Question

A random sample of 8 fields of barley has a mean yield of 49.6 bushels per acre and standard deviation of 2.33 bushels per acre. Determine the 99% confidence interval for the true mean yield. 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.

Answer 1

Answer:

99% confidence interval for the true mean yield is [46.718 , 52.482].

Step-by-step explanation:

We are given that a random sample of 8 fields of barley has a mean yield of 49.6 bushels per acre and standard deviation of 2.33 bushels per acre.

Firstly, the pivotal quantity for 99% confidence interval for the true mean yield is given by;

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

where, $\bar X$ = sample mean yield = 49.6 bushels per acre

             s = sample standard deviation = 2.33 bushels per acre

             n = sample of fields of barley = 8

             $\mu$ = true population mean

Here for constructing 99% confidence interval we have used t statistics because we don't know about population standard deviation.

So, 99% confidence interval for the true mean, $\mu$ is ;

P(-3.499 < $t_1_5$ < 3.499) = 0.99  {As the critical value of t at 7 degree of

                                                freedom are -3.499 & 3.499 with P = 0.5%}

P(-3.499 < $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ < 3.499) = 0.99

P( $-3.499 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $3.499 \times {\frac{s}{\sqrt{n} } }$ ) = 0.99

P( $\bar X -3.499 \times {\frac{s}{\sqrt{n} }$ < $\mu$ < $\bar X +3.499 \times {\frac{s}{\sqrt{n} }$ ) = 0.99

99% confidence interval for $\mu$ = [ $\bar X -3.499 \times {\frac{s}{\sqrt{n} }$ , $\bar X +3.499 \times {\frac{s}{\sqrt{n} }$ ]

                                                 = [ $49.6 -3.499 \times {\frac{2.33}{\sqrt{8} }$ , $49.6 +3.499 \times {\frac{2.33}{\sqrt{8} }$ ]

                                                 = [46.718 , 52.482]

Therefore, 99% confidence interval for the true mean yield is  [46.718 , 52.482].