Question

In: Statistics and Probability

A random sample of ten households in College Park revealed they generated a mean of 10.91...

A random sample of ten households in College Park revealed they generated a mean of 10.91 pounds of garbage per week with a standard deviation of 4.736 pounds. Construct the 80% confidence interval to estimate the mean amount of garbage all College Park households generate per week

Expert Solution

Solution :

Given that,

= 10.91

s = 4.736

n = 10Degrees of freedom = df = n - 1 = 10- 1 =9

At 80% confidence level the t is ,

= 1 - 80% = 1 - 0.80 = 0.20

/ 2= 0.10

t /2,df = t0.10,9 =1.383 ( using student t table)

Margin of error = E = t/2,df * (s / $\sqrt$ n)

= 1.383* ( 4.736 / $\sqrt$ 10)

=0.6904

The 80% confidence interval estimate of the population mean is,

- E < < + E

10.91- 0.6904 < < 10.91+ 0.6904

10.2196< < 11.6004

( 10.2196, 11.6004)

orchestra answered 2 years ago

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