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NemiM [27]
3 years ago
8

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. Determine the 90% confidence interval for the population mean iron concentration. Assume the population is approximately normal. Step 1 of 2 : Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.
Mathematics
1 answer:
Delvig [45]3 years ago
4 0

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

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

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

<u>90% confidence interval for</u> \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].

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