at 0.1 level of significance, we can conclude that these data are sufficient to indicate that the mean for population 2 is larger than that for population 1.
For the null hypothesis
H0: μd ≥ 0
For the alternative hypothesis
H1: μd < 0
The distribution is a students t. Therefore, degree of freedom, df = n - 1 = 10 - 1 = 9
The formula for determining the test statistic is
t = (xd - μd)/(sd/√n)
t = (- 3.7 - 0)/(2.71/√10)
t = - 4.32
We would determine the probability value by using the t test calculator.
p = 0.00097
Since alpha, 0.1 > than the p value, 0.00097, then we would reject the null hypothesis. Therefore, at 0.1 level of significance, we can conclude that these data are sufficient to indicate that the mean for population 2 is larger than that for population 1.
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