Question

Listed below are the amounts of mercury​ (in parts per​ million, or​ ppm) found in tuna sushi sampled at different stores. the sample mean is 1.066 ppm and the sample standard deviation is 0.233 ppm. use technology to construct a 90​% confidence interval estimate of the mean amount of mercury in the population. 0.99 1.35 0.77 0.76 1.14 1.26 1.19 what is the confidence interval estimate of the mean amount of mercury in the​ population?

Answer 1

The confidence interval is (0.9234, 1.2086)

Explanation:

Given that:

Mean (μ)= 1.066 ppm

Standard deviation (σ)= 0.233 ppm

Confidence = 90% = 0.9

Since the amount of mercury are 0.99, 1.35, 0.77, 0.76, 1.14, 1.26, 1.19, therefore the amount in the population (n) = 7

α = 1 - 0.9 = 0.1

`(\alpha )/(2) = 0.05`

From the probability table,
`Z_{(\alpha )/(2) }=1.64`

Therefore the error (e) is given by the equation:

`e=Z_{(\alpha )/(2) }*(\sigma)/(√(n) )`

Substituting values:

`e=Z_{(\alpha )/(2) }*(\sigma)/(√(n) ) = 1.64 *(0.23)/(√(7) ) = 0.1426`

The confidence interval = (μ - e, μ + e) = (1.066 - 0.1426, 1.066 + 0.1426) = (0.9234, 1.2086)

The confidence interval is (0.9234, 1.2086)