Question

A biologist examines 17 seawater samples for magnesium concentration. The mean magnesium concentration for the sample data is 0.252 cc/cubic meter with a standard deviation of 0.074. Determine the 80% confidence interval for the population mean magnesium concentration. 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

Solution :

Point estimate = sample mean = = 0.252

sample standard deviation = s = 0.074

sample size = n = 17

Degrees of freedom = df = n - 1 = 17 -1 = 16

At 80% confidence level

= 1-0.80% =1-0.80 =0.20

/2 =0.20/ 2= 0.10

t/2,df = t0.10,16 = 1.34

t /2,df = 1.34

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

= 1.34 * (0.074 / 17)

Margin of error = E = 0.024

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

- E < <  + E

0.252 -0.024 < < 0.252 + 0.024

0.228 < < 0.276

(0.228,0.276)