Question

observations are (in milligrams per liter): 5.0, 3.4, 3.9, 1.3, 0.2, 0.9, 2.7, 3.7, 3.8, 4.1, 1.0, 1.0, 0.8, 0.4, 3.8, 4.5, 5.3, 6.1,6.9, and 6.5. Water quality engineers are interested in knowing whether these data support a claim that mean dissolved oxygen concentration is 2.5 milligrams per liter. (a) Formulate an appropriate hypothesis testing procedure to investigate this claim. O In order to use t statistics in hypothesis testing, we need to assume that the underlying distribution is normal. Use the t-test to test the hypothesis that the true mean is 2.5 mg/L. 1) State the parameter of interest: The parameter of interest is the true mean dissolved oxygen level, 4. 2) State the null hypothesis Ho: u= 2.5. 3) State the alternative hypothesis H u + 2.5. 4) Give the statistic to = s/n 5) Reject Ho if tol <ta/2,n-1 for a = 0.05. 6) Sample statistic = 3.265, s = 2.127, n = 20. ... I- t-statistic to = s/n 7) Draw your conclusion and find the P-value. In order to use t statistics in hypothesis testing, we need to assume that the underlying distribution is normal. Use the t-test to test the hypothesis that the true mean is 2.5 mg/L. 1) State the parameter of interest: The parameter of interest is the true mean dissolved oxygen level, 2) State the null hypothesis Hou > 2.5 3) State the alternative hypothesis H u 2.5 1+ 4) Give the statistic to = - 5) Reject Ho if (tol > 14/2,n-1 for a = 0.05 6) Sample statistic = 3.265, 5 = 2.127, n= 20 +H t-statistic to = " 7) Draw your conclusion and find the P-value. In order to use t statistics in hypothesis testing, we need to assume that the underlying distribution is normal. Use the t-test to test the hypothesis that the true mean is 2.5 mg/L. 1) State the parameter of interest: The parameter of interest is the true mean dissolved oxygen level, 2) State the null hypothesis Hou > 2.5 3) State the alternative hypothesis H u # 2.5 1-u 4) Give the statistic to = - stva 5) Reject Ho if to 6) Sample statistic tal2.-i for a = 0.05 = 3.265, s= 2.127, n = 20

Answer 1

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@ option - @ option - ③ P-value = P(H1> 1.608) = 0.124.