Answer:
The t value for 99% CI for 21 df is 2.831.
The critical value that should be used in constructing the confidence interval is (64.593, 86.407).
Step-by-step explanation:
Now the sample size is less than 30 and also population standard deviation is not known.
Then we will use t distribution to find CI
t value for 99% CI for 21 df is TINV(0.01,21)=2.831
The margin of error is ![E=t\times\frac{s}{\sqrt{n}}\\\\=2.831\times\frac{18.07}{\sqrt{22}}\\\\=10.907](https://tex.z-dn.net/?f=E%3Dt%5Ctimes%5Cfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D%5C%5C%5C%5C%3D2.831%5Ctimes%5Cfrac%7B18.07%7D%7B%5Csqrt%7B22%7D%7D%5C%5C%5C%5C%3D10.907)
Hence CI is![CI=\overline{x} \pm E\\\\ =75.50 \pm 10.907\\\\=(64.593,86.407 )](https://tex.z-dn.net/?f=CI%3D%5Coverline%7Bx%7D%20%5Cpm%20E%5C%5C%5C%5C%20%3D75.50%20%5Cpm%2010.907%5C%5C%5C%5C%3D%2864.593%2C86.407%20%29)
Answer:
Give me a second I know this one.
Step-by-step explanation:
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Multiply 30 and 12 and your answer will be 360
Answer:
I say 20% but I may be wrong
Step-by-step explanation: