Answer:
a
The estimate is [tex]- 0.0265\le K \le 0.0465[/tex]
b
Method B this is because the faulty breaks are less
Step-by-step explanation:
The number of microchips broken in method A is [tex]n_1 = 400[/tex]
The number of faulty breaks of method A is [tex]X_1 = 32[/tex]
The number of microchips broken in method B is [tex]n_2 = 400[/tex]
The number of faulty breaks of method A is [tex]X_2 = 32[/tex]
The proportion of the faulty breaks to the total breaks in method A is
[tex]p_1 = \frac{32}{400}[/tex]
[tex]p_1 = 0.08[/tex]
The proportion of the faulty to the total breaks in method B is
[tex]p_2 = \frac{28}{400}[/tex]
[tex]p_2 = 0.07[/tex]
For this estimation the standard error is
[tex]SE = \sqrt{ \frac{p_1 (1 - p_1)}{n_1} + \frac{p_2 (1- p_2 )}{n_2} } }[/tex]
substituting values
[tex]SE = \sqrt{ \frac{0.08 (1 - 0.08)}{400} + \frac{0.07 (1- 0.07 )}{400} } }[/tex]
[tex]SE = 0.0186[/tex]
The z-values of confidence coefficient of 0.95 from the z-table is
[tex]z_{0.95} = 1.96[/tex]
The difference between proportions of improperly broken microchips for the two breaking methods is mathematically represented as
[tex]K = [p_1 - p_2 ] \pm z_{0.95} * SE[/tex]
substituting values
[tex]K = [0.08 - 0.07 ] \pm 1.96 *0.0186[/tex]
[tex]K = - 0.0265 \ or \ K = 0.0465[/tex]
The interval of the difference between proportions of improperly broken microchips for the two breaking methods is
[tex]- 0.0265\le K \le 0.0465[/tex]
