Answer:
![9.9676 - 2.326*0.5904 =8.594](https://tex.z-dn.net/?f=%209.9676%20-%202.326%2A0.5904%20%3D8.594)
![9.9676 + 2.326*0.5904 =11.341](https://tex.z-dn.net/?f=%209.9676%20%2B%202.326%2A0.5904%20%3D11.341)
Step-by-step explanation:
Notation
represent the sample mean
population mean (variable of interest)
s represent the sample standard deviation
n represent the sample size
Solution to the problem
The confidence interval for the mean is given by the following formula:
(1)
For this case the 9% confidence interval is given by:
![8.8104 \leq \mu \leq 11.1248](https://tex.z-dn.net/?f=%208.8104%20%5Cleq%20%5Cmu%20%5Cleq%2011.1248)
We can calculate the mean with the following:
![\bar X = \frac{8.8104 +11.1248}{2}= 9.9676](https://tex.z-dn.net/?f=%5Cbar%20X%20%3D%20%5Cfrac%7B8.8104%20%2B11.1248%7D%7B2%7D%3D%209.9676)
And we can find the margin of error with:
![ME= \frac{11.1248- 8.8104}{2}= 1.1572](https://tex.z-dn.net/?f=%20ME%3D%20%5Cfrac%7B11.1248-%208.8104%7D%7B2%7D%3D%201.1572)
The margin of error for this case is given by:
![ME = t_{\alpha/2}\frac{s}{\sqrt{n}} = t_{\alpha/2} SE](https://tex.z-dn.net/?f=%20ME%20%3D%20t_%7B%5Calpha%2F2%7D%5Cfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D%20%3D%20t_%7B%5Calpha%2F2%7D%20SE)
And we can solve for the standard error:
![SE = \frac{ME}{t_{\alpha/2}}](https://tex.z-dn.net/?f=%20SE%20%3D%20%5Cfrac%7BME%7D%7Bt_%7B%5Calpha%2F2%7D%7D)
The critical value for 95% confidence using the normal standard distribution is approximately 1.96 and replacing we got:
![SE = \frac{1.1572}{1.96}= 0.5904](https://tex.z-dn.net/?f=%20SE%20%3D%20%5Cfrac%7B1.1572%7D%7B1.96%7D%3D%200.5904)
Now for the 98% confidence interval the significance is
and
the critical value would be 2.326 and then the confidence interval would be:
![9.9676 - 2.326*0.5904 =8.594](https://tex.z-dn.net/?f=%209.9676%20-%202.326%2A0.5904%20%3D8.594)
![9.9676 + 2.326*0.5904 =11.341](https://tex.z-dn.net/?f=%209.9676%20%2B%202.326%2A0.5904%20%3D11.341)
C I think................
Answer:
Step-by-step explanation:
1. 6 L = 6000 mL
2. 54000 mL = 54 L
3. 11000 mL = 11 L
4. 13 L = 13000 mL
5. 23500 mL = 23.5 L
6. 0.201 L = 201 mL
17 sleeping bags and 85 flashlights were ordered.