Question 4.DS.13: You are asked to evaluate the statements of a manufacturer o......

Approach:

1. H_{a} : σ^{2} = 2500    H_{a} : σ^{2} > 2500
2. \alpha = 0.01    S = 60   n = 10
3. From the Table 4.6, we find for
(n −1) = 9 and \alpha = 0.01  \Longrightarrow  χ^{2} = 21.67
Displaying the above data on a sketch reveals that the null hypothesis is rejected, when χ^{2} > 21.67.

4. Given χ^{2} =\left[\left(n-1\right)s^{2} \right]/\left(\sigma _{0}\right)^{2} where   n = 10,   S = 60   ,   σ_{0} = 50

We find χ^{2}=\frac{\left(10-1\right) \left(60\right)^{2} }{\left(50\right)^{2} }=12.96

5. Therefore, since the value of χ^{2} = 12.96 is within the bounds of χ^{2} = 21.67 we do not reject H_{0}.

You find that there is insufficient data to reject the hypothesis that the standard deviation is 50 weeks. Therefore, we accept H_{0} and we can conclude that the standard deviation of the adhesive shelf life is 50 weeks.