.94 -62 }{\dfrac{4}{\sqrt{25}}}$

$= P ( \dfrac{-4.06 }{0.8}}$

= P ( -5.08 < Z < 0.08 )

= P ( Z < 0.08) - P ( Z < - 5.08)

Using Excel Function: [ (=NORMDIST (0.08)) - (=NORMDIST(-5.08)) ] ; we have:

= 0.531881372 - 0.00000001887

= 0.531881183

≅ 0.5319

What is the required sample size to satisfy and the type II error probability of b(62) = 0.1

Recall that:

The critical value of ∝ = 2.575 ( i. e $Z_{1 - \alpha/2 } = 2.575$ )

Now ;

the critical value of β is :

$Z _{1- \beta} = 1.28$

The required sample size to satisfy and the type II error probability is therefore determined as :

$n = [\dfrac{(Z_{1 - \alpha/2} + Z_{1 - \beta} ) \sigma }{\delta}]^2$

$n = [\dfrac{(2.575+1.28 ) 4 }{2}]^2$

$n = [\dfrac{(3.855 ) 4 }{2}]^2$

$n = [\dfrac{(15.42 ) }{2}]^2$

n = 7.71 ²

n= 59.4441

Thus; the required sample size to satisfy and the type II error probability is 59.4441

Answer from: Quest

chad should not have included the values (1, 2) and (3, 6) because that would mean he only had half a healing card.

Answer from: Quest

answer:

given the linear inequality : $x-2y\geq -12$

add 2y to both sides we get;

$x-2y+2y\geq -12+2y$

simplify:

$x\geq -12+2y$

add 12 to both sides of an equation;

$x+12\geq -12+2y+12$

simplify:

$x+12\geq 2y$

divide both sides by 2 we get;

$\frac{x}{2} + 6 \geq y$

$y\leq \frac{x}{2} + 6$

the related equation for the inequality equation [1] is;

$y =\frac{x}{2} +6$

y-intercepts defined as the graph crosses the y-axis.

substitute x =0 in [2] to solve for y;

$y =\frac{0}{2} +6$

y = 6

x-intercepts defined as the graph crosses the x-axis.

substitute value y = 0 to solve for x;

$0 =\frac{x}{2} +6$

subtract 6 from both sides we get

$0-6 =\frac{x}{2} +6-6$

$-6 =\frac{x}{2}$

multiply both sides by 2 we get;

$-6 \times 2=x$

x = -12

now, plot these point (0, 6) and (-12, 0) on graph as shown below and shade the graph below the solid line.

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