Question

The FDA regulates that fish that is consumed is allowed to contain 1.0 mg/kg of mercury. In Florida, bass fish were collected in 53 different lakes to measure the amount of mercury in the fish. The data for the average amount of mercury in each lake is in the given table ("Multi-disciplinary niser activity," 2013). Do the data provide enough evidence to show that the fish in Florida lakes has more mercury than the allowable amount? Test at the 10% level. Use the framework below to guide your work. Hypotheses:

H0 : u = 1.0 mg/kg
HA: ul > 1.0 mg/kg
Test statistic = -10.09 p-value is approximately 1, would report 0.9999. Since this is not less than or equal to 0.10, we do not favor Ha. We would conclude that there is not enough evidence to show that the mean amount of mercury in fish in Florida lakes is more than the allowable amount Why is the p-value so high when the test statistic seems extreme?
A. The alternative is > so the p-value matches the area to the left. Since the TS is negative, this results in shading most of the curve.
B. The TS is negative so the p-value matches the area to the left and results in a very small area. This p-value reported is not correct.
C. The alternative is > so the p-value matches the area to the right. Since the TS is negative, this results in shading most of the curve.
D. The TS should be positive so the p-value matches the area to the left and results in shading most of the curve.

Answer 1

Explanation:

H0 : u = 1.0 mg/kg

HA: u > 1.0 mg/kg

Test statistic = -10.09

p-value is approximately 1, would report 0.9999

α = 10% ; 0.1

Using the Pvalue, we can make a decision pattern ;

Recall ; H0 is rejected If Pvalue < α

Here,

Pvalue Given is ' 0.99999 α = 0.1

Pvalue > α ; Hence, we fail to reject the Null ;

The actual Pvalue calculated using the test statistic will be :

Pvalue(-10.09) with test statistic value using a Pvalue calculator

Pvalue < 0.00001