Periodically, a town water department tests the the drinking water of homeowners for contaminants such as lead. The lead levels in water specimens collected for a sample of 10 residents of the town had a mean of 3.4mg/L and a standard deviation of 2.1 mg/L. Complete parts a throughM c.
a.Construct a 90% confidence interval for the mean lead level in water specimens from the town.
(Round to three decimal places as needed.)
Solution :
Given that,
Point estimate = sample mean =
= 3.4
sample standard deviation = s = 2.1
sample size = n = 10
Degrees of freedom = df = n - 1 = 10 - 1 =9
At 90% confidence level the t is ,
= 1 - 90% = 1 - 0.90 = 0.10
/ 2 = 0.10 / 2 = 0.05
t
/2,df = t0.05,9 = 1.833
Margin of error = E = t/2,df
* (s /
n)
= 1.833 * (2.1 /
10)
= 1.217
The 90% confidence interval estimate of the population mean is,
- E <
<
+ E
<
<
3.4 - 1.217 <
< 3.4 + 1.217
(2.183 , 4.617)
Periodically, a town water department tests the the drinking water of homeowners for contaminants such as...
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