Mathematics, 13.12.2019 04:31 lee1677

Arandom sample of size n1equals25, taken from a normal population with a standard deviation sigma1equals6, has a mean x overbar 1equals87. a second random sample of size n2equals33, taken from a different normal population with a standard deviation sigma2equals4, has a mean x overbar 2equals36. find a 96% confidence interval for mu1minusmu2.

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Answer from: SmokeyRN

The 96% confidence interval would be given by $48.156 \leq \mu_1 -\mu_2 \leq 53.844$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X_1 =87$ represent the sample mean 1

$\bar X_2 =36$ represent the sample mean 2

n1=25 represent the sample 1 size

n2=33 represent the sample 2 size

$\sigma_1 =6$ sample standard deviation for sample 1

$\sigma_2 =4$ sample standard deviation for sample 2

$\mu_1 -\mu_2$ parameter of interest.

Solution to the problem

The confidence interval for the difference of means is given by the following formula:

$(\bar X_1 -\bar X_2) \pm z_{\alpha/2}\sqrt{\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2}}$ (1)

The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =87-36=51$

Since the Confidence is 0.96 or 96%, the value of $\alpha=0.04$ and $\alpha/2 =0.02$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.02,0,1)".And we see that $z_{\alpha/2}=2.05$

The standard error is given by the following formula:

$SE=\sqrt{\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2}}$

And replacing we have:

$SE=\sqrt{\frac{6^2}{25}+\frac{4^2}{33}}=1.387$

Confidence interval

Now we have everything in order to replace into formula (1):

$51-2.05\sqrt{\frac{6^2}{25}+\frac{4^2}{33}}=48.156$

$51+2.05\sqrt{\frac{6^2}{25}+\frac{4^2}{33}}=53.844$

So on this case the 96% confidence interval would be given by $48.156 \leq \mu_1 -\mu_2 \leq 53.844$

Answer from: Quest

it's d.

step-by-step explanation:

because: -1(4x + 1y) +2 (1x-3y) distribute each number into the parenthesis.

then you would get: -4x - 1y + 2x -6y combine like terms

and get: 2x -7y

hope my answer has you if not then i'm sorry.

Which expression is the simplest form of -(4x + y) + 2(x – 3y)?

Answer from: Quest

$x=\frac{3}{4}$

step-by-step explanation:

we are given

$-6+2x=6x-9$

since, we have to isolate x on anyone side

so, we can move all x terms on one side

subtract both sides by 2x

$-6+2x-2x=6x-9-2x$

$-6=4x-9$

add both sides by 9

$-6+9=4x-9+9$

$3=4x$

divide both sides by 4

we get

$x=\frac{3}{4}$

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