In a random sample of 10 residents of the state

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22.01.2022

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].

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