Answer :
Answer:
The 90% confidence interval is [tex] 0.0097 < p_1 -p_2 < 0.0773 [/tex]
Step-by-step explanation:
From the question we are told that
The first sample size is [tex]n_1 = 620[/tex]
The number of chips that failed is k = 125
The second sample size is [tex]n_2 = 820[/tex]
The number of chips that fail is [tex]u = 130[/tex]
Generally the confidence level is 90% , hence the level of significance is
[tex]\alpha = (100 - 90)\% [/tex]
=> [tex]\alpha = 0.10 [/tex]
Generally from the normal distribution table the critical value of [tex]\frac{\alpha }{2}[/tex] is
[tex]Z_{\frac{\alpha }{2} } = 1.645[/tex]
Generally the first sample proportion is
[tex]\r p _1 = \frac{125}{620}[/tex]
=> [tex]\r p _1 = 0.202[/tex]
Generally the second sample proportion is
[tex]\r p _1 = \frac{130}{820}[/tex]
=> [tex]\r p _1 = 0.1585[/tex]
Generally the standard error is mathematically represented as
[tex]SE = \sqrt{\frac{\r p_1 (1 - \r p_1)}{n_1} +\frac{\r p_2 (1 - \r p_2)}{n_2} }[/tex]
=> [tex]SE = \sqrt{\frac{0.202(1 - 0.202)}{620} +\frac{0.1585 (1 - 0.1585)}{820} }[/tex]
=> [tex]SE = 0.0206[/tex]
Generally the margin of error is mathematically represented as
[tex]E = Z_{\frac{\alpha }{2} } * SE[/tex]
=> [tex]E = 1.645 * 0.0206[/tex]
=> [tex]E =0.0338[/tex]
Generally 95% confidence interval is mathematically represented as
[tex](\r p_1 -\r p_2) -E < p_1 - p_2 < (\r p_1 -\r p_2) +E [/tex]
=> [tex](0.202 -0.1585) -0.0338 < p_1 - p_2 < (0.202 -0.1585) +0.0338 [/tex]
=> [tex] 0.0097 < p_1 - p_2 < 0.0773 [/tex]