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lbvjy [14]
3 years ago
13

Which one is it?? because I really I dont under​

Mathematics
1 answer:
Rus_ich [418]3 years ago
8 0

Hello from MrBillDoesMath!

Answer:

Choice D.

Discussion:

Substitute the values (5,7), (8,13) into the equations and see what we get..

A.  (5,7). Does 3(5) -7 = 8. Yes  

    (8,13). Does 3(8)-13 = 8 ? No. Points  (8.13) does NOT lie on the line

B.  (5.7). Does 7 = 5 + 2. Yes.

    (8,13). Does 13 = 8 +2 ? No. Points  (8.13) does NOT lie on the line

C.  (5,7). Does 7 -5 = 5 ? . No

    (8,13). Does 13-8 = 5? Yes

D.  (5.7). Does 7 = 2(5) - 3  ?  Yes.

    (8,13). Does 13 = 2(8) - 3?  Yes

Only choice D has both points lying on the line.

Thank you,

MrB

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A company wants to find out if the average response time to a request differs across its two servers. Say µ1 is the true mean/ex
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a) The null and alternative hypothesis are:

H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2\neq 0

Test statistic t=0.88

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This relate to the previous conclusion as there is not enough evidence to support that there is significant difference between the response time, as the hypothesis that there is no difference is not an unusual value for the true difference.

Step-by-step explanation:

This is a hypothesis test for the difference between populations means.

The claim is that there is significant difference in the time response for the two servers.

Then, the null and alternative hypothesis are:

H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2\neq 0

The significance level is 0.05.

The sample 1, of size n1=196 has a mean of 12.5 and a standard deviation of 3.

The sample 2, of size n2=225 has a mean of 12.2 and a standard deviation of 4.

The difference between sample means is Md=0.3.

M_d=M_1-M_2=12.5-12.2=0.3

The estimated standard error of the difference between means is computed using the formula:

s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{3^2}{196}+\dfrac{4^2}{225}}\\\\\\s_{M_d}=\sqrt{0.046+0.071}=\sqrt{0.117}=0.342

Then, we can calculate the t-statistic as:

t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{0.3-0}{0.342}=\dfrac{0.3}{0.342}=0.88

The degrees of freedom for this test are:

df=n_1+n_2-1=196+225-2=419

This test is a two-tailed test, with 419 degrees of freedom and t=0.88, so the P-value for this test is calculated as (using a t-table):

\text{P-value}=2\cdot P(t>0.88)=0.381

As the P-value (0.381) is greater than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that there is significant difference in the time response for the two servers.

<u>Confidence interval </u>

We have to calculate a 95% confidence interval for the difference between means.

The sample 1, of size n1=196 has a mean of 12.5 and a standard deviation of 3.

The sample 2, of size n2=225 has a mean of 12.2 and a standard deviation of 4.

The difference between sample means is Md=0.3.

The estimated standard error of the difference is s_Md=0.342.

The critical t-value for a 95% confidence interval and 419 degrees of freedom is t=1.966.

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MOE=t\cdot s_{M_d}=1.966 \cdot 0.342=0.672

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LL=M_d-t \cdot s_{M_d} = 0.3-0.672=-0.372\\\\UL=M_d+t \cdot s_{M_d} = 0.3+0.672=0.972

The 95% confidence interval for the difference in the two servers population expectations is (-0.372, 0.972).

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