It's a half of an equilateral triangle because their angles measure 30° and 60° (an equilateral triangle has 3 angles of 60°)
So x=(2h/√3)/2=h√3=27√3≈15.6

8 0

3 years ago

A researcher examines 27 water samples for iron concentration. The mean iron concentration for the sample data is 0.802 cc/cubic

Delvig [45]

Answer:

90% confidence interval for the population mean iron concentration is [0.771 , 0.832].

Step-by-step explanation:

We are given that a researcher examines 27 water samples for iron concentration. The mean iron concentration for the sample data is 0.802 cc/cubic meter with a standard deviation of 0.093.

Firstly, the pivotal quantity for 90% confidence interval for the population mean iron concentration is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean iron concentration = 0.802 cc/cubic meter

s = sample standard deviation = 0.093

n = number of water samples = 27

$\mu$ = population mean

Here for constructing 90% confidence interval we have used t statistics because we don't know about population standard deviation.

So, 90% confidence interval for the population mean, $\mu$ is ;

P(-1.706 < $t_2_6$ < 1.706) = 0.90 {As the critical value of t at 26 degree of

freedom are -1.706 & 1.706 with P = 5%}

P(-1.706 < $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ < 1.706) = 0.90

P( $-1.706 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $1.706 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

P( $\bar X -1.706 \times {\frac{s}{\sqrt{n} }$ < $\mu$ < $\bar X +1.706 \times {\frac{s}{\sqrt{n} }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X -1.706 \times {\frac{s}{\sqrt{n} }$ , $\bar X +1.706 \times {\frac{s}{\sqrt{n} }$ ]

= [ $0.802 -1.706 \times {\frac{0.093}{\sqrt{27} }$ , $0.802 +1.706 \times {\frac{0.093}{\sqrt{27} }$ ]

= [0.771 , 0.832]

Therefore, 90% confidence interval for the population mean iron concentration is [0.771 , 0.832].

4 0

3 years ago