Question

A country environmental agency suspects that the fish in a polluted lake have elevated mercury level. To confirm that suspicion, five striped bass in that lake were caught and their tissues were tested for mercury. For the purpose of comparison, four striped bass in an unpolluted lake were also caught and tested. The fish tissue mercury levels in mg/kg are given below. Answer the following questions. Use the rounded answers from previous questions in your calculations. Sample 1 (from polluted lake) Sample 2 (from unpolluted lake) 0.580 0.382 0.711 0.276 0.571 0.570 0.666 0.366 0.508

5) Construct the 95% confidence interval for the difference of the population means based on these data.

Lower endpoint (round to four decimal places) =  

Upper endpoint (round to four decimal places) =  

6) Test, at 5% level of significance, whether the data provide sufficient evidence to conclude that fish in the polluted have elevated levels of mercury in their tissue.

H0:

Ha:

t-Test Statistic (round to three decimal places) =

Critical t-score (LaTeX: t_{\alpha} t α or LaTeX: t_{\alpha/2} t α / 2 ) =

Conclusion (write "R" for reject H0, and "F" for fail to reject H0):

Answer 1

Solution:-

5) 95% confidence interval for the difference of the population means based on these data is C.I = (0.0395,0.3779).

Lower endpoint = 0.0395

Upper endpoint = 0.3779

6)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u₁ = u₂
Alternative hypothesis: u₁ > u₂

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the mean difference between sample means is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = 0.07154
DF = 7

t = [ (x₁ - x₂) - d ] / SE

t = 2.92

t_Critical = = + 1.895

Rejection region is t > 1.895

where s₁ is the standard deviation of sample 1, s₂ is the standard deviation of sample 2, n₁ is thesize of sample 1, n₂ is the size of sample 2, x₁ is the mean of sample 1, x₂ is the mean of sample 2, d is the hypothesized difference between population means, and SE is the standard error.

Interpret results. Since the t-value (2.92) lies in the rejection region, hence we have to reject the null hypothesis.

Conclusion:

R, Reject H0.