Please help me explain and answer this

A newspaper reports the median house price is $250,000. From that standard, tell what you know about housing prices.

Alik [6]

You know that the average house price is $250,000 dollars
Median means medium

8 0

3 years ago

A sociologist was investigating the ages of grandparents of high school students. From a random sample of 10 high school student

umka2103 [35]

Answer:

$(69.8-71.6) -2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= -8.908$

$(69.8-71.6) +2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= 5.308$

So then the confidence interval for the difference of means is given by:

$-8.908 \leq \mu_M -\mu_F \leq 5.308$

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".

$\bar X_{M}= 69.8$ represent the mean for the age of grandmother

$\bar X_{F}= 71.6$ represent the mean for the age of grandfather

$s_{M}= 8.38$ represent the sample deviation for the age of grandmother

$s_{F}= 6.65$ represent the sample deviation for the age of grandfather

$n_M = n_F= 10$

Solution to the problem

For this case the confidence interval is given by:

$(\bar X_{M} -\bar X_F) \pm t_{\alpha/2} \sqrt{\frac{s^2_M}{n_M} +\frac{s^2_F}{n_F}}$

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n_M +n_F-2=10+10-2=18$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,18)".And we see that $t_{\alpha/2}=2.101$

And replacing we got:

$(69.8-71.6) -2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= -8.908$

$(69.8-71.6) +2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= 5.308$

So then the confidence interval for the difference of means is given by:

$-8.908 \leq \mu_M -\mu_F \leq 5.308$

5 0

3 years ago

Gcf of 56f^3 g^2 and 70fg^3

yulyashka [42]

<span>56f^3 g^2 = 7fg^2(8f^2)
and
70fg^3 = 7fg^2(10g)

</span>Gcf of 56f^3 g^2 and 70fg^3 = 7fg^2

7 0

3 years ago

R these correct? Miles per hour transferred to kilometers per hour, show ur work and explain, thanks

Arlecino [84]

They are correct (if rounded up)

Newberry park: 16.0934km

Thousand Oaks: 20.9215km

Los Angeles: 88.5139km

4 0

3 years ago

Which of the following sums would be under the radical symbol to find the distance between the points (7, -1) and (-8, -9)?

fomenos

The "distance formula" an extension of the Pythagorean Theorem is:

d^2=(x2-x1)^2+(y2-y1)^2 in this case:

d^2=(-8-7)^2+(-9--1)^2

d^2=(-15)^2+(-8)^2

d^2=15^2+8^2

d=√(15^2+8^2)

So the answer is 15^2+8^2

5 0

3 years ago