A researcher examines 27 water samples for mercury concentration. The mean mercury concentration for the sample data is 0.097 cc/cubic meter with a standard deviation of 0.0074. Determine the 90% confidence interval for the population mean mercury 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.

Step-by-step explanation:

Confidence interval is written in the form,

(Sample mean - margin of error, sample mean + margin of error)

The sample mean, x is the point estimate for the population mean.

Margin of error = z × s/√n

Where

s = sample standard deviation = 0.0074

n = number of samples = 27

From the information given, the population standard deviation is unknown and the sample size is small, hence, we would use the t distribution to find the z score

In order to use the t distribution, we would determine the degree of freedom, df for the sample.

df = n - 1 = 27 - 1 = 26

Since confidence level = 90% = 0.90, α = 1 - CL = 1 – 0.9 = 0.1

α/2 = 0.1/2 = 0.05

the area to the right of z0.05 is 0.05 and the area to the left of z0.05 is 1 - 0.05 = 0.95

Looking at the t distribution table,

z = 1.706

The critical value is 1.706

Margin of error = 1.706 × 0.0074√27

= 0.0024

The 90% confidence interval is

0.097 ± 0.0024

i used to know but i learned it a long time ago sorry

step-by-step explanation:

1.true 2.false? ? i think. .

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