area ibi 21 60 29 58 31 74 32 92 34 81 34 52 49 94 52 73 2 65 70 70 6 46 28 31 2
ID: 3180288 • Letter: A
Question
area ibi
21 60
29 58
31 74
32 92
34 81
34 52
49 94
52 73
2 65
70 70
6 46
28 31
21 52
59 90
69 55
47 51
8 35
8 67
58 85
54 77
10 53
57 79
18 85
19 19
39 89
49 68
9 40
5 24
14 52
9 34
23 80
31 73
18 80
16 32
21 69
32 80
10 57
26 1
9 74
54 58
12 57
21 69
27 60
23 61
26 77
16 61
26 78
26 72
28 52
6.-n The index of biotic integrity (IBI) is a measure of the water quality in streams. IBI and land-use measures for a collection of streams in the ozark Highland ecoregion of Arkansas were collected as part of a study. The data data226 dat gives the data for IBI and the area of the watershed in square kilometers for streams in the original sample with area less than or equal to 7okm2 (a) Use numerical and graphical methods to describe the variable IBI. Do the same for area (Round your answers for K to two decimal places and your answers for s to three decimal places.) IBI (b) Run the simple linear regression and summarize the results. (Let x area and y IBI. Round your slope, intercept, and r to three decimal places. Round Fto two decimal places and your P-value to four decimal places.) Is area of watershed a good predictor for IBI at the 5% significance level? Yes, the area of watershed is a good predictor for IBI No, there is insufficient evidence to say the area of watershed is a good predictor for IBI. Interpret the intercept. The value of IBI when watershed area is o 0 The increase in IBI when the watershed area is increased by one unit The area of watershed when IBI value is 0 The increase in the watershed area when IBI is increased by one unit Interpret the slope. The increase in IBI when the watershed area is increased by one unit The area of watershed when IBI value is 0 The increase in the watershed area when IBI is increased by one unitExplanation / Answer
Problem. 1
From the question, we get
Condition
Sample size
Mean
Standard deviation
1(control)
4()
53.5
2(experimental)
4()
In the given problem the director will be interested in checking whether the motivational seminar increases the performance of his employees. Therefore, the set up of the hypotheses with level of significance
(i.e. the motivational seminar did not brought any increase in employees performance)
or
( i.e. the motivational seminar increased in employees performance)
This is nothing but a test for differences of sample means with 95% Confidence Intervals. As both samples are small(<30) we will use t-test. As the motivational seminar is expected to increase the performance of employees, therefore, it will be one-sided right-tailed t-statistics test.
i.e.
where
It is a right-tailed t-statistic test with 5% level of significance (critical value will be found t-Table (can be found from any statistics standard book). To determine the critical value of t-statistics with degrees of freedom defined n1+n2-2 = 4+4-2=6. The critical value for a right-tailed t-test with 6 degree of freedom and =0.05 is 1.943 and the decision rule is: Reject H0 if t >1.943.
Now
Condition
Sample size
Mean
Standard deviation
1(control)
4()
53.5
5.686241
2(experimental)
4()
69.25
7.588368
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