A pollution-control inspector suspected that a river side community was releasin
ID: 3388717 • Letter: A
Question
A pollution-control inspector suspected that a river side community was releasing semi-treated sewage into a river and this, as a consequence, was changing the level of dissolved oxygen of the river. To check this, he drew 15 randomly selected specimens of river water at a location above the town and another 15 specimens below. The sample information and the QQplots by groups are given blow. A. Construct a 95% confidence interval for the difference between means. Does the data provide evidence to indicate a difference in the true acerage dissolved oxygen between locations above and below town? B. Run a hypothesis test at significance level alpha=0.05. State the null and alternative hypotheses. Give the test statistic, p-value and conclusion. C. Use the Wilcoxon Rank-Sum method to test for a difference between the distributions at alpha=0.05. Give the p-value and conclusion. D. Which test (two-sample t or Wilcoxon Rank-Sum test) is appropriate? Justify your response. 1. A pollution-control inspector suspected that a riverside community was releasing semi-treated sewage into a river and this, as a consequence, was changing the level of dissolved oxygen of the river. To check this, he drew 15 randomly selected specimens of river water at a location above the town and another 15 specimens below. The sample information and the QQplots by groups are given blow mean sd n Above 4.92 0.1567528 15 Below 4.74 0.3202677 15 Above Below norm quantiles norm quantiles A. Construct a 95% confidence interval for the difference between means Does the data provide evidence to indicate a difference in the true average B. Run a hypothesis test at significance level =0.05. State the null and C. Use the Wilcoxon Rank-Sum method to test for a difference between the D. Which test (two-sample t or Wilcoxon Rank-Sum test) is appropriate? dissolved oxygen between locations above and below town? alternative hypotheses. Give the test statistic, p-value and conclusion. distributions at =0.05. Give the p-value and conclusion Justify your responseExplanation / Answer
A)
Sample N Mean StDev SE Mean
1 15 4.920 0.157 0.040
2 15 4.740 0.320 0.083
Difference = mu (1) - mu (2)
Estimate for difference: 0.1800
95% CI for difference: (-0.0086, 0.3686)
B) T-Test of difference = 0 (vs not =): T-Value = 1.96 P-Value = 0.061 DF = 28
as P-value > 0.05 we accept H0