Complete Question
The average zinc concentrations recovered from a random sample of zinc measurements in 22 different locations was found to be 2.6 grams per milliliter. Assume that the variance for zinc measurements is known to be 0.09 grams per milliliter squared. Calculate and interpret a 90% confidence interval for the overall average zinc concentration.
Answer:
The 90% confidence interval for the overall average zinc concentration [2.59, 2.61]
Step-by-step explanation:
From the question, we are given that following:
Mean = 2.6g/ml
Variance = 0.09(g/ml)²
Standard deviation = √variance
= √0.09
= 0.3g/ml
n = number of sample = 22
Note that our sample size is small because it is less than 30, so instead of using the z score of a 90% confidence interval, we apply the t score of a 90% confidence interval.
First we find the degrees of freedom = n - 1
= 22 - 1 = 21
T score using degrees of freedom from 90% confidence interval = 1.721
Hence, the formula for Confidence Interval =
Mean ± t × standard deviation /√n
2.6 ± 1.721 × 0.3/√22
2.6 ± 1.721 × 0.0639602149
2.6 ± 0.011007553
= 2.6 - 0.011007553
= 2.588992447
≈ 2.59
= 2.6 + 0.011007553
= 2.611007553
≈ 2.61
Therefore, the 90% confidence interval for the overall average zinc concentration [2.59, 2.61]
