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72. Hard water by region Recall from Chapter 7, Exercise 75, that data were collected on the annual mortality rate (deaths pe

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Answer #1

(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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