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natka813 [3]
3 years ago
13

Jshsbdbfbbbfffbbdvdvsgssggdhdh

Chemistry
1 answer:
11Alexandr11 [23.1K]3 years ago
8 0
Very nice english.......
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During a routine check of the fluoride content of Gotham City\'s water supply, the following results were obtained from replicat
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Answer:

The mean is x=0.503\frac{mg}{L}

The 90% confidence interval is:

i_{0.90}=[0.492\frac{mg}{L},0.514\frac{mg}{L}]

Explanation:

1. First organize the data:

x_{1}=0.487

x_{2}=0.487

x_{3}=0.511

x_{4}=0.511

x_{5}=0.519

As there are 5 data, the sample size (n) is n=5

2. Calculate the mean x:

The mean is calculated adding up all the data and divide them between the sample size.

x=\frac{0.511+0.487+0.511+0.487+0.519}{5}

x=0.503\frac{mg}{L}

3. Find 90% confidence interval.

The formula to find the confidence interval is:

i_{0.90}=[x+/-z_{\frac{\alpha}{2}}*(\frac{d}{\sqrt{n}})] (Eq.1)

where x is the mean, d is the standard deviation and n is the sample size.

And

1-\alpha=0.90

\alpha=0.10

\frac{\alpha}{2}=0.05

z_{0.05}=1.645

4. Find the standard deviation

d=\sqrt{\frac{(x_{1}-x)^{2}+(x_{2}-x)^{2}+(x_{3}-x)^{2}+(x_{4}-x)^{2}+(x_{5}-x)^{2}}{n-1}}

d=\sqrt{\frac{(0.487-0.503)^{2}+(0.487-0.503)^{2}+(0.511-0.503)^{2}+(0.511-0.503)^{2}+(0.519-0.503)^{2}}{4}}

d=\sqrt{\frac{(-0.016)^{2}+(-0.016)^{2}+(0.008)^{2}+(0.008)^{2}+(0.016)^{2}}{4}}

d=\sqrt{2.24*10^{-4}}

d=0.015

5. Replace values in (Eq.1):

i_{0.90}=[0.503+/-1.645*(\frac{0.015}{2.236})]

For the addition:

i_{0.90}=[0.503+1.645*(\frac{0.015}{2.236})]

i_{0.90}=0.514

For the subtraction:

i_{0.90}=[0.503-1.645*(\frac{0.015}{2.236})]

i_{0.90}=0.492

The 90% confidence interval is:

i_{0.90}=[0.492,0.514]

4 0
3 years ago
What happens in a chemical reaction with the individual elements?
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Hope it helps! 
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4 years ago
Help asappppppppppppppppppppp
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