The density of the gastropod Terebralia palutris was investigated in four mangrove forests by throwing a quadrat at random points on the sediment.
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The owner of a fishing camp wanted to test the effectiveness of two brands of mosquito repellents (A and B). From the population of visitors he randomly allocated 12 visitors to use Repellent A and another 12 visitors to use Repellent B. All wore short-sleeved shirts and shorts. The table below shows the summary statistics for the number of bites each visitor had after fishing for 4 hours on the lake. Assume that the data are normally distributed. Note: You might not need all of the statistics shown. At the 5% significance level, carry out the most appropriate test to determine whether there is a difference in the mean number of bites between people who used the two brands. Show all steps of the hypothesis test.
Summary statistics | Brand A |
Brand B
|
Difference |
Average | 25.08 | 21.42 | 3.67 |
Standard Deviation | 8.51 | 6.68 | 14.37 |
There is no indication that the two samples are paired or related, therefore the two-mean test for independent samples should be used
H0: mBrand _ A – mBrand _ B = 0 Parameter: mBrand _ A – mBrand _ B
(There is no difference in the mean number of bites between people who used the two brands)
(There is a difference in the mean number of bites between people who used the two brands)
df = n1 + n2 – 2 = 12 +12 – 2 = 22 Thus, (0.15 > P > 0.10) x 2 ==> 0.30 > P-value > 0.20
There is weak evidence against H0. Since P-value > α (0.05), do not reject H0.
At the 5% significance level, the data do not provide sufficient evidence to conclude that there is a difference in the mean number of bites between people who used the two brands.
Suppose that analysis of log transformed data results in the estimate of the difference between the logged means of Treatment A and Treatment B to be 0.746 and a 95% confidence interval for the additive effect of treatment is between 0.218 and 1.278.
Define the parameter as: mLnB – mLnA .
(a)(2 marks) Back transform the estimate and the confidence interval to the original scale. Interpretthe meaning of this confidence interval on the original scale.
Back Transformation of the estimate and confidence interval to the original data gives the following: Estimate of the difference = e0.746 = 2.109
Lower endpoint of the confidence interval = e0.218 = 1.244
Therefore, a 95% confidence interval for the ratio of the medians on the original scale is: (e0.218, e1.278 ) = (1.244,3.589)
OR
It is estimated with 95% confidence that the median effectiveness of Treatment B is between 1.244 and 3.589 times the median effectiveness of Treatment A.
(b)(1 mark) Based on the confidence interval you found in part (a), after back transformation, would you conclude that there is a difference between the medians for Treatment A and Treatment B (with 95% confidence)? Explain the logic of your answer.
Since the confidence interval after back transformation, which is (1.244, 3.589), does not contain 1, that means we can be 95% confident that there is a difference between the medians for Treatment A and Treatment B.
Based on a random sample of graduates who had just completed 4 different degree programmes (Bachelor’s degree in Civil Engineering, Bachelor’s degree in Mechanical Engineering, Master’s degree in Civil Engineering, and Master’s degree in Mechanical Engineering), information was gathered concerning their starting salaries. Assume that all required assumptions are satisfied for applying the analyses needed to answer parts (a) and (b) below. Use the parameters defined below and the computer output in Tables 1 – 4.
The parameters are already defined for you as follows:
mB–Civil = Bachelor’s degree in Civil Engineering
mB–Mech = Bachelor’s degree in Mechanical Engineering
mM –Civil = Master’s degree in Civil Engineering
mM –Mech = Master’s degree in Mechanical Engineering
Table 1: Summary statistics of the salaries of the 4 groups.