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8_murik_8 [283]

3 years ago

15

The manager of a computer retails store is concerned that his suppliers have been giving him laptop computers with lower than av

erage quality. His research shows that replacement times for the model laptop of concern are normally distributed with a mean of 4.2 years and a standard deviation of 0.6 years. He then randomly selects records on 38 laptops sold in the past and finds that the mean replacement time is 3.9 years. Assuming that the laptop replacement times have a mean of 4.2 years and a standard deviation of 0.6 years, find the probability that 38 randomly selected laptops will have a mean replacement time of 3.9 years or less.

1 answer:

yanalaym [24]3 years ago

5 0

Answer: 0.0010

Step-by-step explanation:

Given the following :

Population Mean(m) = 4.2 years

Sample mean (s) = 3.9

Standard deviation (sd) = 0.6

Number of samples (n) = 38

Calculate the test statistic (z) :

(sample mean - population mean) / (sd / √n)

Z = (3.9 - 4.2) / (0.6 / √38)

Z = (- 0.3) / (0.6 / 6.1644140)

Z = -0.3 / 0.0973328

Z = - 3.0822086

Z = - 3.08

From the z table :

P(Z ≤ - 3.08) = 0.0010

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xxTIMURxx [149]

Answer:

$t=\frac{669-634}{\frac{732}{\sqrt{1700}}}=1.971$

$p_v =2*P(t_{(1699)}>1.971)=0.049$

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from 634 at 10% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=669$ represent the sample mean

$s=732$ represent the sample standard deviation

$n=1700$ sample size

$\mu_o =68$ represent the value that we want to test

$\alpha=0.$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is different from 634, the system of hypothesis would be:

Null hypothesis: $\mu = 634$

Alternative hypothesis: $\mu \neq 634$

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{669-634}{\frac{732}{\sqrt{1700}}}=1.971$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=1700-1=1699$

Since is a two sided test the p value would be:

$p_v =2*P(t_{(1699)}>1.971)=0.049$

Conclusion

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from 634 at 10% of signficance.

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3 years ago

Need help please explain why its the answer and show work<br><br> Look at picture

bonufazy [111]

Answer:

14.a

15.c

16.d

Step-by-step explanation:

i choose these because they are in the geometry book

6 0

3 years ago

Read 2 more answers

A 1000-liter (L) tank contains 500 L of water with a salt concentration of 10 g/L. Water with a salt concentration of 50 g/L flo

djverab [1.8K]

Answer:

a) $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) $31690.7 g/L$

Step-by-step explanation:

By definition, we have that the change rate of salt in the tank is $\frac{dy}{dt}=R_{i}-R_{o}$ , where $R_{i}$ is the rate of salt entering and $R_{o}$ is the rate of salt going outside.

Then we have, $R_{i}=80\frac{L}{min}*50\frac{g}{L}=4000\frac{g}{min}$ , and

$R_{o}=40\frac{L}{min}*\frac{y}{500} \frac{g}{L}=\frac{2y}{25}\frac{g}{min}$

So we obtain. $\frac{dy}{dt}=4000-\frac{2y}{25}$ , then

$\frac{dy}{dt}+\frac{2y}{25}=4000$ , and using the integrating factor $e^{\int {\frac{2}{25}} \, dt=e^{\frac{2t}{25}$ , therefore $(\frac{dy }{dt}+\frac{2y}{25}}=4000)e^{\frac{2t}{25}$ , we get $\frac{d}{dt}(y*e^{\frac{2t}{25}})= 4000 e^{\frac{2t}{25}$ , after integrating both sides $y*e^{\frac{2t}{25}}= 50000 e^{\frac{2t}{25}}+C$ , therefore $y(t)= 50000 +Ce^{\frac{-2t}{25}}$ , to find $C$ we know that the tank initially contains a salt concentration of 10 g/L, that means the initial conditions $y(0)=10$ , so $10= 50000+Ce^{\frac{-0*2}{25}}$

$10=50000+C\\C=10-50000=-49990$

Finally we can write an expression for the amount of salt in the tank at any time t, it is $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) The tank will overflow due Rin>Rout, at a rate of $80 L/min-40L/min=40L/min$ , due we have 500 L to overflow $\frac{500L}{40L/min} =\frac{25}{2} min=t$ , so we can evualuate the expression of a) $y(25/2)=50000-49990e^{\frac{-2}{25}\frac{25}{2}}=50000-49990e^{-1}=31690.7$ , is the salt concentration when the tank overflows

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3 years ago

15.60 to 11.70 percent of change

Marizza181 [45]

15.6 - 11.7 = 3.9

3.9 ÷ 15.6 = .25

25% of change

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3 years ago

A magazine provided results from a poll of 1000 adults who were asked to identify their favorite pie. Among the 1000 respondent

Gwar [14]

Answer:

The answer to the given problem is given below.

Step-by-step explanation:

What values do p, q,n,E and p represents?

The value of p is the sample proportion.

The value of q is found from evaluating 1− p.

The value of n is the sample size.

The value of E is the margin of error.

The value of p is the population proportion.

If the confidence level is 90%, what is the value of α?

α = 1- 0.90

α = 0.10

8 0

3 years ago