Answer:
Step-by-step explanation:
Hello!
The variable of interest is
X: the amount a 10-11-year-old spends on a trip to the mall.
Assuming that the variable has a normal distribution, you have to construct a 98% CI for the average of the amount spent by the 10-11 year-olds on one trip to the mall.
For this you have to use a Student-t for one sample:
X[bar] ±
* ![\frac{S}{\sqrt{n} }](https://tex.z-dn.net/?f=%5Cfrac%7BS%7D%7B%5Csqrt%7Bn%7D%20%7D)
n= 6
![t_{n-1;1-\alpha /2}= t_{5; 0.99}= 3.365](https://tex.z-dn.net/?f=t_%7Bn-1%3B1-%5Calpha%20%2F2%7D%3D%20t_%7B5%3B%200.99%7D%3D%203.365)
$18.31, $25.09, $26.96, $26.54, $21.84, $21.46
∑X= 140.20
∑X²= 3333.49
X[bar]= ∑X/n= 140.20/6= 23.37
S²= 1/(n-1)*[∑X²-(∑X)²/n]= 1/5[3333.49-(140.2)²/6]= 11.50
S= 3.39
X[bar] ±
* ![\frac{S}{\sqrt{n} }](https://tex.z-dn.net/?f=%5Cfrac%7BS%7D%7B%5Csqrt%7Bn%7D%20%7D)
[23.37 ± 3.365 *
]
[18.66;29.98]
With a confidence level of 98%, you'd expect that the interval $[18.66;29.98] will include the population mean of the money spent by 10-11 year-olds in one trip to the mall.
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Using the Pythagorean therefrom none of these are correct <span />
I don't have a visual I can provide, but the ratio of the areas is the square of the ratio of the perimeters of a 2D figure.
Intuitively, you can imagine that the triangle is simply being "scaled up" by a factor of 7/3, so the base scales up to 7/3 its original length and the height becomes 7/3 its original height.
Therefore, the new ratio is the square of the ratio of the perimeters, which is 9:49.