A random sample of students at a college reported what they
Solution
Consider the table:
Calculating the standard deviation of the differences (third column):
s = 3.237304581
Thus, the standard error of the difference is sD = s/sqrt(n):
sD = 0.835868449
For the 0.95 confidence level,
alpha/2 = (1 - confidence level)/2 = 0.025 As df = n - 1 = 14
t(alpha/2) = 2.144786688
lower bound = [X1 - X2] - t(alpha/2) * sD = -2.387426188
upper bound = [X1 - X2] + t(alpha/2) * sD = 1.198092855
Thus, the confidence interval is
( -2.387426188 , 1.198092855 ) [ANSWER]
If the order has to be the other way around, then please answer (-1.198, 2.387).
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YES, IT INCLUDES 0, AS YOU CAN SEE.
| 165.1 | 168 | 2.9 |
| 170.18 | 173 | 2.82 |
| 185.42 | 183 | -2.42 |
| 172.72 | 164 | -8.72 |
| 190.5 | 189 | -1.5 |
| 170.18 | 171 | 0.82 |
| 177.8 | 174 | -3.8 |
| 190.5 | 193 | 2.5 |
| 177.8 | 177 | -0.8 |
| 177.8 | 178 | 0.2 |
| 182.88 | 181 | -1.88 |
| 177.8 | 176 | -1.8 |
| 167.64 | 170 | 2.36 |
| 172.72 | 170 | -2.72 |
| 182.88 | 186 | 3.12 |
