Question

Question 1 An engineer select 50 units from a production line. The engineer found 8 defectives units. considering a=0.05 Assuming we want to test the following hypothesis Ho: p = 0.19 H1: p < 0.19 z statistic: х- про np.(1 - po Calculate the lower limit of the rejection region in terms of defective units: Zo Z-table

Answer 1

critical Z value corresponding to is

g= 0,05

$Z_c=-1.645$

This corresponds to some p critical value

$\frac{p_c-0.19}{\sqrt{\frac{0.19*(1-0.19)}{50}}}=-1.645$

$p_c=0.0987$

this corresponds to number of defective units $0.0987*50 \approx 5$

lower limit of the rejection region is 5 defective units