his Question: 10 pts 19 of 28 (15 complete) This Test: 280 pts possib Question H
ID: 3327013 • Letter: H
Question
his Question: 10 pts 19 of 28 (15 complete) This Test: 280 pts possib Question Help A researcher studies water clarity at the same location in a lake on the same dates during the course of a year and repeats the measurements on the same dates 5 years later. The researcher immerses a weighted disk painted black and white and measures the depth (in inches) at which it is no longer visible. The collected data is given in the table below. Complete parts (a) and (b) below Observation 1 2 3 4 5 6 Date Initial After five years 67.9 49.5 57.6 64.6 766 68.8 1/25 3/19 5/30 7/3 9/13 11/7 69.1 41.8 56.8 58.2 70.9 60.6 (a) Why is it important to take the measurements on the same date? A. O B. ° C. 0 D. Using the same dates makes the second sample dependent on the first. Using the same dates maximizes the difference in water clarity. Those are the same dates that all biologists use to take water clarity samples. Using the same dates makes it easier to remember to take samples. (b) Does the evidence suggest that the clarity of the lake is improving at the -0.05 level of significance? Note that the normal probability plot and boxplot of the data indicate that the differences are approximately normally distributed with no outliers. Choose the correct conclusion below Reject Ho. Do not reject Ho. °Explanation / Answer
a) option A is correct
b)
here as test statistic is less than critical value ;we
reject HO.
null Hypothesis: d = 0 alternate Hypothesis: d < 0 for 0.05 level with left tailed test and n-1= 5 degree of freedom, critical value of t= 2.5706 Decision rule : reject Ho if test statistic t<-2.571 S. NO intiial after 5 years difference(d)=x1-x2 d2 1 69.1 67.9 1.2 1.44 2 41.8 49.5 -7.7 59.29 3 56.8 57.6 -0.8 0.64 4 58.2 64.6 -6.4 40.96 4 70.9 76.6 -5.7 32.49 5 60.6 68.8 -8.2 67.24 total = d=-27.6 d2=202.06 mean= d = -4.600 degree of freedom =n-1 = 5.000 Std deviaiton SD=(d2-(d)2/n)/(n-1) = 3.8756 std error=Se = SD/n = 1.5822 t stat (d-d)/Se = -2.9074Related Questions
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