In a random sample of 10 residents of the state of Florida, the mean waste recycled per person per day was 2.8 pounds with a standard deviation of 0.64 pounds. Determine the 95% confidence interval for the mean waste recycled per person per day for the population of Florida. Assume the population is approximately normal.

Answer:

Given:

Sample mean $(x \bar{)} = 2.8$

Sample standard deviation $(s) = 0.64$

Sample size $(n) = 10$

Confidence interval level $= 95$

Solution:

→ The significance level $α = 1 - 0.95 = 0.05$

→ The degree of freedom $df = n - 1 = 10 - 1 = 9$

→ The critical value, $t_{c} = t_{\frac{α}{2}, df} = t_{0.05 / 2, 9} = 2.262$

→ The confidence interval:

$C I = \overset{―}{x} \pm t_{c} \times \frac{s}{\sqrt{n}}$ $= 2.8 \pm 2.262 \times \frac{0.64}{\sqrt{10}}$ $= (2.3, 3.3)$

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