Question

In a random sample of 10 residents of the state of Washington, the mean waste recycled per person per day was 2.3 pounds with a standard deviation of 0.39 pounds. Determine the 90% confidence interval for the mean waste recycled per person per day for the population of Washington. 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:

90% confidence interval for the mean waste recycled per person per day for the population of Washington is [2.074 , 2.526].

Step-by-step explanation:

We are given that a a random sample of 10 residents of the state of Washington, the mean waste recycled per person per day was 2.3 pounds with a standard deviation of 0.39 pounds.

Firstly, the pivotal quantity for 90% confidence interval for the mean waste recycled per person per day for the population of Washington is given by;

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

where, $\bar X$ = sample mean waste recycled per person per day = 2.3 pounds

             s = sample standard deviation = 0.39 pounds

             n = sample of residents = 10

             $\mu$ = population mean waste recycled per person per day

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(-1.833 < $t_9$ < 1.833) = 0.90  {As the critical value of t at 9 degree of

                                                freedom are -1.833 & 1.833 with P = 5%}

P(-1.833 < $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ < 1.833) = 0.90

P( $-1.833 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $1.833 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.833 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.833 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X-1.833 \times {\frac{s}{\sqrt{n} } }$  , $\bar X+1.833 \times {\frac{s}{\sqrt{n} } }$ ]

                                          = [ $2.3-1.833 \times {\frac{0.39}{\sqrt{10} } }$ , $2.3-1.833 \times {\frac{0.39}{\sqrt{10} } }$ ]

                                          = [2.074 , 2.526]

Therefore, 90% confidence interval for the mean waste recycled per person per day for the population of Washington is [2.074 , 2.526].