(a) The hypothesis being tested is:
H0: µ1 = µ2
H1: µ1 ≠ µ2
The output is:
North | Spoth | |
1631.59 | 1388.85 | mean |
138.47 | 151.114 | std. dev. |
34 | 27 | n |
59 | df | |
242.74000 | difference (North - Spoth) | |
20,787.48332 | pooled variance | |
144.17865 | pooled std. dev. | |
37.16589 | standard error of difference | |
0 | hypothesized difference | |
6.531 | t | |
1.67E-08 | p-value (two-tailed) |
Since the p-value (0.000) is less than the significance level (0.05), we can reject the null hypothesis.
Therefore, we have sufficient evidence to conclude that there is a significant difference in the two regions.
(b) The normality condition is violated with the presence of an outlier.
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72. Hard water by region Recall from Chapter 7, Exercise 75, that data were collected on...
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