Data were collected on random specimens of tuna from 1991 and 2010. The mercury

Question

Data were collected on random specimens of tuna from 1991 and 2010. The mercury level for each fish specimen was recorded. For the sample of 43 specimens of Albacore tuna, the average mercury level was found to be 0.358 parts per million (ppm) and the standard deviation of mercury level to be 0.138 ppm. The data are not strongly skewed.

1. Identify the observational unit for this study

a. Average mercury levels

b. The years

c. Specimens of Albacore Tuna

d. Mercury level

2. Identify the variable of interest and whether it is categorical or quantitative

a. Mercury level in parts per million, categorical

b. Specimens of Albacore tuna, quantitative

c. Mercury level in parts per million, quantitative

d. Specimens of Albacore tuna, categorical

3. In the context of this study, it is valid to use the theory-based (t-distribution) approach to find a confidence interval? True or False?

4. Using a Theory-Based Inference applet, find the 95% confidence interval for the average ppm of mercury in Albacore tunas. Round to four decimal places.

Explanation / Answer

1. Identify the observational unit for this study

a. Average mercury levels

2.

Mercury level in parts per million, quantitative

3.

True as population standard dev. is not given

4.

Standard error of the mean = SEM = S/N = 0.021

t(, N-1) = 2.018

Confidence interval = m +/- (t(, N-1)*SEM)

Mean = 0.358

Lower bound: 0.32

Upper bound: 0.40

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