Question

In a random sample of 9 residents of the state of Florida, the mean waste recycled per person per day was 2.4 pounds with a standard deviation of 0.75 pounds. Determine the 80% confidence interval for the mean waste recycled per person per day for the population of Florida. 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 with explanation:

When the sample size is small (< 30) and the population standard deviation is unknown , then we use t-test.

The confidence interval for population mean will be :

$\overline{x}\pm t^*\dfrac{s}{\sqrt{n}}$   (1)

, where $\overline{x}$ = sample mean

t* = Critical value (based on degree of freedom and significance level).

s= sample standard deviation

n= sample size.

As per given we have

n= 9

Degree of freedom = n-1 = 8

$\overline{x}=2.4$

s= 0.75

Significance level =$\alpha=1-0.80=0.20$

Using students' t distribution table ,

Critical value : $t^*=t_{\alpha/2,df}=t_{0.10,8}=1.3304$

We assume that the population is approximately normal.

Then, a 80% confidence interval for the mean waste recycled per person per day for the population of Florida will be :

$2.4\pm (1.3304)\dfrac{0.75}{\sqrt{8}}$   (Substitute the values in (1))

$2.4\pm (1.3304)\dfrac{0.75}{2.82842712475}$

$2.4\pm (1.3304)(0.265165042945)$

$2.4\pm 0.352775573134\approx2.4\pm0.353=(2.4-0.353,\ 2.4+0.353)=(2.047,\ 2.753)$

Hence, the 80% confidence interval for the mean waste recycled per person per day for the population of Florida. = (2.047, 2.753)