Question

A physicist examines 25 water samples for nitrate concentration. The mean nitrate concentration for the sample data is 0.165 cc/cubic meter with a standard deviation of 0.0783. Determine the 80% confidence interval for the population mean nitrate 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

Answer:

The 80% confidence interval for the the population mean nitrate concentration is (0.144, 0.186).

Critical value t=1.318

Step-by-step explanation:

We have to calculate a 80% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=0.165.

The sample size is N=25.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

$s_M=\dfrac{s}{\sqrt{N}}=\dfrac{0.078}{\sqrt{25}}=\dfrac{0.078}{5}=0.016$

The degrees of freedom for this sample size are:

$df=n-1=25-1=24$

The t-value for a 80% confidence interval and 24 degrees of freedom is t=1.318.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_M=1.318 \cdot 0.016=0.021$

Then, the lower and upper bounds of the confidence interval are:

$LL=M-t \cdot s_M = 0.165-0.021=0.144\\\\UL=M+t \cdot s_M = 0.165+0.021=0.186$

The 80% confidence interval for the population mean nitrate concentration is (0.144, 0.186).