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2 years ago

I need help plz real onez plz

Mathematics

2 answers:

melamori03 [73]2 years ago

6 0

The mean is 10.
To find the mean you add up all the numbers and divide by how many sets of numbers there were.
8+10+11+8+13=50
50/5=10.

den301095 [7]2 years ago

6 0

The correct anwser would be 10

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Answer:

Neither Troy nor Keith is correct.

Step-by-step explanation:

Neither Troy nor Keith is correct.

The value 0.9 is the same as the value 0.90.

The zeros after a decimal point are always insignificant unless they are followed by a number different from zero.

0.90 = 0.900 = 0.9000 = 0.90000 = ... = 0.9000000000

and all these can be easily written as 0.9, as the zeros after the decimal point do not change the value.

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And we are 95% confident that the true difference means are between $1.422 \leq \mu_1 -\mu_2 \leq 2.778$

Step-by-step explanation:

We know the following info:

$\bar X_1 = 21.9$ sample mean for group 1

$\bar X_2 = 19.8$ sample mean for group 2

$s_1 = 3.4$ sample standard deviation for group 1

$s_2 = 3.5$ sample standard deviation for group 2

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$n_2 = 200$ sample size group 2

We want to find a confidence interval for the difference of means and the correct formula to do this is:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2} \sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}$

Now we just need to find the critical value. The confidence level is 0.95 then the significance is $1-0.95 =0.05$ and $\alpha/2 =0.025$ . The degrees of freedom are given by:

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The critical value for this case would be : $t_{\alpha/2}=1.966$

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$(21.9-19.8) -1.966\sqrt{\frac{3.4^2}{200} +\frac{3.5^2}{200}}= 1.422$

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And we are 95% confident that the true difference means are between $1.422 \leq \mu_1 -\mu_2 \leq 2.778$

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